Category: Class 11 Physics

NCERT Exemplar Class 11 Physics Chapter 11 Thermodynamics are part of NCERT Exemplar Class 11 Physics. Here we have given NCERT Exemplar Class 11 Physics Chapter 11 Thermodynamics.

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NCERT Exemplar Class 11 Physics Chapter 11 Thermodynamics

Multiple Choice Questions
Single Correct Answer Type

Q1. An ideal gas undergoes four different processes from the same initial state (figure). Four processes are adiabatic, isothermal, isobaric and isochoric. Out of 1, 2, 3 and 4 which one is adiabatic?
(a) 4
(b) 3
(c) 2
(d) 1

Solution:

Q2. If an average person jogs, he produces 14.5 x 10³ cal/min. This is removed by the evaporation of sweat. The amount of sweat evaporated per minute (assuming 1 kg requires 580 x 10³ cal for evaporation) is
(a) 0.25 kg (b) 2.25 kg (c) 0.05 kg (d) 0.20 kg
Sol:(a) Rate of bum calories is equivalent to sweat produced. Then, Amount of sweat evaporated/minute

Q3. Consider P-Vdiagram for an ideal gas is shown in figure. Out of the following diagrams, which figure represents the T-P diagram?

i.e., V 1/p or PV =  Constant

Hence, we can say that the gas is going through an isothermal process. Clearly, from the graph that between process 1 and 2 temperature is constant and the gas expands and pressure decreases, i.e., P₂<P_l. So, we have to keep in mind while drawing the T-P graph, that temperature (T) is constant and pressure at point 2 is greater than the pressure at 1, which corresponds to diagram (iii).

 

Q4. An ideal gas undergoes cyclic process ABCDA as shown in given P-V diagram. The amount of work done by the gas is
(a) 6P_gV₀
(b) -2P₀V₀
(c) +2 P₀V_o
(d) +4Po V₀

Sol:

Important point: In a cyclic process work done is
1. positive if the cycle is clockwise.
2. negative if the cycle is anticlockwise.

Q5. Consider two containers A and B containing identical gases at the same pressure, volume and temperature. The gas in container A is compressed to half of its original volume isothermally while the gas in container B is compressed to half of its original value adiabatically. The ratio of final pressure of gas in B to that of gas in A is 

Sol: (a) According to the P-V diagram shown for the container A (which is going through isothermal process) and for container B (which is going through adiabatic process).

Q6. Three copper blocks of masses M₁ M₂ and M₃ kg respectively are brought into thermal contact till they reach equilibrium. Before contact, they were at T₁, T₂, T₃ (T₁ > T₂> T₃). Assuming there is no heat loss to the surroundings, the equilibrium temperature T is (s is specific heat of copper)

Sol: (b) According to question, since there is no net loss to the surroundings and the equilibrium temperature of the system is T.
Let us assume that T_l,T₂< T< T₃.
Heat lost by M₃ = Heat gained by M₁+ Heat gained by M₂


More Than One Correct Answer Type
Q7. Which of the processes described below are irreversible?
(a) The increase in temperature of an iron rod by hammering it.
(b) A gas in a small container at a temperature T₁, is brought in contact with a big reservoir at a higher temperature T₂ which increases the temperature of the gas.
(c) A quasi-static isothermal expansion of an ideal gas in cylinder fitted with a frictionless piston.
(d) An ideal gas is enclosed in a piston cylinder arrangement with adiabatic walls. A weight w is added to the piston, resulting in compression of gas.
Sol: (a, b, d)
Key concept: Reversible process: A reversible process is one which can be reversed in such a way that all changes occurring in the direct process are exactly repeated in the opposite order and inverse sense and no change is left in any of the bodies taking part in the process or in the surroundings.
The conditions for reversibility are:
• There must be complete absence of dissipative forces such as friction, viscosity, electric resistance etc. ~
• The direct and reverse processes must take place infinitely slowly.
• The temperature of the system must not differ appreciably from its surroundings.
Irreversible process: Any process which is not reversible exactly is an irreversible process. All natural processes such as conduction, radiation, radioactive decay etc. are irreversible. All practical processes such as free expansion, Joule-Thomson expansion, electrical heating of a wire are also irreversible.
(a) In this case internal energy of the rod is increased from external work done by hammer which in turn increases its temperature. So, the process cannot be retraced itself.
(b) In this process energy in the form of heat is transferred to the gas in the small container by big reservoir at temperature T₂.
(c) In a quasi-static isothermal expansion, the gas is ideal, this process is reversible because the cylinder is fitted with frictionless piston.
(d) As the weight is added to the cylinder arrangement in the form of external pressure hence, it cannot be reversed back itself.

Q8. An ideal gas undergoes isothermal process from some initial state i to final state f Choose the correct alternatives

Sol: (a, d)
Key concept: First Law of Thermodynamics:
It is a statement of conservation of energy in thermodynamical process.
According to it heat given to a system (∆Q) is equal to the sum of increase in its internal energy (AIT) and the work done (AW) by the system against the surroundings.
∆Q=∆U+∆W
According to the first law of thermodynamics. ∆AQ = ∆U + ∆Wbut
∆U ∝∆T
∆U=0 [As ∆T= 0]
∆Q = ∆W, i.e., heat supplied in an isothermal change is used to do work against external surrounding.
or if the work is done on the system then equal amount of heat energy will be liberated by the system

Q9. Figure shows the P-V diagram of an ideal gas undergoing a change of state from A to B. Four different parts I, II, III and IV as shown in the figure may lead to the same change of state.
(a) Change in internal energy is same in IV and III cases, but not in I and II.
(b) Change in internal energy is same in all the four cases.
(c) Work done is maximum in case I.
(d) Work done is minimum in case II.

Sol: (b, c)

Key concept: Internal energy (U): Internal energy of a system is the energy possessed by the system due to molecular motion and molecular configuration.
The energy due to molecular motion is called internal kinetic energy U_K and that due to molecular configuration is called internal potential energyUp.
i.e., Total internal energy U= U_K+ U_P
(i) For an ideal gas, as there is no molecular attraction U_P = 0
i.e., internal energy of an ideal gas is totally kinetic and is given by
U = U_K = 3/2 RT

Change in internal energy does not depend on the path of the process. So it is called a point function, i.e. it depends only on the initial and final states (A and B) of the system, i.e. ∆U = Uf – U_i
Hence internal energy is same for all four paths I, II, III and IV.
The work done by an ideal gas is equal to the area bounded between P-V curve.
Work done from A to B, ∆W_A→B = Area under the P-V curve which is maximum for the path I.

Q10. Consider a cycle followed by an engine (figure).
1 to 2 is isothermal
2 to 3 is adiabatic
3 to 1 is adiabatic
Such a process does not exist, because
(a) heat is completely converted to mechanical energy in such a process, which is not possible
(b) mechanical energy is completely converted to heat in this process, which is not possible
(c) curves representing two adiabatic processes don’t intersect
(d) curves representing an adiabatic process and an isothermal process don’t intersect

Sol. (a, c)
(a) The given process is a cyclic process, i.e. it returns to the original state 1. And change in internal energy in a cyclic process is always zero as for cyclic process U_f = U_i So, ∆U = U_f – U_i = 0
Hence, total heat is completely converted to mechanical energy. Such a process is not possible by second law of thermodynamics.
(c) Here, two curves are intersecting, when the gas expands adiabatically from 2 to 3. It is not possible to return to the same state without being heat supplied, hence the process 3 to 1 cannot be adiabatic. So, we conclude that such a process does not exist because curves representing two adiabatic processes do not intersect.

Q11. Consider a heat engine as shown in figure. Q₁ and Q₂ are heat added both to T₁ and heat taken from T₂ in one cycle of engine. W is the mechanical work done on the engine.If W > 0, then possibilities are:

Sol: (a, c)
Key concept: Refrigerator or Heat Pump:
A refrigerator or heat pump is basically a heat engine run in reverse direction. It essentially consists of three parts:
Source: At higher temperature T₁
Working substance: It is called refrigerant liquid ammonia and freon works as a working substance.

Sink: At lower temperature T₂.
The working substance takes heat Q₂ from a sink (contents of refrigerator) at lower temperature, has a net amount of work done W on it by an external agent (usually compressor of refrigerator) and gives out a larger amount of heat Q₁, to a hot body at temperature T₁ (usually atmosphere). Thus, it transfers heat from a cold body to a hot body at the expense of mechanical energy supplied to it by an external agent. The cold body is thus cooled more and more.

We know that the diagram represents the working of a refrigerator. So, we can write

Very Short Answer Type Questions

Q12. Can a system be heated and its temperature remains constant?
Sol: Yes, this is possible when the entire heat supplied to the system is utilised in expansion.
As ∆Q = ∆U + ∆W and ∆U = nC_v∆T
∆Q = nC_v∆T+ ∆W
If temperature remains constant, then ∆T = 0, this implies ∆Q = ∆W. This implies that heat supplied should perform work against the surroundings.

Q13. A system goes from P to Q by two different paths in the P-V diagram as shown in figure.
Heat given to the system in path 1 is 1000 J.
The work done by the system along path 1 is more than path 2 by 100 J. What is the heat exchanged by the system in path 2?

Sol: According to the first law of thermodynamics,
∆Q = AU + ∆W. Let us apply this for each path.
For path 1: Heat given Q₁ = +1000 J
Let work done for path 1 = W₁                                          .
For path 2:
Work done (W₂) = (W_I– 100) J
Heat given Q₂ – ?
As change in internal energy between two states for different path is same.
∆ U=Q_i-W₁ = Q₂-W₂
1000– W_!=Q₂-(W₁ – 100)
=> Q₂= 1000- 100 = 900 J

Q14. If a refrigerator’s door is kept open, will the room become cool or hot? Explain.
Sol: A refrigerator is a heat engine it extracts heat from low temperature reservoir and transfer it to high temperature. If a refrigerator’s door is kept open, then room will become hot, because then refrigerator exhaust more heat into the room than earlier. In this way, temperature of the room increases and room becomes hot. No refrigerator is efficient. Thus it exhaust more heat into the room than it extract from it. Thus, a room cannot be cooled by keeping the door of a refrigerator open.

Q15. Is it possible to increase the temperature of a gas without adding heat to it? Explain.
Sol: Yes, it is possible to increase the temperature of a gas without adding heat to it, during adiabatic compression the temperature of a gas increases while no heat is given to it.
For an adiabatic compression, no heat is given or taken out in adiabatic process.
Therefore, ∆Q = 0
According to the first law of thermodynamics,
∆Q=∆U+∆W
∆U = -∆W ( ∆Q =0)
In compression work is done on the gas, i.e. work done is negative. Therefore, ∆U = Positive
Hence, internal energy of the gas increases due to which its temperature increases.

Q16 Air pressure in a car tyre increases during driving. Explain.
Sol: Volume of a car tyre is fixed. During driving, temperature of the gas increases while its volume remains constant. So, according to Charle’s law, at constant volume (V),
Pressure (P) ∝Temperature (T)
Therefore, pressure of gas increases

Short Answer Type Questions
Q17. Consider a Carnot’s cycle operating between T₁ = 500 K and T₂ = 300 K producing 1 kJ of mechanical work per cycle. Find the heat transferred to the engine by the reservoirs.
Sol: Key concept: Carnot theorem: The efficiency of Carnot’s heat engine depends only on the temperature of source (T₁) and temperature of sink(T₂), and heat supplied (Q₁) i.e., η= W/ Q₁  = 1 – T₂/ T₁
(The efficiency of engine is defined as the ratio of work done to the heat supplied.)
Carnot stated that no heat engine working between two given temperatures of source and sink can be more efficient than a perfectly reversible engine (Carnot engine) working between the same two temperatures. Carnot’s reversible engine working between two given temperatures is considered to be the most efficient engine.

Q18. A person of mass 60 kg wants to lose 5 kg by going up and down a 10 m high stairs. Assume he bums twice as much fat while going up than coming down. If 1 kg of fat is burnt on expending 7000 kcal, how many times must he go up and down to reduce his weight by 5 kg?

Q19. Consider a cycle tyre being filled with air by a pump. Let Vbe the volume of the tyre (fixed) and at each stroke of the pump ∆V(<< V) of air is transferred to the tube adiabatically. What is the work done when the pressure in the tube is increased from P_l to P₂
Sol: Since the process is adiabatic, there is no exchange of heat in the process, Let, pressure is increased by AP and volume is increased by AV at each stroke.
For just before and after an stroke, we can write

Q20. In a refrigerator one removes heat, from a lower temperature and deposits to the surroundings at a higher temperature. In this process, mechanical work has to be done, which is provided by an electric motor. If the motor is of 1 kW power and heat transferred from -3°C to 27°C, find the heat taken out of the refrigerator per second assuming its efficiency is 50% of a perfect engine.
Sol: Carnot designed a theoretical engine which is free from all the defects of a practical engine. The Carnot engine is the most efficient heat engine operating between two given temperatures. The efficiency of Carnot engine is

Q21. If the coefficient of performance of a refrigerator is 5 and operates at the room temperature (27°C), find the temperature inside the refrigerator.
Sol:
Key concept: The performance of a refrigerator is expressed by means of “coefficient of performance” β which is defined as the ratio of the heat extracted from the cold body to the work needed to transfer it to the hot body.

Q22. The initial state of a certain gas is (P_i,V_i T_i). It undergoes expansion till its volume becomes V_f Consider the following two cases.
a)the expansion takes place at constant temperature.
b)the expansion takes place at constant pressure.
Plot the P-V diagram for each case. In which of the two cases, is work done by the gas more?
Sol:

The situation is shown in the given P-V graph, where variation is shown for each process.
It is clear from the graph that Process 1 is isobaric and Process 2 is isothermal.
Since, work done is equal to the area under the P-V curve. Here, area under the P-V curve 1 is more. So, work done is more when the gas expands in isobaric process as in comparison of gas expands in isothermal.

Long Answer Type Questions

Q23. Consider a P-V diagram in which the path followed by one mole of perfect gas in a cylindrical container is shown in figure.
(a) Find the work done when the gas is taken from state 1 to state 2.
(b) What is the ratio of temperature T₁/T₂, if V₂ = 2V₁?




This is the amount of heat supplied.

Q24. A cycle followed by an engine (made of one mole of perfect gas in a cylinder with a piston) is shown in figure.
A to B: volume constant
B to C: adiabatic
C to D: volume constant
D to A: adiabatic

V_c  =V_D  =2V_A = 2V_B

(a) In which part of the cycle heat is supplied to the engine from outside?
(b) In which part of the cycle heat is being given to the surrounding by the engine?
(c) What is the work done by the engine in one cycle? Write your answer in term of P_A, P_B, V_A?
(d) What is the efficiency of the engine?

Sol: (a) For the process AB (which is isochoric process), volume is constant. So,
dV= 0 => dW= 0
dQ = dU + dW = dU
=> dQ = dU = Change in internal energy
Hence, in this process heat supplied is utilised to increase, internal  energy of the system.

(b) For the process CD (which is also isochoric process), volume is constant but pressure decreases.
Hence, temperature also decreases (because Pα T) so heat is given to the surroundings.
(c) To calculate work done by the engine in one cycle, we calculate work done in each part separately.

Q25. A cycle followed by an engine (made of one mole of an ideal gas in a cylinder with a piston) is shown in figure. Find heat exchanged by the engine, with the surroundings for each section of the cycle. [Cv = (3/2)/?]
(a) AB: constant volume
(b) BC: constant pressure
(c) CD: adiabatic
(d) DA : constant pressure

Sol: (a)

 By using first law of thermodynamics, we can find amount of heat associated with each process

For process AB

Volume is constant, hence work done dW = 0

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NCERT Exemplar Class 11 Physics Chapter 12 Kinetic Theory are part of NCERT Exemplar Class 11 Physics. Here we have given NCERT Exemplar Class 11 Physics Chapter 12 Kinetic Theory.

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NCERT Exemplar Class 11 Physics Chapter 12 Kinetic Theory

Q1. A cubic vessel (with face horizontal + vertical) contains an ideal gas at NTP. The vessel is being carried by a rocket which is moving at a speed of500 ms’¹ in vertical direction. The pressure of the gas inside the vessel as observed by us on the ground
(a) remains the same because 500 ms’¹ is very much smaller than v_rms of the gas.
(b) remains the same because motion of the vessel as a whole does not affect the relative motion of the gas molecules and the walls.
(c) will increase by a factor equal to (v²_rms + (500)²)/v²_rms where v_rms was the original mean square velocity of the gas.
(d) will be different on the top wall and bottom wall of the vessel.

Sol: (b) According to the ideal gas law,

P=nRT/V, here temperature of the vessel remain unchanged hence, the
pressure remains same from that point of view.
Now, let us discuss the phenomenon inside the vessel. The gas molecules keep on colliding among themselves as well as with the walls of containing vessel. These collisions are perfectly elastic.
The number of collisions per unit volume in a gas remains constant. So, the pressure of the gas inside the vessel remains the same because motion of the vessel as a whole does not affect the relative motion of the gas molecules with respect to the walls.

Q2. 1 mole of an ideal gas is contained in a cubical volume V, ABCDEFGH at 300 K (figure). One face of the cube (EFGH) is made up of a material which totally absorbs any gas molecule incident on it. At any given time,
(a) the pressure on EFGH would be zero
(b) the pressure on all the faces will be equal
(c) the pressure of EFGH would be double the pressure on ABCD
(d) the pressure on EFGH would be half that on ABCD

Sol: (d) In an ideal gas, the gas molecules keep on colliding among themselves as well as with the walls of containing vessel. These collisions are perfectly elastic. So, their kinetic energy and momentum remains conserved.
So, the momentum transferred to the face ABCD = 2mv And the gas molecule is absorbed by the face EFGH. Hence it does not rebound. So, momentum transferred to the face EFGH = mv.
And the pressure on the faces is due to the total momentum to the faces. So, pressure on EFGH would be half that on ABCD.

Q3. Boyle’s law is applicable for an
(a) adiabatic process
(b) isothermal process
(c) isobaric process
(d) isochoric process
Sol: (b)
Key concept: Boyle’s law: For a given mass of an ideal gas at constant temperature, the volume of a gas is inversely proportional to its pressure.

So we can say that when temperature is constant, Boyle’s law is applicable.

i.e.,           PV= nRT= constant
=>PV = constant (at constant temperature)
i.e.. p ∝ 1/V— [where, P = pressure. V= volume]

So, this law is applicable for an isothermal process, in which temperature remain constant.

Q4. A cylinder containing an ideal gas is in vertical position and has a piston of mass M that is able to move up or down without friction ( figure). If the temperature is increased
(a) both P and V of the gas will change
(b) only P will increase according to Charles’ law
(c) V will change but not P
(d) P will change but not V

Q5. Volume versus temperature graphs for a given mass of an ideal gas are shown in figure. At two different values of constant pressure. What can be inferred about relation between P_l and P₂ ?

(a)P₁ > P₂
(b) P_{1 =} P₂
(c) P₁ < P₂
(d) Data is insufficient

Q6. 1 mole of H₂ gas is contained in a box of volume V = 1.00 m³ at T = 300 K. The gas is heated to a temperature of T= 3000 K and the gas gets converted to a gas of hydrogen atoms. The final pressure would be (considering all gases to be ideal)

(a) same as the pressure initially
(b) 2 times the pressure initially
(c) 10 times the pressure initially
(d) 20 times the pressure initially

Sol: (d) The situation is shown in the diagram, H₂ gas is contained in a box is heated and gets converted to a gas of hydrogen atoms. Then the number of moles would become twice.
According to gas equation,
PV= nRT

P = Pressure of gas, n = Number of moles
R = Gas constant, T = Temperature PV=nRT
As volume (V) of the container is constant. Hence, when temperature (T) becomes 10 times, (from 300 K to 3000 K) pressure (P) also becomes 10 times, asP∝ T.
Pressure is due to the bombardment of particles and as gases break, the number of moles becomes twice of initial, so n₂ = 2n₁

Q7. A vessel of volume V contains a mixture of 1 mole of hydrogen and 1 mole of oxygen (both considered as ideal). Let f₁(v)dv denotes the fraction of molecules with speed between v and (v + dv) with f₂(v)dv, similarly for oxygen. Then,
(a) f₁(v) + f₂(v) = f (v) obeys the Maxwell’s distribution law
(b) f₁(v), f₂(v) will obey the Maxwell’s distribution law separately
(c) neither f₁(v)nor f₂(v)will obey the Maxwell’s distribution law
(d) f₂(v) and f₁(v)will be the same

Q8. An inflated rubber balloon contains one mole of an ideal gas, has a pressure P. volume V and temperature T. If the temperature rises to 1.1 T, and the volume is increased to 1.05 V, the final pressure will be
(a) 1.1 P
(b) P
(c) less than P
(d) between P and 1.1
Sol:(d) According to the equation of ideal gas, PV= nRT
P = pressure
V = volume
n = number of moles of gases
R = gas constant
T = temperature
Thus we have to rewrite this equation in such a way that no. of moles is given by,

More Than One Correct Answer Type

Q9. ABCDEFGH is a hollow cube made of an insulator (figure) face A BCD has positive charge on it. Inside the cube, we have ionised hydrogen.
The usual kinetic theory expression for pressure
(a) will be valid
(b) will not be valid, since the ions would experience forces other than due to collisions with the walls
(c) will not be valid, since collisions with walls would not be elastic
(d) will not be valid because isotropy is lost

Sol: (b, d) According to the problem, ionized hydrogen is present inside the cube, they are having charge. Now, due to the presence of positive charge on the surface A BCD hydrogen ions would experience forces other than the forces due to collision with the walls of container. So, these forces must be of electrostatic nature. Hence, Isotropy of system is lost at only one face ABCD because of the presence of external positive charge. The usual expression for pressure on the basis of kinetic theory will be valid.

Q10. Diatomic molecules like hydrogen have energies due to both translational as well as rotational motion. From the equation in kinetic theory PV = 2/3 E,E is
(a) the total energy per unit volume
(b) only the translational part of energy because rotational energy is very small compared to the translational energy
(c) only the translational part of the energy because during collisions with the wall pressure relates to change in linear momentum
(d) the translational part of the energy because rotational energies of molecules can be of either sign and its average over all the molecules is zero

Sol: (c) According to kinetic theory equation, PV = 2/3 E [where P= Pressure V = volume]
E is representing only translational part of energy. Internal energy contains all types of energies like translational, rotational, vibrational etc. But the molecules of an ideal gas is treated as point masses in kinetic theory, so its kinetic energy is only due to translational motion. Point mass does not have rotational or vibrational motion. Here, we assumed that the walls only exert perpendicular forces on molecules. They do not exert any parallel force, hence there will not be any type of rotation present. The wall produces only change in translational motion.

Q11. In a diatomic molecule, the rotational energy at a given temperature
(a) obeys Maxwell’s distribution
(b) have the same value for all molecules
(c) equals the translational kinetic energy for each molecule
(d) is (2/3)rd the translational kinetic energy for each molecule
Sol: (a, d)
Key concept: Kinetic Energy of Ideal Gas:
Molecules of ideal gases possess only translational motion. So they possess only translational kinetic energy.

According to the problem we have to find the rotational energy of a diatomic molecule in the terms of translation kinetic energy.
First let us check the options by picking them one by one.
(a) Translational kinetic energy and rotational kinetic energy both obey Maxwell’s distribution independent of each other.
(b) Rotational kinetic energy is different for different molecule.
(c) Molecules of ideal gases possess only translational motion. So they possess only translational kinetic energy. But in case of non-ideal gas there is a smaller rotational energy.
(d) Here, 2 rotational and 3 translational energies are associated with each molecule. Translation kinetic energy of each molecule,

Important points: Kinetic energy per molecule of a gas does not depend upon the mass of the molecule but only depends upon the temperature of the gas.

Kinetic energy per mole of gas depends only upon the temperature of gas.
Kinetic energy per gram of gas depend upon the temperature as well molecular weight (or mass of one molecule) of the gas.

From the above expressions it is clear that higher the temperature of the gas, more will be the average kinetic energy possessed by the gas molecules at T= 0, E = 0, i.e. at absolute zero the molecular motion stops.

Q12. Which of the following diagrams (figure) depicts ideal gas behaviour ?

So, graph of PV versus T will be a straight line parallel to the temperature axis (x-axis).
i.e., slope of this graph will be zero.
So, (d) is not correct.

Q13. When an ideal gas is compressed adiabatically, its temperature rises the molecules on the average have more kinetic energy than before. The kinetic energy increases,
(a) because of collisions with moving parts of the wall only
(b) because of collisions with the entire wall
(c) because the molecules gets accelerated in their motion inside the volume
(d) because of redistribution of energy amongst the molecules
Sol:(a) Since the gas is ideal and the collisions of the molecules are elastic. When the molecules collides with the moving parts of the wall, its kinetic energy increases. But the total kinetic energy of the system will remain conserved. When the gas is compressed adiabatically, the total work done on the gas increases, its internal energy which in turn increases the KE of gas molecules and hence, the collisions between molecules also increases.

Very Short Answer Type Questions
Q14. Calculate the number of atoms in 39.4 g gold. Molar mass of gold is 197 g mole^-1

Q15. The volume of a given mass of a gas at 27°C, 1 atm is 100 cc. What will be its volume at 327°C?
Sol:
Key concept: Here the temperatures are given in Celsius. To apply ideal gas equation, we must convert the given temperature in kelvin. So, to convert them in kelvin we use the relation

Q16. The molecules of a given mass a gas have root mean square speeds of 100 ms^-1 at 27°C and 1.00 atmospheric pressure. What will be the root mean square speeds of the molecules of the gas at 127°C and 2.0 atmospheric pressure?
Sol: Key concept:Root Mean Square Speed: It is defined as the square root of mean of squares of the speed of different molecules

Q17. Two molecules of a gas have speeds of 9 x 10¹⁶ ms^-1 and 1 x I0⁶ ms^-1 respectively. What is the root mean square speed of these molecules? 

Q18. A gas mixture consists of 2.0 moIes of oxygen and 4.0 moles of neon at temperature T. Neglecting all vibrational modes, calculate the total internal energy of the system. (Oxygen has two rotational modes.)

Sol: Key concept: Degree of Freedom:
The term degree of freedom of a system refers to the possible independent motions, systems can have or
The total number of independent modes (ways) in which a system can possess energy is called the degree of freedom (f).
The independent motions can be translational, rotational or vibrational or any combination of these.
So the degree of freedom are of three types:
(i) Translational degree of freedom
(ii) Rotational degree of freedom
(iii) Vibrational degree of freedom
General expression for degree of freedom
f=3A- B; where A = Number of independent particles.
B = Number of independent restriction

Diatomic gas: Molecules of diatomic gas are made up of two atoms joined rigidly to one another through a bond. This cannot only move bodily, but also rotate about one of the three co-ordinate axes. However its moment of inertia about the axis joining the two atoms is negligible compared to that about the other two axes.
Hence it can have only two rotational motion. Thus a diatomic molecule has 5 degree of freedom: 3 translational and 2 rotational.
Monoatomic gas: Molecules of
monoatomic gas can move in any direction in space so it can have three independent motions and hence 3 degrees of freedom (all translational).

Neon (Ne) is a monoatomic gas having 3 degrees of freedom.
Energy per mole = 3/2RT
Hence, Energy = 4 x3/2 RT = 6RT ….(ii)
[Using Eqs. (i) and (ii)]
Total energy = 5RT = 6RT= 11RT

Q19. Calculate the ratio of the mean free paths of the molecules of two gases having molecular diameters 1 A and 2 A. The gases may be considered under identical conditions of temperature, pressure and volume.

Short Answer Type Questions

When the partition is removed, the gases get mixed without any loss of energy. The mixture now attains a common equilibrium pressure and the total volume of the system is sum of the volume of individual chambers V1 and V2. Let P be the pressure after the partition is removed.

Q21. A gas mixture consists of molecules of A, B and C with masses m_A > m_B > m_c. Rank the three types of molecules in decreasing order of (a) average KE, (b) rms speeds.

Q22. We have 0.5 g of hydrogen gas in a cubic chamber of size 3 cm kept at NTP. The gas in the chamber is compressed keeping the temperature constant till a final pressure of 100 atm. Is one justified in assuming the ideal gas law, in the final state? (Hydrogen molecules can be consider as spheres of radius 1 A).

Q23. When air is pumped into a cycle tyre the volume and pressure of the air in the tyre both are increased. What about Boyle’s law in this case?
Sol: Here, according to the question, when air is pumped, more molecules are pumped and Boyle’s law is stated for situation where, mass of molecules remains constant.

In this case, when air is pumped into a cycle tyre, mass of air in it increases as the number of air molecules keep increasing. Hence, this is a case of variable mass, Boyle’s law (and even Charle’s law) is only applicable in situations, where mass of gas molecules remains fixed. Hence, Boyle’s law is not applicable in this case.

Q24. A balloon has 5.0 mole of helium at 7°C. Calculate
(a) the number of atoms of helium in the balloon.
(b) the total internal energy of the system.

Important point: The above degrees of freedom are shown at room temperature. Further at high temperature, in case of diatomic or polyatomic molecules, the atoms with in the molecule may also vibrate with respect to each other. In such cases, the molecule will have an additional degrees of freedom, due to vibrational motion.
An object which vibrates in one dimension has two additional degrees of freedom. One for the potential energy and one for the kinetic energy of vibration. Helium is a mono atomic gas and It has only 3 degrees of freedom. But after addition its degree of freedom will be 5.

Q25. Calculate the number of degrees of freedom of molecules of hydrogen in 1 cc of hydrogen gas at NTP.

Q26. An insulated container containing monoatomic gas of molar mass m is moving with a velocity v₀. If the container is suddenly stopped, find the change in temperature.
Sol: Since, the container is suddenly stopped which is initially moving with velocity v₀, there is no time for exchange of heat in the process. Then total KE of the container is transferred to gas molecules in the form of translational KE, thereby increasing the absolute temperature.
Let n be the no. of moles of the monoatomic gas in the container. Since molar mass of the gas is m.
Total mass of the container, M = mn
KE of molecules due to velocity v₀,
KE = 1/2(mn) v₀²

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NCERT Exemplar Class 11 Physics Chapter 1 Units and Measurements are part of NCERT Exemplar Class 11 Physics. Here we have given NCERT Exemplar Class 11 Physics Chapter 1 Units and Measurements.

NCERT Exemplar Class 11 Physics Chapter 1 Units and Measurements

Single Correct Answer Type

Q1. The number of significant figures in 0.06900 is
(a) 5 (b) 4 (c) 2 (d) 3
Sol: (b)
Key concept: Significant figures in the measured value of a physical quantity tell the number of digits in which we have confidence. Larger the number of significant figures obtained in a measurement, greater is the accuracy of the measurement. The reverse is also true.
The following rules are observed in counting the number of significant figures in a given measured quantity.
1. All non-zero digits are significant.
2. A zero becomes significant figure if it appears between two non¬zero digits.
3. Leading zeros or the zeros placed to the left of the number are never significant.
4. Trailing zeros or the zeros placed to the right of the number are significant.
5. In exponential notation, the numerical portion gives the number of significant figures.
Leading zeros or the zeros placed to the left of the number are never

Hence, number of significant figures are four.

Q2. The sum of the numbers 436.32, 227.2 and 0.301 inappropriate significant figures is
(a) 663.821 (b) 664 (c) 663.8 (d) 663.82
Sol: (b) The result of an addition or subtraction in the number having different precisions should be reported to the same number of decimal places as present in the number having the least number of decimal places.

The final result should, therefore, be rounded off to one decimal place, i.e. 664.

Q3. The mass and volume of a body are 4.237 g and 2.5 cm³, respectively. The density of the material of the body in correct significant figures is
(a) 1. 6048 g cm^-3
(b) 1.69 g cm^-3
(c) 1.7 g cm ³                                           
(d) 1.695 g cm^-3

Sol: (c) The answer to a multiplication or division is rounded off to the same number of significant figures as possessed by the least precise term used in the calculation. The final result should retain as many significant figures as are there in the original number with the least significant figures. In the given question, density should be reported to two significant figures

After rounding off the number, we get density =1.7

Q4. The numbers 2.745 and 2.735 on rounding off to 3 significant figures will give
(a) 2.75 and 2.74
(b) 2.74 and 2.73
(c) 2.75 and 2.73
(d) 2.74 and 2.74
Sol: (d)
Key concept: While rounding off measurements, we use the following rules by convention:
1. If the digit to be dropped is less than 5, then the preceding digit is left unchanged.
2. If the digit to be dropped is more than 5, then the preceding digit is raised by one.
3. If the digit to be dropped is 5 followed by digits other than zero, then the preceding digit is raised by one.
4. If digit to be dropped is 5 or 5 followed by zeros, then preceding digit is left unchanged, if it is even.
5. If digit to be dropped is 5 or 5 followed by zeros, then the preceding digit is raised by one, if it is odd.
Units and Measurements
Let us round off 2.745 to 3 significant figures.
Here the digit to be dropped is 5, then preceding digit is left unchanged, if it is even.
Hence on rounding off 2.745, it would be 2.74.
Now consider 2.737, here also the digit to be dropped is 5, then the preceding digit is raised by one, if it is odd. Hence on rounding off 2.735 to 3 significant figures, it would be 2.74.

Q5. The length and breadth of a rectangular sheet are 16.2 cm and 10.1 cm, re­spectively. The area of the sheet in appropriate significant figures and error is
(a) 164 ±3 cm²
(b) 163.62 ± 2.6 cm²
(c) 163.6 ±2.6 cm²                                  
(d) 163.62 ±3 cm²
Sol: (a)

Q6. Which of the following pairs of physical quantities does not have same dimensional formula?
(a) Work and torque
(b) Angular momentum and Planck’s constant
(c) Tension and surface tension
(d) Impulse and linear momentum

Q7. Measure of two quantities along with the precision of respective measuring instrument is
A = 2.5 ms^-1 ± 0.5 ms^-1, B = 0.10 s ± 0.01 s. The value of AB will be
(a) (0.25 ± 0.08) m
(b)  (0.25       ± 0.5) m
(c) (0.25 ± 0.05) m
(d)  (0.25    ± 0.135) m

Q8. You measure two quantities as A = 1.0 m ± 0.2 m, B = 2.0 m ± 0.2 m. We should report correct value for √AB  as
(a) 1.4 m± 0.4 m
(b) 1.41 m± 0.15 m
(c) 1.4 m + 0.3 m
(d) 1.4 m± 0.2 m

Q9. Which of the following measurements is most precise?
(a) 5.00 mm
(b) 5.00 cm
(c) 5.00 m
(d) 5.00 km
Sol:(a)
Key concept: Precision is the degree to which several measurements provide answers very close to each other. It is an indicator of the scatter in the data. The lesser the scatter, higher the precision.
Let us first check the units. In all the options magnitude is same but units of measurement are different. As here 5.00 mm has the smallest unit. All given measurements are correct upto two decimal places. However, the absolute error in (a) is 0.01 mm which is least of all the four. So it is most precise.

Q10. The mean length of an object is 5 cm. Which of the following measurements is most accurate?
(a) 4.9 cm
(b) 4.805 cm
(c) 5.25 cm
(d) 5.4 cm
Sol: (a)
Key concept: Accuracy describes the nearness of a measurement to the standard or true value, i.e. a highly accurate measuring device will provide measurements very close to the standard, true or known values.
Example: In target shooting, a high score indicates the nearness to the bull’s eye and is a measure of the shooter’s accuracy.

Q11. Young’s modulus of steel is 1.9 x 10¹¹ N/m². When expressed in CGS units of dyne/cm², it will be equal to (1 N = 10⁵ dyne, 1 m² = 10⁴ cm²)                     .
(a) 1.9 xlO¹⁰                                            
(b) 1.9×10¹²
(c) 1.9 xlO¹²                                           
(d) 1.9 xlO¹³

Q12. If momentum (p), area (A) and time (T) are taken to be fundamental quantities, then energy has the dimensional formula

More Than One Correct Answer Type
Q13. On the basis of dimensions, decide which of the following relations for the displacement of a particle undergoing simple harmonic motion is not correct?

Hence, (c) is not the correct option.
=> LHS ≠ RHS.
So, option (b) is also not correct.

Q14. If P, Q, R are physical quantities, having different dimensions, which of the following combinations can never be a meaningful quantity?
(a) (P-Q)/R          
(b) PQ-R
(c) PQ/R                                  
(d) (PR-Q²)/R
(e)(R + Q)/P
Sol: (a, e)
Key concept: Principle of Homogeneity of dimensions: It states that in a correct equation, the dimensions of each term added or subtracted must be same. Every correct equation must have same dimensions on both sides of the equation.
According to the problem P, Q and R are having different dimensions, since, sum and difference of physical dimensions, are meaningless, i.e., (P – Q) and (R + Q) are not meaningful.
So in option (b) and (c), PQ may have the same dimensions as those of R and in option (d) PR and Q² may have same dimensions as those of R.
Hence, they cannot be added or subtracted, so we can say that (a) and (e) are not meaningful.

Q15. Photon is a quantum of radiation with energy E = hv, where v is frequency and h is Planck’s constant. The dimensions of h are the same as that of
(a) Linear impulse
(b) Angular impulse
(c) Linear momentum                          
(d) Angular momentum

Q16. If Planck’s constant (h) and speed of light in vacuum (c) are taken as two fundamental quantities, which one of the following can, in addition, be taken to express length, mass and time in terms of the three chosen fundamental quantities?
(a) Mass of electron (m_e)             
(b) Universal gravitational constant (G)
(c) Charge of electron (e)             
(d) Mass of proton (m_p)

Q17. Which of the following ratios express pressure?
(a) Force/Area                                       
(b) Energy/Volume
(c) Energy/Area                                     
(d) Force/Volume

Sol: (a, b) Let us first express the relation of pressure with other physical quantities one by one with the help of dimensional analysis.

Q18. Which of the following are not a unit of time?
(a). Second
(b) Parsec
(c) Year
(d) Lightyear
Sol: (b, d) Parsec and light year are those practical units which are used to measure large distances. For example, the distance between sun and earth or other celestial bodies. So they are the units of length not time. Here, second and year represent time.
Important point: 1 light year (distance that light travels in 1 year with speed = 3 x 10⁸ m/s.) = 9.46 x 10¹¹ m And 1 par see = 3.08 x 10¹⁶ m

Very Short Answer Type Questions

Q19. Why do we have different units for the same physical quantity?
Sol: Magnitude of any given physical quantity may vary over a wide range, therefore, different units of same physical quantity are required.
For example:
1.Mass ranges from 10^-30 kg (for an electron) to 10⁵³ kg (for the known universe). We need different units to measure them like miligram, gram, kilogram etc.
2.The length of a pen can be easily measured in cm, the height of a tree can be measured in metres, the distance between two cities can be measured in kilometres and distance between two heavenly bodies can be measured in light year.

Q20. The radius of atom is of the order of 1 A and radius of nucleus is of the order of fermi. How many magnitudes higher is the volume of atom as compared to the volume of nucleus?

Q21. Name the device used for measuring the mass of atoms and molecules.
Sol: A mass spectrograph is a device which is used for measuring the mass of atoms and molecules.

Q22. Express unified atomic mass unit in kg.
Sol: The unified atomic mass unit is the standard unit that is used for indicating mass on an atomic or molecular scale (atomic mass). One unified atomic mass unit is approximately the mass of one nucleon (either a single proton or neutron) and is numerically equivalent to 1 g/mol. It is defined as one- twelfth of the mass of an unbound neutral atom of carbon-12 in its nuclear and electronic ground state.

Q24. Why length, mass and time are chosen as base quantities in mechanics?
Sol: Normally each physical quantity requires a unit or standard for its specification, so it appears that there must be as many units as there are physical quantities. However, it is not so. It has been found that if in mechanics we choose arbitrarily units of any three physical quantities we can express the units of all other physical quantities in mechanics in terms of these. So, length, mass and time are chosen as base quantities in mechanics because
(i) Length, mass and time cannot be derived from one another, that is these quantities are independent.
(ii) All other quantities in mechanics can be expressed in terms of length, mass and time.

Short Answer Type Questions
25. (a) The earth-moon distance is about 60 earth radius. What will be the . diameter of the earth (approximately in degrees) as seen from the moon?
(b) Moon is seen to be of (1/2)° diameter from the earth. What must be the relative size compared to the earth?
(c) From parallax measurement, the sun is found to be at a distance of about 400 times the earth-moon distance. Estimate the ratio of sun-earth diameters.

Q26. Which of the following time measuring devices is most precise?
(a) A wallclock                                        
(b) A stop watch
(c) A digital watch                                  
(d) An atomic clock
Given reason for your answer.
Sol: Option (d) is correct because a clock can measure time correctly up to one second. A stop watch can measure time correctly up to a fraction of a second. A digital watch can measure time up to a fraction of second whereas an atomic clock is the most accurate timekeeper and is based on characteristic frequencies of radiation emitted by certain atoms having precision of about 1 second in 300,000 years. So an atomic clock can measure time most precisely as precision of this clock is about 1 s in 10¹³ s.

Q27. The distance of a galaxy is of the order of 10²⁵ Calculate the order of magnitude of time taken by light to reach us from the galaxy.
Sol: According to the problem, distance of the galaxy = 10²⁵m.
Speed of light = 3 x 10⁸ m/s
Hence, time taken by light to reach us from galaxy is

Q28. The Vernier scale of a travelling microscope has 50 divisions which coincide with 49 main scale divisions. If each main scale division is 0.5 mm, calculate the minimum inaccuracy in the measurement of distance.

Q29. During a total solar eclipse the moon almost entirely covers the sphere of the sun. Write the relation between the distances and sizes of the sun and moon.
Sol: Key point: In geometry, a solid angle (symbol: Ω or w) is the two­dimensional angle in three-dimensional space that an object subtends at a point. It is a measure of how large the object appears to an observer looking from that point. In the International System of Units (SI), a solid angle is expressed in a dimensionless unit called a steradian (symbol: sr).

Q30. If the unit of force is 100 N, unit of length is 10 m and unit of time is 100 s, what is the unit of mass in this system of units?

Q31. Give an example of
(a) a physical quantity which has a unit but no dimensions
(b) a physical quantity which has neither unit nor dimensions
(c) a constant which has a unit
(d) a constant which has no unit

Q32. Calculate the length of the arc of a circle of radius 31.0 cm which subtends an angle of π/6 at the centre.

Q33. Calculate the solid angle subtended by the periphery of an area of 1 cm² at a point situated symmetrically at a distance of 5 cm from the area.

Important point: Please keep in mind that solid angle is for 3-D figure like sphere, cone etc and plane angle is for plane objects or 2-D figures like circle, arc etc.

Q34. The displacement of a progressive wave is represented by y =A sin(wt– kx), where x is distance and / is time. Write the dimensional formula of (i) w and (ii) k 
Sol: We have to apply principle of homogeneity to solve this problem. Principle of homogeneity states that in a correct equation, the dimensions of each term added or subtracted must be same, i.e., dimensions of LHS and RHS should be equal.
According to the problem

Q35. Time for 20 oscillations of a pendulum is measured as t₁ =39.6 s; t₂ = 39.9 s and t₃ = 39.5 s. What is the precision in the measurements? What is the accuracy of the measurement?
Sol: According to the problem, time for 20 oscillations of a pendulum,
t₁ = 39.6 s, t₂ = 39.9 s and t₃ = 39.5 s
It is quite obvious from these observations that the least count of the watch is 0.1 s. As measurements have only one decimal place. Precision in the measurement = Least count of the measuring instrument= 0.1 s
Precision in 20 oscillations = 0.1

Long Answer Type Questions

Q36. A new system of units is proposed in which unit of mass is α kg, unit of length β m and unit of time γ s). How much will 5J measure in this new system?
Sol: For solving this problem, dimensions of physical quantity will remain same whatever be the system of units of its measurement.

Q40. If velocity of light c, Planck’s constant h and gravitational constant G are taken as fundamental quantities, then express mass, length and time in terms of dimensions of these quantities.
Sol: We have to apply principle of homogeneity to solve this problem. Principle of homogeneity states that in a correct equation, the dimensions of each term added or subtracted must be same, i.e., dimensions of LHS and RHS should be equal,
We know that, dimensions of

Q41. An artificial satellite is revolving around a planet of mass M and radius R, in a circular orbit of radius r. From Kepler’s third law about the period of a satellite around a common central body, square of the period of revolution T is proportional to the cube of the radius of the orbit r. Show using dimensional

Q42. In an experiment to estimate ‘the size of a molecule of oleic acid, 1 mL of oleic acid is dissolved in 19 mL of alcohol. Then 1 mL of this solution is diluted to 20 mL by adding alcohol. Now, 1 drop of this diluted solution is placed on water in a shallow trough. The solution spreads over the surface of water forming one molecule thick layer. Now, lycopodium powder is sprinkled evenly over the film and its diameter is measured. Knowing the volume of the drop and area of the film we can calculate the thickness of the film which will give us the size of oleic acid molecule.

Read the passage carefully and answer the following questions.

• Why do we dissolve oleic acid in alcohol?
• What is the role of lycopodium powder?
• What would be the volume of oleic acid in each mL of solution prepared?
• How will you calculate the volume of n drops of this solution of oleic
• What will be the volume of oleic acid in one drop of this solution?

Sol: (a) Since Oleic acid does not dissolve in water, hence it is dissolved in alcohol.

(b)Lycopodium powder spreads on the entire surface of water when it is sprinkled evenly. When a drop of prepared solution of oleic acid and alcohol is dropped on water, oleic acid does not dissolve in water. Instead it spreads on the water surface pushing the lycopodium powder away to clear a circular area where the drop falls. We can thus be able to measure the area over which oleic acid spreads.

(c)Since 20 mL (1 mL oleic acid + 19 mL alcohol) contains 1 mL of oleic acid, oleic acid in each mL of the solution =1/20 mL. Further, as this 1 mL is diluted to 20 mL by adding alcohol. In each mL of solution prepared, volume of oleic acid = 1/20 mL x 1/20  = 1/400 mL

(d) Volume of n drops of this solution of oleic acid can be calculated by means of a burette (used to make solution in the form of countable drops) and measuring cylinder and measuring the number of drops.

(e) As 1 mL of solution contains n number of drops, then the volume of oleic acid in one drop will be = 1/(400)n mL

Q43. (a) How many astronomical units (AU) make 1 parsec?
(b) Consider the sun like a star at a distance of 2 parsecs. When it is seen through a telescope with 100 magnification, what should be the angular size of the star? Sun appears to be (1/2)° from the earth. Due to atmospheric fluctuations, eye cannot resolve objects smaller than 1 arc minute.
(c) Mars has approximately half of the earth’s diameter. When it is closest to the earth it is at about 1/2 AU from the earth. Calculate at what size it will appear when seen through the same telescope.

Q44. Einstein’s mass-energy relation emerging out of his famous theory of relativity relates mass (m) to energy (E) as E= mc², where c is speed of light in vacuum. At the nuclear level, the magnitudes of energy are very small. The energy at nuclear level is usually measured in MeV, where 1 MeV = 1.6 x 10^-13 J; the masses are measured in unified atomic mass unit (u) where, 1 u = 67 x 10^-27 kg.
(a) Show that the energy equivalent of 1 u is 931.5 MeV.
(b) A student writes the relation as 1 u = 931.5 MeV. The teacher points out that the relation is dimensionally incorrect. Write the correct relation.
Sol: (a) We can apply Einstein’s mass-energy relation in this problem, E = mc², to calculate the energy equivalent of the given mass.
Here

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We hope the NCERT Exemplar Class 11 Physics Chapter 1 Units and Measurements help you. If you have any query regarding NCERT Exemplar Class 11 Physics Chapter 1 Units and Measurements, drop a comment below and we will get back to you at the earliest.

NCERT Exemplar Class 11 Physics Chapter 2 Motion in a Straight Line are part of NCERT Exemplar Class 11 Physics. Here we have given NCERT Exemplar Class 11 Physics Chapter 2 Motion in a Straight Line.

NCERT Exemplar Class 11 Physics Chapter 2 Motion in a Straight Line

Multiple Choice Questions
Single Correct Answer Type
Motion In A Straight Line Numericals NCERT Class 11

Q1. Among the four graphs shown in the figure, there is only one graph for which average velocity over the time interval (0, 7) can vanish for a suitably chosen T. Which one is it?

Key concept: Average velocity : It is defined as the ratio of displacement to time taken by the body.
Displacement /Time taken
According to this problem, we need to identify the graph which is having same displacement for two timings. When there are two timings for same displacement, the corresponding velocities should be in opposite directions.
As shown in graph (b), the first slope is decreasing that means particle is going in one direction and its velocity decreases, becomes zero at highest point of curve and then increasing in backward direction. Hence the particle
return to its initial position. So, for one value of displacement there are two different points of time and we know that slope of x, x-t graph gives us the average velocity. Hence, for one time, slope is positive then average velocity is A also positive and for other time slope is negative then average velocity is also negative.

As there are opposite velocities in the interval 0 to T, hence average velocity can vanish in (b).
This can be seen in the figure given alongside.
As shown in the graph, OA = BT (same displacement) for two different points of time.

Important points:
Various position-time graphs and their interpretation

1. Graph: Line parallel to time axis

Interpretation: It represents that the particle is at rest.

2. Graph: Line perpen¬dicular to time axis

Interpretation: It represents that particle is changing its position but time does not change, it means the particle possesses infinite velocity.
This situation is practically not possible.

3. Graph: Line with constant slope

Interpretation: It represents uniform velocity of the particle.

4. Graph: Parabola bending towards position axis

Interpretation: It represents increasing velocity of the particle. It means the particle possesses acceleration.
Hence slope of position-time graph goes on increasing.

5. Graph: Parabola bending towards time axis

Interpretation: It represents decreasing velocity of particle. It means the particle possesses retardation.
Hence slope of position-time graph goes on decreasing.

6. Graph: Line with negative slope

Interpretation: It represents that the particle returns towards the point of reference (negative displacement) with uniform velocity.

Problems On Motion In A Straight Line NCERT Class 11 

2. A lift is coming from 8th floor and is just about to reach 4th floor. Taking ground floor as origin and positive direction upwards for all quantities, which one of the following is correct?
(a) x < 0, v < 0, a > 0 (b) x > 0, v < 0, a < 0
(c) x > 0, v < 0, a > 0 (d) x > 0, v > 0, a < 0
Sol:(a)
Key concept: The time rate of change of velocity of an object is called acceleration of the object.
It is a vector quantity. Its direction is same as that of change in velocity (Not of the velocity).
In the table: Possible ways of velocity change

When only direction of velocity changes	When only magnitude of velocity changes	When both magnitude and direction of velocity change
Acceleration perpendicular to velocity	Acceleration parallel or anti­parallel to velocity	Acceleration has two components—one is perpendicular to velocity and another parallel or anti­parallel to velocity
E.g.: Uniform circular motion	E.g.: Motion under gravity	E.g: Projectile motion

Here we will take upward direction positive. As. the lift is coming in downward direction, the displacement will be negative. We have to see whether the motion is accelerating or retarding.
We know that due to downward motion displacement will be negative. When the lift reaches 4th floor and is about to stop velocity is decreasing with time, hence motion is retarding in nature. Thus, x < 0; a > 0. Asdisplacementisinnegativedirection, velocity will also be negative, i.e. v < 0.
The motion of lift will be shown like this.

Motion In A Straight Line Problems NCERT Class 11

Q3. In one dimensional motion, instantaneous speed v satisfies 0 < v < v₀
(a )The displacement in time T must always take non-negative values.
(b) The displacement x in time T satisfies -v₍₎T < x < v₀
(c) The acceleration is always a non-negative number.
(d) The motion has no turning points.
Sol: (b) .
Key concept: Instantaneous speed: It is the speed of a particle at a particular instant of time. When we say “speed”, it usually means instantaneous speed. The instantaneous speed is average speed for infinitesimally small time interval (i.e., ∆ –> 0).

As instantaneous speed is less than maximum speed. Then either the velocity is increasing or it is decreasing. For maximum and minimum displacement we have to keep in mind the magnitude and direction of maximum velocity.
As maximum velocity in positive direction is v₀, magnitude of maximum velocity in opposite direction is also v₀.
Maximum displacement in one direction = v₀T Maximum displacement in opposite directions = -v₀T Hence,-v₀T<x< v₀T.
Important point: We should not confuse with direction of velocities, i.e., in one direction it is taken as positive and in another direction it is taken as negative.

Motion In Straight Line Numericals NCERT Class 11 

Q4. A vehicle travels half the distance L with speed V₁ and the other half with speed v₂, then its average speed is

Let the vehicle travels from A to B. Distances, velocities and time taken are shown. To calculate average speed we will calculate total distance covered and will divide by time interval in which it covers that total distance.

Motion In A Straight Line Class 11 Numericals NCERT

Q5. The displacement of a particle is given by x = (t- 2)² where x is in metres and t in seconds. The distance covered by the particle in first 4 seconds is
(a) 4 m
(b) 8 m                    
(c) 12 m                  
(d) 16 m
Sol: (b)
Key concept: Instantaneous velocity : Instantaneous velocity is defined as the rate of change of position vector of particles with time at a certain instant of time.

i.e., if x  is given as a function of time, second time derivative of displacement gives acceleration.
In such type of problems we have to analyze whether the motion is accelerating or retarding. When acceleration is parallel to velocity, velocity of particle increases with time, i.e. motion is accelerated. And when acceleration is anti-parallel to velocity, velocity of particle decreases with time, i.e. motion is retarded. During retarding journey, particle will stop in between.
According to the problem, displacement of the particle is given as a function of time.
x = (t-2)²
By differentiating this equation w.r.t. time we get velocity of the particle as a function of time.
v = dx/ dt = d/dt ( t-2)² = 2(t – 2) m/s
If we again differentiate this equation w.r.t. time we will get acceleration of the particle as a function of time.

Class 11 Motion In A Straight Line Important Questions

Q6. At a metro station, a girl walks up a stationary escalator in time t₁ If she remains stationary on the escalator, then the escalator takes her up in time t₂. The time taken by her to walk up on the moving escalator will be
(a) (t_l+t₂)/2 (b) t
(b) t₁t₂ /(t₂ – t₁)
(c) t₁t₂ /(t₂ +t₁)                                     
(d) t₁ —t₂
Sol: (c)

Key concept: Net velocity when object is moving on the moving frame in One Dimension:
We will define this concept by taking an example.
River-Man problem in one dimension:
Velocity of river water current is u and velocity of man in still water is v, i.e. man can swim in water with velocity v.

Problem-Solving Tips for Relative Velocity

• If the velocity is mentioned without specifying the frame, assume it is with respect to the ground.
• In many cases, a body travels on water or in air. Depending on the context you will have to figure out whether the velocity is with respect to the water/air or with respect to the ground.
• In some situations you have to presume the velocities. For example, if the problem says that a man can walk at a maximum of 8 kmlf¹ and if it asks you to find the velocity on a train, then you have to assume that the velocity of the man with respect to the surface he is on (in this case the tram is 8 kmh^-1). Similarly the velocity of a bullet is always measured with respect to the gun. If the gun is mounted on a truck, the bullet will have a different velocity.

If particle is moving with constant velocity towards right (+x-axis):

Equation to be used: x = x_Q + vt. Graph will be a straight line.

Let the particle be at some point P initially at time t – 0 which is at a distance of x₀ from origin. Since the particle is moving towards right so its distance from origin goes on increasing. Hence position-time graph for a particle moving with constant velocity towards right will be a straight line inclined to time axis making an acute angle α.
Recall that tan α is slope of position-time graph which is equal to velocity of the particle.

For uniform motion velocity is constant, hence slope will be positive. Hence quantity A is displacement.
If particle is moving with a constant positive acceleration:
Equation to be used: v = u + at
As the time passes velocity goes on increasing. Hence velocity-time graph for a particle moving with constant positive acceleration is a straight line inclined to time axis making an acute angle a. Here tan a is the slope of velocity-time graph (Figure).

For uniformly accelerated motion, slope will be positive and A will represent velocity.

Q9. A graph of x versus t is shown in figure.

Choose the correct alternatives given below.
(a) The particle was released from rest at t =0
(b) At B, the acceleration a >0
(c) At C, the velocity and the acceleration vanish.
(d) Average velocity for the motion between Aand D is positive.
(e) The speed at D exceeds that at E

Sol: (a, c, e)
Key concept: We know that velocity v = dx/dt and slope of x-t graph gives ‘
us velocity. This implies slope = dx/dt for the graph.
As per the diagram, at point A the graph is parallel to time axis, hence dx
v = dx/dt = 0. As the starting point is A, hence we can say that the particle is starting from rest. Thus option (a) is correct.
At C, the graph changes slope, hence velocity also changes. As graph at C is almost parallel to time axis, hence we can say that velocity vanishes. Hence option (c) is correct.
As direction of acceleration changes, hence we can say that it may be zero in between.
From the graph it is clear that | slope at D| > | slope at E |
Hence, speed at D will be more than at E. Hence option (e) is correct.
Important point: Here, negative slope does not mean less value. It represents change in direction of velocity.

Q9. For the one-dimensional motion, described by x = t – sin t.
(a) x(t) > 0 for all t > 0 (b) v(t) > 0 for all r > 0
(c) a(t) > 0 for all t > 0 (d) v(t) lies between 0 and 2

Sol: (a, d) Position of the particle is given as a function of time i.e. x = t – sint By differentiating this equation w.r.t. time we get velocity of the particle as a function of time.

Important points:
(i) When sinusoidal function is involved in an expression we should be careful about sine and cosine functions.
(ii) We should be very careful when calculating maximum and minimum value of velocity because it is in inverse relation with cost in the given expression.

Q10. A spring with one end attached to a mass and the other to a rigid support is stretched and released.
(a) Magnitude of acceleration, when just released is maximum.
(b) Magnitude of acceleration, when at equilibrium position, is maximum.
(c) Speed is maximum when mass is at equilibrium position.
(d) Magnitude of displacement is always maximum whenever speed is minimum
Sol:(a, c)

Q11. A ball is bouncing elastically with a speed 1 m/s between walls of a railway compartment of size 10 m in a direction perpendicular to the walls. The train is moving at a constant velocity of 10 m/s parallel to the direction of motion of the ball. As seen from the ground,
(a) the direction of motion of the ball changes every 10 seconds.
(b) speed of ball changes every 10 seconds.
(c) average speed of ball over any 20 second interval is fixed.
(d) the acceleration of ball is the same as from the train.
Sol: (b, c, d) In this problem, we have to observe the motion from different frames. Here the problem can be solved by the frame of the observer but here we must be clear that we are considering the motion from the ground so we just keep in mind the motion from frame of observer. Compared to the velocity of trains (10 m/s) speed of ball is less (1 m/s). (b, c, d) In this problem, we have to observe the motion from different frames. Here the problem can be solved by the frame of the observer but here we must be clear that we are considering the motion from the ground so we just keep in mind the motion from frame of observer. Compared to the velocity of trains (10 m/s) speed of ball is less (1 m/s).
The speed of the ball before collision with side of train is 10 + 1 = 11 m/s Speed after collision with side of train =10-1=9 m/s As speed is changing after travelling 10 m and speed is 1 m/s, hence time duration of the changing speed is 10 s.
Since, the collision of the ball is perfectly elastic there is no dissipation of energy, hence total momentum and kinetic energy are conserved.
Since, the train is moving with a constant velocity, hence it will act as an inertial frame of reference as that of Earth and acceleration will be same in both frames.
Remember: We should not confuse with non-inertial and inertial frame of reference. A frame of reference that is not accelerating will be inertial.

Very Short Answer Type Questions

Ans. (a) (iii); (b) (ii); (c) (iv);(d) -(i)
Sol: Let us pick graphs one by one.

In graph (a),
There is a point (B) on the curve for which displacement is zero. So curve, (a) matches with (iii).

In graph (b),
In this graph, x is positive (> 0) throughout and at point B the highest point of curve the slope of curve is zero. It means at
this point v = dx/dt = 0 . Also at point C the dt
curvature changes, it means at this point the acceleration of the particle should be zero or a = 0, So curve (b) matches with (ii).

In graph (c),
In this graph the slope is always negative, hence velocity will be negative or v < 0. Also x-t graph opens up, it represents positive acceleration. So curve (c) matches with (iv).

In graph (d),
In this graph the slope is always positive, hence velocity will be positive or v > 0. Also x-t graph opens down, it represents negative acceleration. So curve (d) matches with (i).

13. A uniformly moving cricket ball is turned back by hitting it with a bat for a very short time interval. Show the variation of its acceleration with time (Take acceleration in the backward direction as positive).
Sol: Impulsive Force is generated by the bat: If we ignore the effect of gravity just by analyzing the motion of ball in horizontal direction only, then ball moving uniformly will return back with the same speed when a bat hits it.
Acceleration of the ball is zero just before it strikes the bat. When the ball strikes the bat, it gets accelerated due to the applied impulsive force by the bat.

Q14. Give examples of a one-dimensional motion where
(a) the particle moving along positive x-direction comes to rest periodically and moves forward.
(b) the particle moving along positive x-direction comes to rest periodically and moves backward.
Sol: The equation which contains sine and cosine functions is periodic in nature.
(a) The particle will be moving along positive x-direction only if t > sin t We have displacement as a function of time, x(t) = t – sin t By differentiating this equation w.r.t. time we get velocity of the particle as a function of time.

At t = 2π, x = 0, v = 1 (positive) and a = 0
Hence the particle moving along positive x-direction comes to rest periodically and moves backward.
As displacement and velocity is involving sin t and cos t, hence these equations represent periodic nature.

Q15. Give example of a motion where x > 0, v < 0, a > 0 at a particular instant.
Sol: Let the motion is represented by

Q16. An object falling through a fluid is observed to have acceleration given by a = g – bvwhere g= gravitational acceleration and b is constant After a long time of release, it is observed to fall with constant speed. What must be the value of the constant speed?
Sol: Key concept: If a spherical body of radius r is dropped in a viscous fluid, it is first accelerated and then its acceleration becomes zero and it attains a constant velocity called terminal velocity.
According to the problem, acceleration of object is given by the relation
a=g-bv
When speed becomes constant acceleration a = dv/dt = 0 (uniform motion).
where, g = gravitational acceleration

Clearly, from above equation as speed increases acceleration will decrease. At a certain speed say v₀, acceleration will be zero and speed will remain constant. Hence, a = g- bv₀ = 0 => v₀ = g/b

Short Answer Type Questions

Q17. A ball is dropped and its displacement versus time graph is as shown (Displacement x from ground and all quantities are positive upwards).
(a) Plot qualitatively velocity versus time graph.
(b) Plot qualitatively acceleration versus time graph.

Q19. A bird is tossing (flying to and fro) between two cars moving towards each other on a straight road. One car has a speed of 18 km/h while the other has the speed of 27 km/h. The bird starts moving from first car towards the other and is moving with the speed of 36 km/h and when the two cars were separated by 36 km. What is the total distance covered by the bird?
Sol:
Concept of relative velocity (for 1-D): If two objects are moving along the same straight line and we are observing the motion from the frame of one object. Then for the relative velocity, it will be subtracted for velocities in same direction and added for velocities in opposite directions. (Remember: add or subtract them with proper sign conventions).

Q20. A man runs across the roof-top of a tall building and jumps horizontally with the hope of landing on the roof of the next building which is at a lower height than the first. If his speed is 9 m/s, the (horizontal) distance between the two buildings is 10 m and the height difference is 9 m, will he be able to land on the next building? (Take g = 10 m/s²)
Sol: Key concept: Horizontal Projectile:
When a body is projected horizontally from a certain height ‘y’ vertically above the ground with initial velocity u. If friction is considered to be absent, then there is no other horizontal force which can affect the horizontal motion. The horizontal velocity therefore remains constant and so the object covers equal distance in horizontal direction in equal intervals of time.

Time of flight: If a body is projected horizontally from a height h with velocity u and time taken by the body to reach the ground is T, then

Horizontal range: Let R be the horizontal distance travelled by the body

We will apply kinematic one by one along downward and along horizontal. We first consider motion along horizontal and there is no horizontal force which can affect the horizontal motion. The horizontal velocity therefore remains constant and so the object covers equal distance in horizontal direction in equal intervals of time.
According the problem, horizontal speed of the man (u _x) = 9 m/s Horizontal distance between the two buildings = 10 m
Height difference between the two buildings = 9 m and g =10 m/s²

Horizontal distance travelled by the man is greater than 10 m, therefore, he will land on the next building.

Q21. A ball is dropped from a building of height 45 m. Simultaneously another ball is thrown up with a speed of 40 m/s. Calculate the relative speed of the balls as a function of time.
Sol: In motion under gravity, if the ball is released or dropped that means its initial velocity is zero. In this problem as ball is dropped, so its initial velocity will be taken as zero. We will apply kinematic equations.

Important point: Sign Convention:
Any vector quantity directed upward will be taken as positive and directed downward will be taken as negative. According to this sign convention:

(i) Displacement will be taken as positive if final position lies above initial position and negative if final position lies below initial position.
(ii) Velocity(initial or final) will be taken as positive if it is upward and negative if it is downward.
(iii) Acceleration a is always taken to be -g.
In equations of motions we replace a by -g (minus sign, because acceleration is always directed downward)
Q22. The velocity-displacement graph of a particle is shown in figure.
(a) Write the relation between v and x.
(b) Obtain the relation between acceleration and displacement and plot it.

Long Answer Type Questions

Q23. It is a common observation that rain clouds can be at about a kilometer altitude above the ground.

• If a rain drop falls from such a height freely under gravity, what will be its speed? Also calculate in km/h (g = 10 m/s²).
• A typical rain drop is about 4 mm diameter. Momentum is mass x speed in magnitude. Estimate its momentum when it hits the ground.
• Estimate the time required to flatten the drop.
• Rate of change of momentum’s force. Estimate how much force such a drop would exert on you.
• Estimate the order of magnitude force on umbrella. Typical lateral separation between two rain drops is 5 cm. (Assume that umbrella is circular and has a diameter of 1 m and cloth is not pierced through)

Sol: Key concept: This problem can be solved by kinematic equations of
motion and Newton’s second law that F_ext = dp/dt will be used, where dp is change in momentum over time dt.
(a) According to the problem (h) =1 km = 1000 m and we know that the initial velocity of the ball is zero. And displacement covered by rain drop in downward direction, so we will taking h as negative. (We are neglecting the air resistance.)

(c) Time required to flatten the drop = Time taken by the drop to travel the distance equal to the diameter of the drop near the ground

Q24. A motor car moving at a speed of 72 km/h cannot come to a stop in less than 3.0 s while for a truck this time interval is 5.0 s. On a highway, the car is behind the truck both moving at 72 km/h. The truck gives a signal that it is going to stop at emergency. At what distance the car should be from the truck so that it does not bump onto (collide with) the truck. Human response time is 0.5 s.

Q25. A monkey climbs up a slippery pole for 3 and subsequently slips for 3 seconds. Its velocity at time t is given by v(t) = 2t(3 – t); 0 < t < 3 and v(t) = -(t – 3)(6 – t) for 3 < t < 6s in m/s. It repeats this cycle till it reaches the height of 20 m.

(a) At what time is its velocity maximum?
(b) At what time is its average velocity maximum?
(c) At what time is its acceleration maximum in magnitude?
(d) How many cycles (counting fractions) are required to reach the top? Sol. We have to calculate time corresponding to maximum velocity. So we first need to find the maximum velocity in this problem. To calculate maximum dv velocity we will use dv/dt=0

Q26. A man is standing on top of a building 100 m high. He throws two balls vertically, one at t = 0 and after a time interval (less than 2 seconds). The later ball is thrown at a velocity of half the first. The vertical gap between first and second ball is + 15 m at t = 2 s. The gap is found to remain constant. Calculate the velocity with which the balls were thrown and the exact time interval between their throw.
Sol: We solve this problem by using kinematic equations with proper sign convention and to calculate time interval we will take We solve this problem by using kinematic equations with proper sign convention and to calculate time interval we will take difference of displacements.

Important note: We should be very careful when we are applying the equation of rectilinear motion. These equations are applicable only in case of constant acceleration.

Some important observations for motion under gravity:

• The motion is independent of the mass of the body, as in any equation of motion, mass is not involved. That is why a heavy and light body when released from the same height, reach the ground simultaneously and with same velocity, i.e., t = √(2h/g) and v = √2gh.
• In case of motion under gravity time taken to go up is equal to the time taken to fall down through the same distance.

Time of descent (t₁) = time of ascent (t₂) = u/g
Total time of flight T=t_x + t₂ = .2u / g

• In case of motion under gravity, the speed with which a body is projected up is equal to the speed with which it comes back to the point of projection.
  As well as the magnitude of velocity at any point on the path is same whether the body is moving in upwards or downward direction.

• A body is thrown vertically upwards. If air resistance is to be taken into account, then the time of ascent is less than the time of descent t₂^>t₁

where g is acceleration due to gravity and a is retardation by air resistance and for upward motion both will act vertically downward.
For downward ‘motion a and g will act in opposite direction because a always act in direction opposite to motion and g always act vertically downward.

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We hope the NCERT Exemplar Class 11 Physics Chapter 2 Motion in a Straight Line help you. If you have any query regarding NCERT Exemplar Class 11 Physics Chapter 2 Motion in a Straight Line, drop a comment below and we will get back to you at the earliest.

NCERT Exemplar Class 11 Physics Chapter 3 Motion in a Plane are part of NCERT Exemplar Class 11 Physics. Here we have given NCERT Exemplar Class 11 Physics Chapter 3 Motion in a Plane.

NCERT Exemplar Problems Class 11 Physics Chapter 3 Motion in a Plane

Multiple Choice Questions
Single Correct Answer Type

Motion In A Plane Class 11 Numericals With Solutions 

Sol: (b)
Key concept: Scalar Product of Two Vectors:
(1) Definition : The scalar product (or dot product) of two vectors is defined as the product of the magnitude of two vectors with cosine of angle between them.





Q2. Which one of the following statements is true?
(a) A scalar quantity is the one that is conserved in a process.
(b) A scalar quantity is the one that can never take negative values.
(c) A scalar quantity is the one that does not vary from one point to another in space.
(d) A scalar quantity has the same value for observers with different orientation of the axes.
Sol: (d) A scalar quantity is independent of direction hence has the same value for observers with different orientations of the axes.
For example, a car is traveling along +x axis, it travels 10 m. If the same car is moving with the same speed for the same time interval along -x axis, then the distance meter of car shows the same travelled distance. The path length is same in both the cases.

Q3. Figure shows the orientation of two vectors u and v in the TY-plane.

which of the following is correct?
(a) a and p are positive while b and q are negative
(b) a, p and b are positive while q is negative
(c) a, q and b are positive while p is negative
(d) a, b, p and q are all positive

Here it is worthy to note once a vector is resolved into its components, the components themselves can be used to specify the vector as:

In such type of problems we have to resolve the rectangular components according to the diagram.

Q4. The component of a vector r along X-axis will have maximum value if
(a) r is along positive Y-axis
(b) r is along positive X-axis
(c) r makes an angle of 45° with the X-axis
(d) r is along negative Y-axis

Q5. The horizontal range of a projectile fired at an angle of 15° is 50 m. If it is fired with the same speed at an angle of 45°, its range will be
(a) 60 m (b) 71m (c) 100 m (d) 141m

Q6. Consider the quantities pressure, power, energy, impulse, gravitational potential, electrical charge, temperature, area. Out of these, the only vector quantities are
(a) Impulse, pressure and area
(b) Impulse and area
(c) Area and gravitational potential
(d) Impulse and pressure
Sol: (b) We know that impulse J = F. ∆t = ∆p, where F is force, At is time duration and Ap is change in momentum. As ∆p is a vector quantity, hence impulse is also a vector quantity. Sometimes area can also be treated as vector direction of area vector is perpendicular to its plane.
Q7. In a two dimensional motion, instantaneous speed ==vo== is a positive constant. Then, which of the following are necessarily true?
(a) The average velocity is not zero at any time
(b) Average acceleration must always vanish
(c) Displacements in equal time intervals are equal
(d) Equal path lengths are traversed in equal intervals

Sol :(d) Speed (Instantaneous Speed): The magnitude of the velocity at any instant of time is known as Instantaneous Speed or simply speed at that instant of time. It is denoted by v.
Quantitatively: Speed = distance/ time
Mathematically, it is the time rate at which distance is being travelled by the particle.
• Speed is a scalar quantity. It can never be negative (as shown by speedometer of our vehicle).
• Instantaneous speed is the speed of a particle at a particular instant of time.
Hence, Total distance travelled = Path length = (speed) x time taken Important point: We should be very carefttl with the fact that speed is related with total distance covered not with displacement.

Q8. In a two-dimensional motion, instantaneous speed v₀ is a positive constant. Then, which of the following are necessarily true?
‘(a) The acceleration of the particle is zero.
(b) The acceleration of the particle is bounded.
(c) The acceleration of the particle is necessarily in the plane of motion.
(d) The particle must be undergoing a uniform circular motion.
Sol. (c) This motion is two dimensional and given that instantaneous speed v₀  is positive constant. Acceleration is defined as the rate of change of velocity (instantaneous speed), hence it will also be in the plane of motion.

Sol : (c) These types of problems can be solved by hit and trial method by picking up options one by one






More Than One Correct Answer type

Q11. Two particles are projected in air with speed v₀ at angles θ₁ and θ ₂ (both acute) to the horizontal, respectively. If the height reached by the first particle is greater than that of the second, then tick the right choices.

(a) Angle of projection: θ₁ > θ₂
(b) Time of flight: T₁ > T₂
(c) Horizontal range: R_x> R₂                  
(d) Total energy: U₁> U₂

Q12. A particle slides down a frictionless parabolic (y = x²) track (A – B – C)
starting from rest at point A (figure). Point B is at the vertex of parabola and point C is at a height less than that of point A. After C, the particle moves freely in air as a projectile. If the particle reaches highest point at P, then
(a) KE at P = KE at B
(b) height at P = height at A
(c) total energy at P = total energy at A
(d) time of travel from AtoB


Solution : (c)

Key concept: In such type of problems, we have to observe the nature of track that if there is a friction or not, as friction is not present in this track, total energy of the particle will remain constant throughout the journey.
According to the problem, the path traversed by the particle on a frictionless track is parabolic, is given by the equation y = x², thus total energy (KE + PE) will be same throughout the journey.
Hence, total energy at A = total energy at P
At B the particle is having only KE but at P some KE is converted to PE.
So, (KE)_B  > (KE)_P
Total energy at A = PE = Total energy at B = KE = Total energy at P = PE + KE
The potential energy at A is converted to KE and PE at P, hence (PE)P < (PE)A
Hence, (Height) P < (Height) A
As, Height of P < Height of A
Hence, path length AB > path length BP
Hence, time of travel from A to B ≠ Time of travel from B to P.

Q14. For a particle performing uniform circular motion, choose the correct
statement(s) from the following.
(a) Magnitude of particle velocity (speed) remains constant.
(b) Particle velocity remains directed perpendicular to radius vector.
(c) Direction of acceleration keeps changing as particle moves.
(d) Angular momentum is constant in magnitude but direction keeps changing.

Sol: (a, b, c) While a particle is in uniform circular motion. Then the following statements are true.

(i) speed will be always constant throughout.
(ii) velocity will be always tangential in the direction of motion at a particular point.
(iii) the centripetal acceleration a = v²/r and its direction will always towards centre of the circular trajectory.
(iv) angular momentum (mvr) is constant in magnitude and direction. And its direction is perpendicular to the plane containing r and v.
Important point: In uniform circular motion, magnitude of linear velocity and centripetal acceleration is constant but direction changes continuously.

Very Short Answer Type Questions

Q16 .A cyclist starts from centre O of a circular park of radius 1 km and moves along the path OPRQO as shown in figure. If he maintains constant speed of 10 ms”¹, what is his acceleration at point R in magnitude and direction?

Ans: According to the problem the path of the cyclist is O-P- R-Q-O.

The cyclist is in uniform circular motion and it is given that linear velocity = 10 m/s, R = 1 km = 1000 m. As we know whenever an object is performing circular motion, acceleration is called centripetal acceleration and is always directed towards the centre.So cyclist experiences a centripetal force (acceleration) at point R towards centre.

Q17. A particle is projected in air at some angle to the horizontal, moves along parabola as shown in figure where x and y indicate horizontal and vertical directions, respectively. Show in the diagram, direction of velocity and acceleration at points A, B and C.

Sol: In projectile motion horizontal component of velocity will always be constant and acceleration is always vertically downward and is equal to g. Direction of velocity will always be tangential to the curve in the direction of motion.
As shown in the diagram in which a particle is projected at an angle 0.


Q18. A ball is thrown from a roof top at an angle
of 45° above the horizontal. It hits the ground a few seconds later. At what point during its motion, does the ball have
(a) greatest speed
(b) smallest speed
(c) greatest acceleration Explain.

Ans: In this problem total mechanical energy of the ball is conserved. As the ball is projected from point O, and covering the path OABC.
At point A it has both kinetic and potential energy.
But at point C it have only kinetic energy, (keeping the ground as reference where PE is zero.)
(a) At point B, it will gain the same speed u and after that speed increases and will be maximum just before reaching C.
(b) During upward journey from OtoA speed decreases and smallest speed attained by it is at the highest point, i.e., at point A.
(c) Acceleration is always constant throughout the journey and is vertically downward equal to g.

Q19. A football is kicked into the air vertically upwards. What is its (a) acceleration and (b) velocity at the highest point?

Sol: (a) The situation is shown in the diagram below in which a football is kicked into the air vertically upwards. Acceleration of the football will always be vertical downward and is called acceleration due to gravity (g).
(b) When the football reaches the highest point it is momentarily at rest and at that moment its velocity will be zero as it is continuously retarded by acceleration due to gravity (g).

Key concept: Collinear vectors: When the vectors under consideration can share the same support or have a common support then the considered vectors are collinear.
Coplanar vectors: Three (or more) vectors are called coplanar vector if they lie in the same plane. Two (free) vectors are always coplanar.

Q21. A boy travelling in an open car moving on a levelled road with constant speed tosses a ball vertically up in the air and catches it back. Sketch the motion of the ball as observed by a boy standing on the footpath. Give explanation to support your diagram.
Sol: With respect to the observer standing on the footpath ball is thrown with velocity u at an angle θ with the horizontal, hence it seems as a projectile. So path of the ball will be parabolic. The horizontal speed of the ball is same as that of the car, therefore, ball as well car travels equal horizontal ult distance. Due to its vertical speed, the ball follows a parabolic path.



Important point: We must be very clear that we are working with respect to ground. When we observe with respect to the car, motion will be along vertical direction only.

Q22. A boy throws a ball in air at 60° to the horizontal along a road with a speed of 10 m/s (36 km/h). Another boy sitting in a passing by car observes the ball. Sketch the motion of the ball as observed by the boy in the car, if car has a speed of 18 km/h. Give explanation to support your diagram.
Sol: The situation is shown in the below diagram.

According to the problem the boy standing on ground throws the ball at an angle of 60° with horizontal at a speed of 10 m/s.

Speed of the car = 18 km/h = 5 m/s
As horizontal speed of ball and car is same, hence relative velocity of ball w.r.t car in the horizontal direction will be zero.
Only vertical motion of the ball will be observed by the boy in the car, as shown in above diagram.

Q23. In dealing with motion of projectile in air, we ignore effect of air resistance on motion. This gives trajectory as a parabola as you have studied. What would the trajectory look like if air resistance is included? Sketch such a trajectory and explain why you have drawn it that way.
Sol:

When we are dealing with projectile motion generally we neglect the air resistance. But if air resistance is included the horizontal component of velocity will not be constant and obviously trajectory will change.
Due to air resistance, particle energy as well as horizontal component of velocity keep on decreasing making the fall steeper than rise as shown in the figure.
When we are neglecting air resistance path was symmetric parabola (OAC). When air resistance is considered path is asymmetric parabola (OAB).

Short Answer Type Questions
Q24. A fighter plane is flying horizontally at an altitude of 1.5 km with speed 720 km/h. At what angle of sight (w.r.t. horizontal) when the target is seen, should the pilot drop the bomb in order to attack the target ?


2. When an object is dropped/released by any moving vehicle. Then initial velocity of the object is same as the moving vehicle.

When the bomb is dropped from Plane the plane which is moving horizontally. So, the bomb will have same initial velocity as that of plane along horizontal direction.
The situation is shown in the diagram below. Let a fighter plane, when it be plane at position P, drops a bomb to hit a target T.
Let the target is seen at an angle θ with horizontal.

Important point: Angle is with respect to target. As seen by observer in the plane, motion of the bomb will be vertically downward below the plane. .

Q25. (a) Earth can be thought of as a sphere of radius 6400 km. Any object (or a person) is performing circular motion around the axis of the earth due to the earth rotation (period 1 day). What is the acceleration of object on the surface of the earth (at equator) towards its centre? What is it at latitude 9? How does these accelerations compare with g = 9.8 m/s²? (b) Earth also moves in circular orbit around the sun once every year with an orbital radius of 1.5 x 10¹¹ What is the acceleration of the earth (or any object on the surface of the earth) towards the centre of the sun? How does this acceleration compare with g = 9.8 m/s²?

Q26. Given below in Column I are the relations between vectors a, b and c and in Column II are the orientations of a, b and c in the AY-plane. Match the relation in Column I to correct orientations in Column II.


Sol: We apply triangulr law of addition
Triangular law of vector addition: Two vectors are considered as two sides of a triangle taken in the same order. The third side or completing side of the triangle is the resultant taken in the opposite order.
or
We can say that vectors are arranged head to tail, this graphical method is called the head-to-tail method. The two vectors and their resultant form three sides of a triangle, so this method is also known as triangle method of vector addition.

As shown in the diagram below in which vectors A and B are corrected by head and tail. Resultant vector C = A + B
(a) from (iv), it is clear that c = a + b
(b) from (iii), c + b = a => a- c = b
(c) from (i), b = a + c => b-a = c
(d) from (ii), -c = a + b => a + b + c = 0

Long Answer Type Questions

Q29. A hill is 500 m high. Supplies are to be sent across the hill using a canon that can hurl packets at a speed of 125 m/s over the hill. The canon is located at a distance of 800 m from the foot of hill and can be moved on the ground at a speed of 2 m/s, so that its distance from the hill can be adjusted. What is the shortest time in which a packet can reach on the ground across the hill? Take g= 10 m/s².
Sol.According to the problem, speed of packets =125 m/s , height of the hill = 500 m, distance between the cannon and the foot of the hill, d = 800 m

To cross the hill in shortest time, then the vertical component of the velocity should be minimum so that it just crosses the height of hill.

Distance through which canon has to be moved = 800 – 750 = 50 m Speed with which canon can move = 2 m/s

Important point: We should not confuse with the positive direction of motion. May be vertically upward direction or vertically downward direction is taken as positive according to convenience. And this problem can also be solved by taking motion from point P to T. From point P in the diagram projection at speed v0 at an angle θ below horizontal with height h and horizontal range ∆x. Then this will be analyzed from the alongside diagram

Q31. A particle is projected in air at an angle β to a surface which itself is inclined at an angle α to the horizontal (figure).
(a) Find an expression of range on the plane surface (distance on the plane from the point of projection at which particle will hit the surface).
(b) Time of flight.
(c) β at which range will be maximum

Q32. A particle falling vertically from a height hits a plane surface inclined to horizontal at an angle θ with speed v0 and rebounds elastically. Find the distance along the plane where it will hit second time.

Q33. A girl riding a bicycle with a speed of 5 m/s towards north direction, observes rain falling vertically down. If she increases her speed to 10 m/s, rain appears to meet her at 45° to the vertical. What is the speed of the rain? In what direction does rain fall as observed by a ground based observer?
Sol: V_rgis the velocity of rain appears to the girl.
We must draw all vectors in the reference frame of ground-based observer.

Q34. A river is flowing due east with a speed 3 m/s. A swimmer can swim in still water at a speed of 4 m/s (figure).
(a) If swimmer, starts swimming due north, what will be his resultant velocity (magnitude and direction)?
(b) If he wants to start from point A on south bank –
and reach opposite point B on north bank,
(i) which direction should he swim?
(ii) what will be his resultant speed?
(c) From two different cases as mentioned in (a) and (b) above, in which case will he reach opposite bank in shorter time?

Q32. A cricket fielder can throw the Cricket ball with a speed v₀. If he throws the ball while running with speed u at an angle θ to the horizontal, find
(a) the effective angle to the horizontal at which the ball is projected in air as seen by a spectator.
(b) what will be time of flight?
(c) what is the distance (horizontal range) from the point of projection at which the ball will land?
(d) find θ at which he should throw the ball that would maximize the horizontal range as found in (c).
(e) how does θ for maximum range change if u > v₀, u = v₀, u < v₀?
(f) how does θ in (e) compare with that for u = 0 (i.e., 45°)?
Sol:. The observer on ground (spectator) observes that the x-component of ball is more because of the speed of fielder. As shown in the adjacent diagram,
So, initial velocity in x-direction

Q37. A man wants to reach from A to the opposite comer of the square C. The sides of the square are 100 m. A central square of 50 m x 50 m is filled with sand. Outside this square, he can walk at a speed 1 m/s. In the central square, he can walk only at a speed of v m/s (v < 1). What is smallest value of v for which he can reach faster via a straight path through the sand than any path in the square outside the sand?

Solu:

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NCERT Exemplar Class 11 Physics Chapter 4 Laws of Motion are part of NCERT Exemplar Class 11 Physics. Here we have given NCERT Exemplar Class 11 Physics Chapter 4 Laws of Motion.

NCERT Exemplar Class 11 Physics Chapter 4 Laws of Motion

Multiple Choice Questions
Single Correct Answer Type

Q1. A ball is travelling with uniform translatory motion. This means that
(a) it is at rest.
(b) the path can be a straight line or circular and the ball travels with uniform
(c) all parts of the ball have the same velocity (magnitude and direction) and the velocity is constant.
(d) the centre of the ball moves with constant velocity and the ball spins about its centre uniformly.
Sol: (c) When a body moves in such a way that the linear distance covered by each particle of the body is same during the motion, then the motion is said to be translatory or translation motion.
Translatory motion can be, again of two types viz., curvilinear (shown in fig. (a)) or rectilinear (shown in fig. (b)), accordingly as the paths of every constituent particles are similarly curved or straight line paths. Here it is important that the body does not change its orientation. Here we can also define it further in uniform and non-uniform translatory motion. Here figure
(b) is uniformly translatory motion.

Q2. A metre scale is moving with uiiiform velocity. This implies
(a) the force acting on the scale is zero, but a torque about the centre of mass can act on the scale.
(b) the force acting on the scale is zero and the torque acting about centre of mass of the scale is also zero.
(c) the total force acting on it need not be zero but the torque on it is zero.
(d) neither the force nor the torque need to be zero.
Sol: (b)
Key concept: To solve these types of problem we have to apply Newton’s second law of motion!
Newton’s Second Law of Motion
According to this law: The rate of change of linear momentum of a body is directly proportional to the external force applied on the body and this change takes place always in the direction of the force applied.

takes place always in the direction of the force applied.
We know that F = df/ dt
According to the question that the meter scale is moving with uniform velocity, hence, change in momentum will be zero, i.e. dp = 0
This implies momentum will remains same. So, Force = F= 0.
So, we can say that all parts of the meter scale is moving with uniform velocity because total force is zero and if there is any torque acting on the body this means that the body will be in rotational motion which means that the direction of velocity will be changing continuously. So, the torque acting about centre of mass of the scale is also zero.

Important point: Change in velocity = final velocity – initial velocity

Q5. Conservation of momentum in a collision between particles can be understood from
(a) Conservation of energy
(b) Newton’s first law only
(c) Newton’s second law only
(d) Both Newton’s second and third law
Sol: (d)
Key concept: If no external force acts on a system (called isolated) of constant mass, the total momentum of the system remains constant with time.

This equation shows that in absence of external force for a closed system the linear momentum of individual particles may change but their sum remains unchanged with time.
Conservation of linear momentum is equivalent to Newton’s third law of motion.
For a system of two particles in absence of external force by law of conservation of linear momentum.

i.e., for every action there is equal and opposite reaction which is Newton’s third law of motion.
In case of collision between particles equal and opposite forces will act on individual particles by Newton’s third law.
Hence total force on the system will be zero.

Important point: We should not confuse with system and individual particles. As total force on the system of both particles is zero but force acts on individual particles.
Law of conservation of linear momentum is independent of frame of reference though linear momentum depends on frame of reference.

Q6. A hockey player is moving northward and suddenly turns westward with the same speed to avoid an opponent. The force that acts on the player is
(a) frictional force along westward (b) muscle force along southward (c) frictional force along south-west (d) muscle force along south-west
Sol: (c)
Key concept: According to Newton’s second law of motion only external forces can change linear momentum of the system. The internal forces cannot change linear momentum of system under consideration. If we take hockey player as a system, the external force which can change the direction of motion of the player is the force must be friction between the ground and shoes of player. The muscle force is the internal force, this cannot change the linear momentum of the player. According to Newton’s Second Law, The rate of change of linear momentum of a body is equal to the external force applied on the body or F = dp/dt .So, the external force must be in the direction of change in momentum.

According to the problem, mass = 2 kg

Position of the particle is given here as a function of time, x(t) =pt + qt² + rt³ By differentiating this equation w.r.t. time we get velocity of the particle as a function of time.

v = dx/dt = p + 2 qt + 3 rt²

If we again differentiate this equation w.r.t. time, we will get acceleration of the particle as a function of time.

a  = dv/dt = 0+ 2q + 6rt

At t = 2 s; a = 2q + 6x2xr
= 2 q+ 12r
= 2×4+12×5
= 8 + 60 = 68 m/s
Force =F=ma
= 2×68= 136 N






Let us assume the eastward
direction as x-axis.
A car is able to move towards due to friction acting between its tyres and the road.
The force of friction of the road on the tyre acts in the forward direction and is equal but in the opposite direction to the force of friction of the tyre on the road.
Mass of the car = m
As car starts from rest, its initial velocity u = 0 Velocity acquired along east = vi
Time interval (in which car acquired that velocity) t = 2 s.
As acceleration is uniform, so by applying kinematic equation (v = u + at), we get

More Than One Correct Answer Type

Q10. The motion of a particle of mass m is given by x – 0 for t < 0 s, x(t) = A sin 4πt for 0 < t< (1/4) s (A > 0),
and x = 0 for / > (1/4) s.

Which of the following statements is true?

• The force at t = (1/8) s on the particle is -16π²A-m.
• The particle is acted upon by an impulse of magnitude 4/rA-m at t = 0 s and t = (1/4) s.
• The particle is not acted upon by any force.
• The particle is not acted upon by a constant force.
• There is no impulse acting on the particle.

Sol:

Q11. In figure the co-efficient of friction between the floor and the body B is 0.1. The co-efficient of friction between the bodies B and A is 0.2. A force F is applied as shown on B. The mass of A is m/2 and of B is m. Which of the following statements are true?

(a) The bodies will move together if F = 0.25 mg.
(b) The body A will slip with respect to B if F = 0.5 mg.
(c) The bodies will move together if F = 0.5 mg.
(d) The bodies will be at rest if F = 0.1 mg.
(e) The maximum value of F for which the two bodies will move together is 0.45 mg.

Sol: (a, b, d, e)

Q12. Mass m, moves on a slope making an angle θ with the horizontal and is attached to mass m₂ by a string passing over a frictionless pulley as shown in figure. The coefficient of friction between m_u and the sloping surface is µ. Which of the following statements are true?

Sol: (b, d)
Key concept: When a mass mx is placed on a rough inclined plane: Another mass m2 hung from the string connected by ftictionless pulley, the tension IT) produced in string will try to start the motion of mass w,.
At limiting condition,

Simplified situation is shown in the diagram.
Let m1 moves up the plane. Different forces involved are shown in the diagram.

Q13. In figure a body A of mass m slides on a plane inclined at angle θ₁ to the horizontal and µ is the coefficient of friction between A and the plane. A is connected by a light string passing over a frictionless pulley to another body 5, also of mass m, sliding on a frictionless plane inclined at an angle θ₂ to the horizontal. Which of the following statements are true?

Q14. Two billiard balls A and B, each of mass 50 g and moving in opposite directions with speed of 5 m s^-1 each, collide and rebound with the same speed. If the collision lasts for 10 ^-3 s, which of the following statements are true?
The impulse imparted to each ball is 0.25 kg-ms ¹ and the force on each ball is 250 N.
(a) The impulse imparted to each ball is 0.25 kg-ms ¹ and the force exerted on each ball is 25 x 10 ⁵
(b) The impulse imparted to each ball is 0.5 N-s.
(c) The impulse and the force on each ball are equal in magnitude and opposite in directions.

Q15. Abody of mass 10 kg is acted upon by two perpendicular forces, 6 N and 8 N. The resultant acceleration of the body is

Sol: (a, c) Recall the concept of resultant of two vectors, when they are perpendicular

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NCERT Exemplar Class 11 Physics Chapter 5 Work, Energy and Power are part of NCERT Exemplar Class 11 Physics. Here we have given NCERT Exemplar Class 11 Physics Chapter 5 Work, Energy and Power.

NCERT Exemplar Class 11 Physics Chapter 5 Work, Energy and Power

Multiple Choice Questions
Single Correct Answer Type

Q1. An electron and a proton are moving under the influence of mutual forces. In calculating the change in the kinetic energy of the system during motion, one ignores the magnetic force of one on another. This is, because
(a) the two magnetic forces are equal and opposite, so they produce no net ‘ effect
(b) the magnetic forces do not work on each particle
(c) the magnetic forces do equal and opposite (but non-zero) work on each particle
(d) the magnetic forces are necessarily negligible
Sol: (b)
Key concept: To calculate the change in kinetic energy of the system during motion we have to apply work-energy, theorem. According to this theorem, Net work done = Final kinetic energy – Initial kinetic energy of the object The above statement shows the connection between work and kinetic energy as: “The work done by the net force acting on an object is equal to the change in the kinetic energy of that object”.
Net work done (IF) on a particle equals change in kinetic energy of the particle.

ΣW=K₂-K₁

According to the problem as the electron and proton are moving under the influence of mutual forces, the magnetic forces will be perpendicular to their motion, hence, it acts as a centripetal force for the particle. In this way the particle performs the uniform circular morion, this implies speed will remain constant. So, there is no change in kinetic energy of the particle. Hence no work is done by these forces.

(magnetic force) will be perpendicular to both B and v, where B is the external magnetic field and v is the velocity of particle. That is why one ignores the magnetic force of one particle on another.

Q2. A proton is kept at rest. A positively charged particle is released from rest at a distance d in its field. Consider two experiments; one in which the charged particle is also a proton and in another, a positron. In the same time t, the work done on the two moving charged particles is
(a) same as the same force law is involved in the two experiments
(b) less for the case of a positron, as the positron moves away more rapidly and the force on it weakens
(c) more for the case of a positron, as the positron moves away a larger distance
(d) same as the work done by charged particle on the stationary proton
Sol: (c) Force between two protons is equal to the force between proton and a
positron because their charges are same. As the mass of positron is much lesser than proton, (1/1840 times) it moves away through much larger distance compared to proton.
Change in their momentum will be same. So, velocity of lighter particle will be greater than that of a heavier particle. So, positron is moved through a larger distance.
As work done = force x distance. As forces are same in case of proton and positron but distance moved by positron is larger, hence, work done will be more.

Q3. A man squatting on the ground gets straight up and stand. The force of reaction of ground on the man during the process is
(a) constant and equal to mg in magnitude
(b) constant and greater than mg in magnitude
(c) ‘ variable but always greater than mg
(d) at first greater than mg and later becomes equal to mg
Sol: (d) In the process of squatting on the ground he gets straight up and stand. Then he is tilted somewhat, the man exerts a variable force on the ground to balance his weight, hence he also has to balance frictional force besides his weight in this case.
N = Normal reaction force = friction + mg => N > mg
Once the man gets straight up that variable force = 0 =>
Normal reaction force = mg

Q4. A bicyclist comes to a skidding stop in 10 m. During this process, the force on the bicycle due to the road is 200 N and is directly opposed to the motion. The work done by the cycle on the road is
(a) +2000 J
(b) -200 J
(c) zero
(d) -20,000 J
Sol: (c) As the friction is present in fhis problem, so mechanical energy is not conserved. So energy will be lost due to dissipation by friction.^Here, work is done by the frictional force on the cycle and is equal to __
200 x 10 = -2000 J
As the road does not move at all, therefore, work done by the cycle on the road is zero.
Important point: We should be aware that here the energy of bicyclist is lost during the motion, but it is lost due to friction in the form of heat.

Q5. A body is falling freely under ‘the action of gravity alone in vacuum. Which of the following quantities remain constant during the fall?
(a) Kinetic energy
(b) Potential energy
(c) Total mechanical energy
(d) Total linear momentum
Sol: (c) As the body is falling freely under gravity, the potential energy decreases continuously and kinetic energy increases continuously as all the conservative forces are doing work. So, total mechanical energy (PE + KE) of the body will be constant.
Let us discuss this in detail:
In the given diagram an object is dropped from-a height H from ground.
At point A total mechanical energy will be EA = K.E + P.E

So, total mechanical energy will remain same (if we neglect the air friction).

Q6. During inelastic collision between two bodies, which of the following quantities always remain conserved?
(a) Total kinetic energy
(b) Total mechanical energy
(c) Total linear momentum
(d) Speed of each body
Sol: (c) If in a collision kinetic energy after collision is not equal to kinetic energy before collision, the collision is said to inelastic.
Coefficient of restitution 0 < e < 1
When we are considering the two bodies as system the total external force on the system will be zero.
Hence, total linear momentum of the system remain conserved.
Here kinetic energy appears in other forms, i.e. energy may be lost in the form of heat and sound etc. In some cases

(KE)_final < (KE)_initial  such as when initial KE is converted into intertial energy of the product (as heat, elastic or excitation) while in other cases (KE)_final > (KE)_initialsuch as when internal energy stored in the colliding particles is released.
Examples’. (1) Collision between two billiard balls.
(2) Collision between two automobiles on a road.
In fact all majority of collisions belong to this category.

Q7. Two inclined ffictionless tracks, one gradual and the other steep meet at A from where two stones are allowed to slide down from rest, one on each track as shown in figure.
Which of the following statement is correct?
(a) Both the stones reach the bottom at the same time but not with the same speed.
(b) Both the stones reach the bottom with the same speed and stone I reaches the bottom earlier than stone II.
(c) Both the stones reach the bottom with the same speed and stone II reaches the bottom earlier than stone I.
(d) Both the stones reach the bottom at different times and with different speeds.

Sol: (c) As shown in diagram AB and AC are two smooth planes inclined to the angle θ₁  and θ₂  respectively. As friction is absent here, hence, mechanical energy will be conserved. As both the tracks having common height h,
From conservation of mechanical energy,

(a) V=0,K = E
(b) V=E,K=0
(c) V< E, K= O
(d) V= 0,K<E

Sol:(b)

Q9. Two identical ball bearings in contact with each other frictionless table are hit head-on by another ball bearing of the same mass moving initially with a speed v as shown in figure.
If the collision is elastic, which of the following (figure) is a possible result after collision?

Sol: (b)
Key concept: In a collision if the motion of colliding particles before and after the collision is along the same line, the collision is said to be head on or one dimensional.
When two bodies of equal masses collides elastically, their velocities are interchanged.
Kinetic energy and linear momentum remains conserved Total kinetic energy of the system before collision

Q10. A body of mass 0.5 kg travels in a straight line with velocity v = ax^3/2 where a = 5 m^-1/2 s^_1. The work done by the net force during its displacement from x = 0 to; x = 2m is

(a) 5 J
(b) 50 J 
(c) 10 J  
(d) 100 J
Sol:

Q11. A body is moving unidirectionally under the influence of a source of constant power supplying energy. Which of the diagrams shown in figure correctly shown the displacement-time curve for its motion?

Q12. Which of the diagrams shown in figure most closely shows the variation in kinetic energy of the earth as it moves once around the sun in its elliptical orbit?

Sol: (d) As the earth moves once around the sun in its elliptical orbit, when the earth is closest to the sun, speed of the earth is maximum, hence KE is maximum. When the earth is farthest from the sun speed is minimum, hence KE is minimum but never zero and negative.
This variation of KE vs t is correctly represented by option (d).

Q13. Which of the diagrams in figure represents varation of total mechanical energy of a pendulum oscillating in air as functon of time?

Sol: (c) When a pendulum oscillates in air, its total mechanical energy decreases continuously in overcoming resistance due to air. Therefore, total mechanical energy of the pendulum decreases exponentially with time.
The variation of E vs t is correctly represented by curve (c) in which the relation between energy and time is shown.

Q14. A mass of 5 kg is moving along a circular path of radius 1 m. If the mass moves with 300 rev/min, its kinetic energy would be

Q15. A raindrop falling from a height h above ground, attains a near terminal velocity when it has fallen through a height (3/4)h. Which of the diagrams shown in figure correctly shows the change in kinetic and potential energy of the drop during its fall up to the ground?

Sol:(b) At height h from ground raindrop have maximum potential energy. And kinetic velocity will be zero (at the instant when it dropped its velocity will be zero), then as the rain drop falls its PE starts decreasing and kinetic energy start increasing.
The total mechanical energy will remain conserved if we neglect the air resistance. If there is some air resistance, there is some force called upthrust (in fluids) which opposes its motion. It depends upon velocity of object as the velocity increases, upthrust also increases. Hence during fall of raindrop first its velocity increases and then become constant after some time.
This constant velocity is called terminal velocity, hence KE also become constant. PE decreases continuously as the drop is falling continuously.
The variation in PE and KE is best represented by (b).

Q16. In a shotput event an athlete throws the shotput of mass 10 kg with an initial speed of 1 m s^-1 at 45° from a height 1.5 m above ground. Assuming air resistance to be negligible and acceleration due to gravity to be 10 m s ^-2, the kinetic energy of the shotput when it just reaches the ground will be
(a) 2.5 J    
(b) 5.0 J                    
(c) 52.5 J                  
(d) 155.0 J

Sol: (d) If air resistance is negligible, total mechanical energy of the system will remain constant. And let us take ground as a reference where potential energy will be zero.
According to the problem, h = 1.5 m, v = 1 m/s, m = 10 kg, g = 10 ms^{– 2
23}

Q17. Which of the diagrams in figure correctly shows the change in kinetic energy of an iron sphere falling freely in a lake having sufficient depth to impart it a terminal velocity?

Sol: (b) When an iron sphere is falling freely in the lake, it will accelerate by the force due to gravity then its velocity increases continuously and resistance due to water cause a force called viscous force or upthrust (in fluids) which opposes its motion. It depends upon velocity of object as the velocity increases upthrust increases. Hence during fall of sphere first its velocity increases and then become constant after some depth.
This constant velocity is called terminal velocity, hence KE become constant beyond this depth, which is best represented by (b).

Q18. A cricket ball of mass 150 g moving with a speed of 126 km/h hits at the middle of the bat, held firmly at its position by the batsman. The ball moves straight back to the bowler after hitting the bat. Assuming  that collision between ball and bat is completely elastic and the two remain    in contact for0.001 s, the force that the batsman had to apply to hold the bat firmly at its place would be

More Than One Correct Answer Type
Q19. A man of mass m, standing at the bottom of the staircase, of height L climbs it and stands at its top.
(a) Work done by all forces on man is equal to the rise in potential energy mgL.
(b) Work done by all forces on man is zero.
(c) Work done by the gravitational force on man is mgL.
(d) The reaction force from a step does not do work because the point of application of the force does not move while the force exists.
Sol: (b, d) When a man of mass m climbs up the staircase of height L, work done by the gravitational force on the man = -mgL
Work done by internal muscular forces = -Work done against gravitational force = mgL
Work done by all the forces = mgL – mgL = 0
As the point of application of the contact forces does not move, hence work done by reaction forces will also be zero.
And work done by friction will also be zero as there a no dissipation or rubbing is involved.

Q20. A bullet of mass m fired at 30° to the horizontal leaves the barrel of the gun with a velocity v. The bullet hits a soft target at a height h above the ground while it is moving downward and emerge out with half the kinetic energy it had before hitting the target.
Which of the following statements are correct in respect of bullet after it emerges out of the target?
(a) The velocity of the bullet will be reduced to half its initial value.
(b) The velocity of the bullet will be more than half of its earlier velocity
(c) The bullet will continue to move along the same parabolic path.
(d) The bullet will move in a different parabolic path.
(e) The bullet will fall vertically downward after hitting the target.
(f) The internal energy of the particles of the target will increase.
Sol:(b, d, f) The given situation is shown in the diagram. Let speed after emerging from the target is v”, v’ is speed of the bullet just before hitting the target, v is the initial speed of bullet

(a) The velocity of the bullet after it emerges out of the target will be reduced to half its initial value.

Compare the kinetic energies before and after hitting the target.Just before hitting the target the mechanical energy will be conserved. After hitting its energy will be loose. So by conservation of mechanical energy between “O” and “A” (A is the point where the target is placed),

Hence, after emerging from the target velocity of the bullet (v”) is more than half of its earlier velocity v’ (velocity before emerging into the target). So option (b) is correct.

(c) Since velocity of the bullet is changed after hitting the target, so it follows a different parabolic path.
(d) As the velocity of the bullet changes to v’which is less than v’ hence path followed will change and the bullet reaches at point B instead of A’, as shown in the figure.
(e) After emerging from the target, bullet follows a parabollic path, the bullet will not fall vertically downward, so it is an incorrect option.
(f) As the bullet is passing through the target, the loss in energy of the • bullet is transferred to particles of the target. Therefore, their internal energy increases.

Q21. Two blocks M₁ and M₂ having equal mass are free to move on a horizontal frictionless surface. M₂ is attached to a massless spring as shown in figure. Initially M₂ is at rest and A/, is moving toward M₂ with speed v and collides head-on with M₂.
(a) While spring is fully compressed all the KE of Myis stored as PE of spring.
(b) While spring is fully compressed the system momentum is not conserved, though final momentum is equal to initial momentum.
(c) If spring is massless, the final state of the M₁ is state of rest.
(d) If the surface on which blocks are moving has friction, then collision cannot be elastic

Sol: (c, d) If there is not specified we always consider the collision elastic.
When two bodies of equal masses collides elastically, their velocities are interchanged in these types of collision.
Kinetic energy and linear momentum remain conserved.
According to the above diagram when m₁ comes in contact with the spring, m₁ is retarded by the spring force and m₂ is accelerated by the spring force.

(a) The spring will continue *to compress until the two blocks acquire common velocity. So some of kinetic energy of block M_x store into P.E and some part of it stores into K.E of block M₂. So option (a) is incorrect.
(b) As surfaces are frictionalless momentum of the system will be conserved. So option (b) is also incorrect.
(c) The two bodies of equal mass exchange their velocities in a head on elastic collision between them. So, if spring is massless, the final state of the M₁ is state of rest.
(d) Since there is a loss of K.E when the blocks collides on the rough surface. Hence, the collision is inelastic.

Very Short Answer Type Questions
Q22. A rough inclined plane is placed on a cart moving with a constant velocity u on horizontal ground. A block of mass M rests on the incline. Is any work done by force of friction between the block and incline? Is there then a dissipation of energy?
Sol: Let us consider the diagram according to the situation. Just imagine the situation that wedge is moving with velocity u. As the block Mis at rest with respect to the inclined plane. There is no pseudo force acting on the block because the wedge is moving with constant velocity. So,
f = frictional force = Mg sinθ (upward as shown in the figure)

As the block rests on the incline, the force of friction acting between the block and the incline opposes the tendency of sliding of the block. Since, block is not in motion with respect to incline, therefore, work done by the force of friction between the block and the inclined plane is zero. Also due to this reason there is no dissipation of energy.
Q23. Why is electrical power required at all when the elevator is descending? Why should there be a limit on the number of passengers in this case?
Sol: When the elevator is descending, then electric motor (mainly induction motor are used) provides some force to overcome the weight of elevator to prevent it from falling freely under gravity, then electric power is required to run electric motor which is holding the elevator via cable to prevent it from falling freely. This cable is able to sustain some limiting value of tension developed in it. Due to this reason, it is needed to limit the number of passengers in the elevator.

Q24. A body is being raised to a height h from the surface of earth. What is the sign of work done by
(a) applied force and  (b) gravitational force?
Sol:(a) External force is applied on the body to lift it in upward direction against its weight, therefore, angle between the applied force and displacement is = 0°

Work done by the applied force
W= F. S =Fs cos = Fs cos 0° = Fs ( cos 0° = 1)

i.e., the sign of work done by applied force is positive.

(b) As shown in figure the gravitational force acts in downward direction and displacement in upward direction, therefore, angle between them is = 180°.
Work done by the gravitational force
W = Fs cos 180° = -Fs (cos 180° = -1)

Q25. Calculate the work done by a car against gravity in moving along a straight horizontal road. The mass of the car is 400 kg and the distance moved is 2 m.
Sol:Weight of the car (mg) vertically downward and car is moving along horizontal road, so displacement of the car is in the along horizontal, i.e. angle between them is 90°.
Work done by weight of the car
W = Fs cos 90° = 0                          (cos 90° = 0)
Hence, the work done by car against gravity will also be zero.

Q26. A body falls towards earth in air. Will its total mechanical energy be conserved during the fall? Justify.
Sol: If we neglect the air resistance then the total mechanical energy of the body is conserved but if there is some air resistance then a small part of its energy is utilized against resistive force of air, which is non-conservative force. So, total mechanical energy of the body falling freely under gravity in this case is not conserved. In this condition, gain in KE < loss in PE.
Let E be the total mechanical energy.
Initial mechanical energy of the body

Q27. A body is moved along a closed loop. Is the work done in moving the body necessarily zero? If not, state the condition under which work done over a closed path is always zero.
Sol:
Key concept: If the work done by a force on a body depends upon the initial and final positions only of that body, then the force is conservative e.g., gravitational, electrostatic, magnetic forces.
If the work done by a force on a body which has moved in closed path and has come back to its initial position is zero, the force is conservative.
Work done in moving along a closed loop is not always zero. Work done in moving a closed path is zero when forces acting on the body are conservative but for non-conservative forces like friction force, viscous force etc. work done in a closed path is not zero.
Q28. In an elastic collision of two billiard balls, which of the following quantities remain conserved during the short time of collision of the balls (i.e., when they are in contact)?
(a) Kinetic energy
(b) Total linear momentum
Give reason for your answer in each case.
Sol: During collision no external force is acting on the balls, therefore total linear momentum of the system of two balls is always conserved.
During collision when the balls are in contact, there may be deformation, hence some kinetic energy of system will be transformed to potential energy of system and consequently kinetic energy will not be conserved.
Important point: Though kinetic energy of system will not be conserved but total energy of the system will be conserved.

Q29. Calculate the power of a crane, in watts, which lifts a mass of 100 kg to a height of 10 m in 20 s.
Sol: According to the problem, mass = m= 100 kg height = h- 10 m, time interval, t = 20 s Power is the rate of doing work with respect to time.

Q30. The average work done by a human heart while it beats once is 0.5 J. Calculate the power used by heart if it beats 72 times in a minute.
Sol: According to the problem, average work done by a human heart per beat = 0.5 J
Total work done during 72 beats in 1 minute
= 72 x 0.5 J =36 J
Power = Work done = 36 J / 60s = 0.6 w

Q31. Give example of a situation in which an applied force does not result in a change in kinetic energy.
Sol: Assume a ball tied to a string and is moving in a vertical circle. Work done by tension force will be zero and hence tension force will not cause any change in KE of ball. Because at any instant of time the displacement is tangential and the force is central in nature, i.e., tension in the string and the small displacement at any instant are perpendicular to each other.

Q32. Two bodies of unequal mass are moving in the same direction with equal kinetic energy. The two bodies are brought to rest by applying retarding force of same magnitude. How would the distance moved by them before coming to rest compare?
Sol: According to work-energy theorem,
Change in KE is equal to work done by all the forces acting on the body. Let us assume that only one force (retarding force) is acting on the body, therefore,
KE of the body = Work done by retarding force KE of the body = Retarding force x Displacement
As KE of the bodies and retarding forces applied on them are same, therefore, both bodies will travel equal distances before coming to rest.

Q33. A bob of mass m suspended by a light string of length L is whirled into a vertical circle as shown in figure. What will be the trajectory of the particle, if the string is cut at

• Point B?
• Point C?
• Point A?

Sol: Key concept: According to the situation shown above that a bob of mass m is whirled into a vertical circle, the required centripetal force is obtained from the net force towards center at any point of time in the string. Tension in the string is variable and it is always towards center. But the gravitational force on the bob is always towards center. The speed of the body will be different at different points. So the equations of dynamical equilibrium (F_c = ma_c, F_t = ma_t) must be satisfied at all the points. Let when the string makes an angle 9 with vertical, the speed of mass is v.
Apply Newton’s law perpendicular to the string:
Mg sin = ma, => a,= g sin
The above equation gives tangential acceleration as a function of angle . At lowest point = 0° and at highest point = 180°. So at both points sin 9=0. Hence a, = 0 at both points L and H.
At point M, = 90°, then a₁ = g. It is the maximum value of a_t
Apply Newton’s law along the string: T – mg cos = ma_c
or            T=mgcos + m v² /r…(i)

As the body goes up, its velocity will go on decreasing and angle θ will go on increasing. Maximum speed of the body will be at lowest point L and minimum at highest point H. Then from above relation we can find that tension will be maximum at lowest point and minimum at highest point.

When string is cut, tension in string becomes zero and centripetal force is not provided. Hence, bob tends to move in along the direction of its velocity.
(a) If the string is cut at any point, then velocity of body of mass m is along the tangent to the circle. Tangent at point B is vertically downward so the trajectory of the particle is the straight line.
(b) Tangent at point C is horizontally towards right.
So the trajectory of the particle is the parabola.
(c) Tangent at point X makes some angle with the horizontal. Again bob will follow a parabolic path with vertex higher than C.

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We hope the NCERT Exemplar Class 11 Physics Chapter 5 Work, Energy and Power help you. If you have any query regarding NCERT Exemplar Class 11 Physics Chapter 5 Work, Energy and Power, drop a comment below and we will get back to you at the earliest.

NCERT Exemplar Class 11 Physics Chapter 6 System of Particles and Rotational Motion are part of NCERT Exemplar Class 11 Physics. Here we have given NCERT Exemplar Class 11 Physics Chapter 6 System of Particles and Rotational Motion.

NCERT Exemplar Class 11 Physics Chapter 6 System of Particles and Rotational Motion

Multiple Choice Questions
Single Correct Answer Type

Q1. For which of the following does the centre of mass lie outside the body?
(a) A pencil (b) Ashotput
(c) A dice (d) A bangle
Sol: (d) .
Key concept: Center of mass of a system (body) is a point that moves as though all the mass were concentrated there and all external forces were applied there.
Important Points about Center of Mass:
(i) The position of center of mass is independent of the co-ordinate system chosen.
(ii) The position of center of mass depends upon the shape of the body and distribution of mass.
Example: The center of mass of a circular disc is within the material of the body while that of a circular ring is outside the material of the body.
(iii) We can imagine a rigid body also as a system of masses and hence every rigid body has a center of mass. In case of a regularly shaped uniform rigid body, center of mass is simply the geometric centre of the body.
A bangle is in the form of a ring as shown in the diagram below. We know that the position of center of mass depends upon the shape of the body and distribution of mass. So, out of four given bodies, the centre of mass lies at the centre, which is outside the body (boundary) whereas in all other three bodies it lies within the body because they are completely solid.

Q2. Which of the following points is the likely position of Hollow
the centre of mass of the system shown in figure? sphere
(a) A
(b) B
(c) C
(d) D
Sol: (c) The position of centre of mass of the system in this problem is closer to heavier mass or masses or we can say that it depends upon distribution of mass. So it is likely to be at C. In the above diagram, lower part of the sphere containing sand is more heavier than upper part containing air. Hence CM of the system lies below the horizontal diameter.

Q3. A particle of mass m is moving in yz-plane with a uniform velocity v with its trajectory running parallel to +ve y-axis and intersecting z-axis at z = a in figure. The change in its angular momentum about the origin as it bounces elastically from a wall at y = constant is

Q4. When a disc rotates with uniform, angular velocity, which of the following is not true?
(a) The sense of rotation remains same.
(b) The orientation of the axis of rotation remains same.
(c) The speed of rotation is non-zero and remains same.
(d) The angular acceleration is non-zero and remains same.
Sol: (d)
Key concept: The rate of change of angular velocity is defined as angular acceleration. If particle has angular velocity ω₁ at time
t₁, and angular velocity ω₂ at time t₂ , then
Angular acceleration α = ω₂ – ω₁ / t₂ – t₁

When the disc is rotated with constant angular velocity, angular acceleration of the disc is zero. Because we know that angular acceleration
α  = ∆ ω/∆ t
Here ω  is constant, so ∆ ω = 0

Q5. A uniform square plate has a small piece Q of an irregular shape removed and glued to the centre of the plate leaving a hole behind in figure. The moment of inertia about the z-axis is then
(a) increased
(b) decreased
(c) the same
(d) changed in unpredicted manner

Q6. In problem 5, the CM of the plate is now in the following quadrant of x-y plane.
(a) I (b) II (c) III (d) IV
Sol: (c) Let us consider the diagram below, which shows the position of the piece which is removed from the plate. First center of mass is at the centre of the plate (only if its mass is uniformly distributed over the surface) when the piece is removed from quadrant I, therefore the centre of mass is shifted to the


The sum of moments of inertia of a laminar object about two mutually perpendicular axes lying in the plane of lamina is equal to the moment of inertia about an axis normal to the plane of the lamina and passing through the two perpendicular axes. This theorem is applicable only for laminar (thin sheet kind of) object.

I_x and I_y both decreases with the hole. Gluing the removed piece at the centre of the square plate does not affect I_z. The mass comes closer to the z-axis, hence, moment of inertia decreases overall about z-axis.

Q6. In problem 5, the CM of the plate is now in the following quadrant of x-y plane.
(a) I (b) II (c) III (d) IV
Sol: (c) Let us consider the diagram below, which shows the position of the piece which is removed from the plate. First center of mass is at the centre of the plate (only if its mass is uniformly distributed over the surface) when the piece is removed from quadrant I, therefore the centre of mass is shifted to the  opposite of the quadrant III.

Q8. A merry-go-round, made of a ring-like platform of radius R and mass M, is revolving with angular speed ω. A person of mass Mis standing on it. At one instant, the person jumps off the round, radially away from the centre of the round (as seen from the round). The speed of the round afterwards is
(a) 2 ω
(b) ω
(c) ω/2
(d) 0
Sol: (b)As no torque is exerted by the person jumping, radially away from the centre of the round (as seen from the round), let the total moment of inertia of the system is 2I (round + Person (because the total mass is 2M) and the round is revolving with angular speed ωSince the angular momentum of the person when it jumps off the round is Iω the actual momentum of round seen from ground is  2 Iω – Iω = Iω
So we conclude that the angular speed remains same, i.e ω

More Than One Correct Answer Type
Q9. Choose the correct alternatives:
(a) For a general rotational motion, angular momentum L and angular velocity ω need not be parallel.
(b) For a rotational motion about a fixed axis, angular momentum L and angular velocity ωare always parallel.
(c) For a general translational motion, momentum p and velocity v are always parallel.
(d) For a general translational motion, acceleration a and velocity v are always parallel.

Sol: (a, c) .
(a) For a general rotational motion where axis of rotation is not symmetric. Angular momentum Z and angular velocity 0) need not be parallel. The wobbly motion of a wheel rotating about an axis inclined at a small angle to the symmetry axis of the wheel represents a situation where angular momentum and angular velocity are not parallel.
(b) Fixed axis should pass through CM of the body, so it is not necessary angular momentum Z and angular velocity ω are always parallel.
(c) As we know in a general translational motion linear momentum is given by, p = mv , hence, direction of p is always along v .
(d) In projectile motion, v and a are not always parallel.

Q10. Figure shows two identical particles 1 and 2, each of mass m, moving in opposite directions with same speed v along parallel lines. At a particular instant r₁ and r₂ are their respective position vectors drawn from point A which is in the plane of the parallel lines. Choose the correct options:

Figure shows two identical particles 1 and 2, each of mass m, moving in opposite directions with same speed v along parallel lines. At a particular instant r₁ and r₂ are their respective position vectors drawn from point A which is in the plane of the parallel lines. Choose the correct options:

Q11. The net external torque on a system of particles about an axis is zero. Which of the following are compatible with it?
(a) The forces may be acting radially from a point on the axis.
(b) The forces may be acting on the axis of rotation.
(c) The forces may be acting parallel to the axis of rotation.
(d) The torque caused by some forces may be equal and opposite to that caused by other forces.
Sol: (a, b, c, d)

Important point: To get the direction where you can use right hand rule: Place the fingers of right hand along r and then curl them into F through the smaller angle between them, n is directed along the (stretched) thumb. For involving two dimensions only, can be replaced by sense of rotation, clockwise or anticlockwise; if the fingers of the right hand curl (while going from clockwise, torque is taken as clockwise and when they curl anticlockwise torque is taken as anticlockwise.)



Sol: (b, d) We can apply the concept of symmetry to calculate the net moment of inertia. Moment of inertia about two symmetrical axes are same.

 

Very Short Answer Type Questions
Q14. The centre of gravity of a body on the earth coincides with its centre of mass for a small object whereas for an extended object it may not. What is the qualitative meaning of small and extended in this regard? For which of the following two coincides—A building, a pond, a lake, a mountain?
Sol:
Key concept: The center of gravity of a body is that point through which the resultant of the system of parallel forces formed by the weights of all the particles constituting the body passes for all positions of the body. It is denoted as “C.G.” or “G”.
Centre of gravity is centre of a given structure but centre of mass is a point where whole mass of the body can be assumed to be concentrated.
An object is said to be small if its vertical height is very small compared to the radius of the earth, otherwise it is extended.
Building and ponds are small objects so their CG coincides with CM, while a deep lake and a mountain can be considered as extended objects, so the CG does not coincide in their CM.

Q15. Why does a solid sphere have smaller moment of inertia than a hollow cylinder of same mass and radius, about an axis passing through their axes of symmetry?

Sol:Key concept: Moment of inertia of a particle l = mr² where r is the perpendicular distance of particle from rotational axis.
Moment of inertia of a body made up of number of particles (discrete distribution)

I = m₁r₁² + m₂r₂² + m₃r₃²
Moment of inertia of a continuous distribution of mass, treating the element of mass dm at position r as particle
dl = dmr²

MI is not constant for a body. It depends on the axis of rotation.
MI depends on the mass of the body. The higher the mass, higher the MI.
MI depends on the distribution of the mass about an axis. The farther the mass is distributed from the axis, higher will be the MI.
Moment of inertia depends on mass, distribution of mass and on the position of axis of rotation.
All the mass in a cylinder lies at distance R from the axis of symmetry but most of the mass of a solid sphere lies at a smaller distance than R. Therefore,
I_{hollowcylinder} > I_sphere

Q16. The variation of angular position , of a point on a rotating rigid body, with time t is shown in figure. Is the body rotating clockwise or anti-clockwise?

Q17. A uniform cube of mass m and side a is placed on a frictionless horizontal surface. A vertical force F is applied to the edge as shown in figure. Match the following (most appropriate choice).

Column I		Column 11
(a)	mg/4 <F< mg/2	(i)	Cube will move up.
(b)	F > mg/2	(ii)	Cube will not exhibit motion.
(c)	F> mg	(iii)	Cube will begin to rotate and slip at A.
(d)	F = mg/4	(iv)	Normal reaction effectively at a/3 from A, no motion.

 

 Sol: Let us first consider the below diagram torque or moment of the force F about point A is given by _l=aF

This is anticlockwise.
Torque of weight mg about A,
₂ = q/2
This is clockwise.
N is acting at point A. So, torque due to normal reaction about A will be zero. There is no motion in cube if ₁ = ₂

(a) → (ii)
(b) → (iii)
(c) → (i)
(d) → (iv)

Q18. A uniform sphere of mass m and radius R is placed  on a rough horizontal surface (figure). The sphere h is struck horizontally at a height h from the floor. ,, Match the following.

Column 1		Column II
(a)	h =R/2	(0	Sphere rolls without slipping with a constant velocity and no loss of energy.
(b)	h = R	(ii)	Sphere spins clockwise, loses energy by friction
(c)	h = 3R/2	(iii)	Sphere spins anti-clockwise, loses energy by friction.
(d)	h = 1R!5	(iv)	Sphere has only a translational motion, looses energy by friction.

 

Sol: Mass of the sphere = m
Radius = R
h = height from the floor
The sphere will roll without slipping when

ω = V/R
where, v is linear velocity and to is angular velocity of the sphere.
Now, angular momentum of sphere, about centre of mass [We are applying conservation of angular momentum just before and after struck.]

Short Answer Type Questions

Q19. The vector sum of a system of non-collinear forces acting on a rigid body is given to be non-zero. If the vector sum of all the torques due to the system of forces about a certain point is found to be zero, does this mean that it is necessarily zero about any arbitrary point?

Q20. A wheel in uniform motion about an axis passing through its centre and perpendicular to its plane is considered to be in mechanical (translational plus rotational) equilibrium because no net external force or torque is required to sustain its motion. However, the particles that constitute the wheel do experience a centripetal acceleration directed towards the centre. How do you reconcile this fact with the wheel being in equilibrium?How would you set a half-wheel into uniform motion about an axis passing through the centre of mass of the wheel and perpendicular to its plane? Will you require external forces to sustain the motion?
Sol: Internal elastic forces gives rise to the centripetal acceleration of its particles in a wheel. These forces are in pairs and cancel each other because they are part of a symmetrical system.
In a half wheel, the distribution of mass about its centre of mass (through which axis of rotation passes) is not symmetrical. Therefore, the direction of angular momentum of the wheel does not coincide with the direction of its angular velocity. Hence, an external torque is required to maintain the motion of the half wheel.

Q21. A door is hinged at one end and is free to rotate about a vertical axis (figure). Does its weight cause any torque about this axis? Give reason for your answer.
Sol: According to the diagram, where weight of the door acts along negative y-axis.
Torque is not produced by weight about y-axis.
Because the direction of weight is parallel to y-axis (axis of rotation).
A force can produce torque only along direction normal to itself because f = r x F. So, when the door is in the xy-plane, the torque produced by gravity can only be along ±z-direction never about an axis passing through y-direction.
Hence, the weight will not produce any torque about y-axis.

Q22. (n – 1) equal point masses each of mass m are placed at the vertices of a regular n-polygon. The vacant vertex has a position vector a with respect to the centre of the polygon. Find the position vector of centre of mass.

Hence the center of mass of n particles is a weighted average of the position vectors of n particles making up the system.
The centre of mass of a regular n-polygon lies at its geometrical centre.
Let position vector of each centre of mass or regular n polygon is r .
(n – 1) equal point masses each of mass m are placed at (n – 1) vertices of the regular n-polygon, therefore, for its centre of mass

Long Answer Type Questions
Q23. Find the centre of mass of a uniform (a) half-disc, (b) quarter-disc.

If Mis mass of the half-disc of radius R, then mass per unit area of the half-disc

Q24. Two discs of moments of inertia I₁, and I₂ about their respective axes (normal to the disc and passing through the centre), and rotating with angular speed ω₁ and ω₂ and are brought into contact face to face with their axes of rotation coincident.

(a) Does the law of conservation of angular momentum apply to the situation? Why?
(b) Find the angular speed of the two discs system.
(c) Calculate the loss in kinetic energy of the system in the process.
(d)Account for this loss.

Q25. A disc of radius R is rotating with an angular co0 about a horizontal axis. It is placed on a horizontal table. The coefficient of kinetic friction is jJ.K.
(a) What was the velocity of its centre of mass before being brought in contact with the table?
(b) What happens to the linear velocity of a point on its rim when placed in contact with the table?
(c) What happens to the linear speed of the centre of mass when disc is placed in contact with the table?
(d) Which force is responsible for the effects in (b) and (c)?
(e) What condition should be satisfied for rolling to begin?
(f) Calculate the time taken for the rolling to begin.
Sol:
Key concept:When the axis of rotation is not fixed (Stationary in space), the motion of a rigid body is considered as combination of motion of centre of mass plus a rotation about centre of mass. The two components of motion are described by

How the translational motion and rotational motion about the centre of mass are superimposed to get the motion of a rigid body (say a disc of radius R) are illustrated in the following figures.

To get the instantaneous velocity of any point on the rigid body, we calculate the instantaneous velocity of that point in pure translation and in pure rotation and add them vectorially.

(a) Disc is rotating only about its horizontal axis before being brought in contact with the table. Hence its CM is at rest; v_CM = 0
(b) When the disc is placed in contact with the table due to friction velocity of a point on the rim
(c) Linear speed of the CM of disc increases when disc is placed in contact with the table, because its acceleration becomes
a_CM = _kg
(d) Friction is responsible for the effects in (b) and (c) because friction is disturbing the velocity of the point which is in contact with table, hence velocity at all the points of disc is disturbed

Important point: In pure rolling motion, frictional force will support. Or we can say that it just opposes the relative motion of point of contact at any instant.

Q26. Two cylindrical hollow drums of radii R and 2R, and of a common height h, are rotating with angular velocities (anti-clockwise) and (clockwise), respectively. Their axes, fixed are parallel and in a horizontal plane separated by (3R + ). They are now brought in contact ( → 0).
(a) Show the frictional forces just after contact.
(b) Identify forces and torques external to the system just after contact.
(c) What would be the ratio of final angular velocities when friction ceases?
Sol: (a) The frictional forces acting between two cylindrical hollow drums are as shown in the diagram below.

Force F upward shows the friction force on left drum.
Force F downward shows the friction force on right drum.

(b) F¹ = F =F” where F¹ and F” are external forces through support.
=> F_net = 0 (one each cylinder)
Net external torque to the system about any axis=Fx3R, anticlockwise

(c) Let _l and ₂ be final angular velocities of smaller and bigger drum respectively (anti­clockwise and clockwise respectively).

Finally, there will be no friction. When friction ceases at the point of contact, then both drums has equal linear velocity at that point.

^VA = ^VB
Hence,  R _l = 2R ₂ ⇒_l / ₂ = 2

Important point: Friction force just opposes the relative motion of point of contacts at any instant. So, we should be very careful while indicating direction of frictional forces.

Q27. A uniform square plate S (side c) and a uniform rectangular plate R (sides b, a) have identical areas and masses.

Q28. A uniform disc of radius R, is resting on a table on its rim. The coefficient of friction between disc and table is μ , (Figure). Now, the disc is pulled with a force F as shown in the figure. What is the maximum value of F for which the disc rolls without slipping?

Sol:In this problem friction force on the disc will act in opposite direction of F at the point which in contact with surface at any instant of time and supports the rotation of the disc in clockwise direction.

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We hope the NCERT Exemplar Class 11 Physics Chapter 6 System of Particles and Rotational Motion help you. If you have any query regarding NCERT Exemplar Class 11 Physics Chapter 6 System of Particles and Rotational Motion, drop a comment below and we will get back to you at the earliest.

NCERT Exemplar  Class 11 Physics Chapter 7 Gravitation are part of NCERT Exemplar Class 11 Physics. Here we have given NCERT Exemplar Class 11 Physics Chapter 7 Gravitation.

NCERT Exemplar Class 11 Physics Chapter 7 Gravitation

Multiple Choice Questions
Single Correct Answer Type
Ncert Exemplar Class 11 Physics Gravitation

Q1. The earth is an approximate sphere. If the interior contained matter which is not of the same density everywhere, then on the surface of the earth, the acceleration due to gravity
(a) ‘ will be directed towards the centre but not the same every where
(b) will have the same value everywhere but not directed towards the centre
(c) will be same everywhere in magnitude directed towards the centre
(d) cannot be zero at any point
Sol: (d) Acceleration due to gravity g = 0, at the centre if we assume the earth as a sphere of uniform density, then it can be treated as point mass placed at its centre.
But on surface of the earth the acceleration due to gravity cannot be zero at any point. .

Gravitation Class 11 Ncert Exemplar Solutions

Q2. As observed from the earth, the sun appears to move in an approximate circular orbit. For the motion of another planet like mercury as observed from the earth, this would
(a) be similarly true
(b) not be true because the force between the earth and mercury is not inverse square law
(c) not be true because the major gravitational force on mercury is due to the sun
(d) not be true because mercury is influenced by forces other than gravitational forces
Sol: (c) As observed from the earth, the sun appears to move in an approximate circular orbit. The gravitational force of attraction between the earth and the sun always follows inverse square law.
All planets move around the sun due to the huge gravitational force of the sun acting on them. The gravitational force on the mercury due to earth is much smaller as compared to that acting on it due to sun and hence it revolves around the sun and not around the earth.

Gravitation Ncert Exemplar Class 11

Q3. Different points in the earth are at slightly different distances from the sun and hence experience different forces due to gravitation. For a rigid body, we know that if various forces act at various points in it, the resultant motion is as if a net force acts on the CM (centre of mass) causing translation and a net torque at the CM causing rotation around an axis through the CM. For the earth-sun system (approximating the earth as a uniform density sphere)
(a) the torque is zero
(b) the torque causes the earth to spin
(c) the rigid body result is not applicable since the earth is not even approximately a rigid body
(d) the torque causes the earth to move around the sun
Sol: (a) As the earth is revolving around the sun in a circular motion (approximately in actual the path of earth around the sun is elliptical) due to gravitational attraction. When we consider the earth-sun as a single system and we are taking earth as a sphere of uniform density. Then the gravitational force (F) will be of radial nature, i.e. angle between position vector r and force F is zero. So, torque

Q4. Satellites orbiting the earth have finite life and sometimes debris of satellites fall to the earth. This is because
(a) the solar cells and batteries in satellites run out
(b) the laws of gravitation predict a trajectory spiralling inwards
(c) of viscous forces causing the speed of satellite and hence height to gradually decrease
(d) of collisions with other satellites
Sol: (c) We know that the total energy of the earth satellite of radius R bounded system is negative
(-GM\ 2 R)
where Mis mass of the earth. –
Due to the atmospheric friction (viscous force) acting on satellite, energy decreases continuously, radius of the orbit or height decreases gradually and the satellite spirals down with increasing speed till it bums in the denser layers of the atmosphere.

Q5. Both the earth and the moon are subject to the gravitational force of the sun. As observed from the sun, the orbit of the moon
(a) will be elliptical
(b) will not be strictly elliptical because the total gravitational force on it is not central
(c) is not elliptical but will necessarily be a closed curve
(d) deviates considerably from being elliptical due to influence of planets other than the earth
Sol: (b) Moon revolves around the earth in a nearly circular orbit. When it is observed from the sun, two types of forces are acting on the moon one is due to gravitational attraction between the sun and the moon and the other is due to gravitational attraction between the earth and the moon. So moon is moving under the combined gravitational pull acting on it due to the earth and the sun. Hence, total force on the moon is not central.

Q6. In our solar system, the inter-planetary region has chunks of matter (much smaller in size compared to planets) called asteroids. They
(a) will not move around the sun, since they have very small masses compared to the sun
(b) will move in an irregular way because of their small masses and will drift away into outer space
(c) will move around the sun in closed orbits but not obey Kepler’s laws
(d) will move in orbits like planets and obey Kepler’s laws
Sol: (d)
Key concept: The Law of Orbits-. Every planet moves around the sun in an elliptical orbit with sun at one of the foci.
The Law of Area-. The, line joining the sun to the planet sweeps out equal areas in equal intervals of time, i.e. areal velocity is constant. According to this law planet will move slowly when it is farthest from sun and more rapidly when it is nearest to the sun. It is similar to law of conservation of angular momentum.

The Law of Periods: The square of period of revolution (7) of any planet around sun is directly proportional to the cube of the semi-major axis of the orbit.

Asteroids are moving in circular orbits like planets because they are being acted upon by central gravitational forces, they must obey Kepler’s laws. You may consider an asteroid and analyze that it satisfies above laws or not.

Q7. Choose the wrong option.
(a) Inertial mass is a measure of difficulty of accelerating a body by an external force whereas the gravitational mass is relevant in determining the gravitational force on it by an external mass.
(b) That the gravitational mass and inertial mass are equal is an experimental result.
(c) That the acceleration due to gravity on the earth is the same for all bodies is due to the equality of gravitational mass and inertial mass.
(d) Gravitational mass of a particle like proton can depend on the presence of neighbouring heavy objects but the inertial mass cannot.
Sol: (d)
Key concept: Inertial mass: It is the mass of the material body, which measures its inertia.
Gravitational Mass: It is the mass of the material body, which determines the gravitational pull acting upon it.
According to the principle of equivalence, Gravitational mass of proton is equivalent to its inertial mass and is independent of presence of neighboring heavy objects.
Important point: Comparison between inertial and gravitational mass:
• Both are measured in the same units.
• Both are scalars.
• Both do not depend on the shape and state of the body.
• Inertial mass is measured by applying Newton’s second law of motion whereas gravitational mass is measured by applying Newton’s law of gravitation.
• Spring balance measures gravitational mass and inertial balance measure inertial mass.
Q8. Particles of masses 2M, m and M are respectively at points A, B and C with AB = 1/2(BC). m is much-much smaller than M and at time t = 0, they are all at rest as given in figure.
At subsequent times before any collision takes place.

(a) m will remain at rest
(b) m will move towards M
(c) m will move towards 2M
(d) m will have oscillatory motion
Sol:(c) The particle m at B will move towards A with the greater force, due to particle 2M at A.

Force on m at B due to 2M at A is

More Than One Correct Answer Type

Q9. Which of the following options are correct?
(a) Acceleration due to gravity decreases with increasing altitude.
(b) Acceleration due to gravity increases with increasing depth (assume the earth to be a sphere of uniform density).
(c) Acceleration due to gravity increases with increasing latitude.
(d) Acceleration due to gravity is independent of the mass of the earth.
Sol: (a, c)
Key concept: Acceleration due to Gravity:
The force of attraction exerted by the earth on a body is called gravitational pull or gravity.
We know that when force acts on a body, it produces acceleration. Therefore, a body under the effect of gravitational pull must accelerate.
The acceleration produced in the motion of a body under the effect of gravity is called acceleration due to gravity. It is denoted by g.
Consider a body of mass m lying on the surface of earth then gravitational force on the body is given by

F = GMm/R²

where M- mass of the earth and R = radius of the earth.
The value of acceleration due to gravity vary due to the following factors:
(a) Shape of the earth, (b) Height above the earth surface, (c) Depth below the earth surface and (d) Axial rotation of the earth.

Q10. If the law of gravitation, instead of being inverse square law, becomes an inverse cube law
(a) planets will not have elliptic orbits
(b) circular orbits of planets is not possible
(c) projectile motion of a stone thrown by hand on the surface of the earth will be approximately parabolic
(d) there will be no gravitational force inside a spherical shell of uniform density
Sol: (a, b, c) If the law of gravitation becomes an inverse cube law instead of inverse square law, then for a planet of mass m revolving around the sun of mass M, we can write



Q11. If the mass of the sun were ten times smaller and gravitational constant G were ten times larger in magnitude. sun Then,
(a) walking on ground would become more difficult
(b) the acceleration due to gravity on the earth will not change
(c) raindrops will fall much faster
(d) airplanes will have to travel much faster


As r »R (radius of the earth) => F will be very small.
So, the effect of the sun will be neglected.
Due to this reason gravity pull on the person will increase. Due to it, walking on ground would become more difficult.
Critical velocity, v_c is proportional to g, i.e., v^cg
As, g’ > g => v_c‘>v_c
Hence, rain drops will fall much faster.
To overcome the increased gravitational force of the earth, the aeroplanes will have to travel much faster.

Q12. If the sun and the planets carried huge amounts of opposite charges,
(a) all three of Kepler’s laws would still be valid
(b) only the third law will be valid
(c) the second law will not change
(d) the first law will still be valid
Sol:(a, c, d) Coulomb’s electric force or Electrostatic force of attraction will produce due to opposite charges.
If the sun and the planets carries huge amount of opposite charges, then electrostatic force of attraction will be large. Gravitational force is also attractive in nature have both forces will be added and both are radial in nature.
Both the forces obey inverse square law and are central forces. As both the forces are of same nature, hence all the three Kepler’s laws will be valid.

Q13. There have been suggestions that the value of the gravitational constant G becomes smaller when considered over very large time period (in billions of years) in the future. If that happens, for our earth,
(a) nothing will change
(b) we will become hotter after billions of years
(c) we will be going around but not strictly in closed orbits
(d) after sufficiently long time we will leave the solar system
Sol: (c, d) We know that gravitational force exists between the earth and the sun.

F _G = G(M_S x m_e)/ r²  Where M_S is mass of the sun and m_e is mass of the earth.

This provides the necessary centripetal force for the circular orbit of the earth around the sun. As G decreases with time, the gravitational force F_G will become weaker with time. As F_G is changing with time due to it, the earth will be going around the sun not strictly in closed orbit and radius also increases, since the attraction force is getting weaker.
Hence, after long time the earth will leave the solar system.

Q14.       Supposing Newton’s law of gravitation for gravitational forces F₁ and F₂ between two masses m₁ and m₂ at positions r₁, and r₂ read

(a)  the acceleration due to gravity on the earth will be different for different objects
(b) none of the three laws of Kepler will be valid
(c)  only the third law will become invalid
(d)  for n negative, an object lighter than water will sink in water

Sol: (a, c, d) According to the problem,

Since, g depends upon position vector and mass of object, hence it will be different for different objects. As g is not constant, hence constant of proportionality will not be constant in Kepler’s third law.
Hence, Kepler’s third law will not be valid.
As the force is of central nature,

Hence, first two Kepler’s laws will be valid. Hence option (b) in incorrect and (c) is correct.

(d) When n is negative,
F = k/mⁿ
Or we can say that F is inversely proportional to mass. This implies that lighter bodies will experience a greater force than the heavier bodies and vice versa. Hence, object lighter than water will sink in water
Q15. Which of the following are true?
(a) A polar satellite goes around the earth’s pole in north-south direction
(b) A geostationary satellite goes around the earth in east-west direction
(c) A geostationary satellite goes around the earth in west-east direction
(d) A polar satellite goes around the earth in east-west direction
Sol. (a, c)
Key concept: The satellite which appears stationary relative to earth is called geostationary or geosynchronous satellite, communication satellite.
A geostationary satellite revolves around the earth with the same angular velocity and in the same sense as done by the earth about its own axis, i.e. west-east direction.
A polar satellite revolves around the earth’s pole in north-south direction. It is independent of earth’s rotation.

Q16. The centre of mass of an extended body on the surface of the earth and its centre of gravity
(a) are always at the same point for any size of the body
(b) are always at the same point only for spherical bodies
(c) can never be at the same point
(d) is close to each other for objects, say of sizes less than 100 m
(e) both can change if the object is taken deep inside the earth
Sol: (d) The center of gravity is based on weight, whereas the center of mass is based on mass. So, when the gravitational field across an object is uniform, the two are identical. However, when the object enters a spatially-varying gravitational field, the COG will move closer to regions of the object in a stronger field, whereas the COM is unmoved.
More practically, the COG is the point over which the object can be perfectly balanced; the net torque due to gravity about that point is zero. In contrast, the COM is the average location of the mass distribution or it is the point where whole mass of the body is supposed to be concentrated. If the object were given some angular momentum, it would spin about the COM.
For small objects, say of sizes less than 100 m placed in uniform gravitational field then centre of mass is very close with the centre of gravity of the body. But when the size of object increases, its weight changes and its CM and CG become far from each other. Like in the case of spherical ball, the CM and the CG are the same, but in case of Mount Everest, its CM lies a bit above its CG.

Very Short Answer Type Questions
Q17. Molecules in air in the atmosphere are attracted by gravitational force of the earth. Explain why all of them do not fall into the earth just like an apple falling from a tree.
Sol: Air molecules in the atmosphere are attracted by gravitational force of the earth but they do not fall into earth due to the reason that the molecules in air has some random motion due to temperature, so their resultant motion is not exactly in the vertical downward direction.
But in case of apple, only vertical motion dominates because of being heavier than air molecules. But due to gravity, the density of air is more near to the earth than the density as we go up.

Q18. Give one example each of central force and non-central force.
Sol: Examples of central force:
(i) Gravitational force due to point mass.
(ii) Electrostatic force on the point charge.
Examples of non-central force:
(i) Nuclear force which depends upon the spin of particles.
(ii) Magnetic systems acting between two current carrying loops

Q19. Draw areal velocity versus time graph for Mars.
Sol: Areal velocity of Mars revolving around the Sun does not change with time according to Kepler’s law, i.e. it is constant with time. Then graph of a real velocity versus time is a straight line parallel to time axis.

Q20. What is the direction of a real velocity of the earth around the sun?
Sol:


Q21. How is the gravitational force between two point masses affected when they are dipped in water keeping the separation between them the same?
Sol: Gravitational force acting between two point masses ml and m2 is given by the relation,
F = G m₁m₂/ r²
It does not depend upon the medium separating the two point masses. Therefore, gravitational force acting between two point masses will remain unaffected when they are dipped in water keeping the separation between them same. Only their apparent weights change, there is no effect on masses.

Q22. Is it possible for a body to have inertia but no weight?
Sol: Key concept:The weight of a body is the force with which it is attracted towards the centre of earth. When a body is stationary with respect to the earth, its weight equals the gravity. This weight of the body is known as its static or true weight.
Inertia is a property of mass. Hence, a body can have inertia (i.e., mass) but no weight. Everybody always have inertia but its weight (mg) can be zero, when it is taken at the centre of the earth or during free fall under gravity or a body placed at a very large distance from earth. Basically weight of a body can zero when acceleration due to gravity is zero, that condition is called weightlessness.

For example, When a satellite revolves in its orbit around the earth:
Weightlessness possess many serious problems to the astronauts. It becomes quite difficult for them to control their movements. Everything in the satellite has to be kept tied down. They can be displaced due to their inertia. Creation of artificial gravity is the answer to this problem.

Q23. We can shield a charge from electric fields by putting it inside a hollow conductor. Can we shield a body from the gravitational influence of nearby matter by putting it inside a hollow sphere or by some other means?
Sol: Gravitational force between two point mass bodies is independent of the intervening medium between them. That’s why a body cannot be shielded from the gravitational influence of nearby matter, because it is independent of the medium (as discussed in the previous problem). So, we cannot shield a body from the gravitational influence of nearby matter by putting it either inside a hollow sphere or by some other means.

Q24. An astronaut inside a small spaceship orbiting around the earth cannot detect gravity. If the space station orbiting around the earth has a large size, can he hope to detect gravity?
Sol: The value of acceleration due to gravity, supposed to be constant inside a small spaceship orbiting around the earth, and hence astronaut feels weightlessness. If the size of the space station orbiting around the earth is very large, then the astronaut inside the spaceship will experience variation in acceleration due to gravity.

Q25. The gravitational force between a hollow spherical shell (of radius R and uniform density) and a point mass is F. Show the nature of F versus r graph where r is the distance of the point from the centre of the hollow spherical shell of uniform density.
Sol: Let us consider the diagram of spherical shell having uniform density (p).

Mass of the shell = (density) x (volume)

Q26. Out of aphelion and perihelion, where is the speed of the earth more arid why?
Sol: Aphelion is the location of the earth where it is at the greatest distance from the sun and perihelion is the location of the earth where it is at the nearest distance from the sun. According to the diagram given below point A represents aphelion and point P perihelion.

Therefore, the speed of the earth is more at the perihelion than at the aphelion.

 

Q27. What is the angle between the equatorial plane and the orbital plane of
(a) polar satellite?
(b) geostationary satellite?

Sol: According to the diagram where plane of geostationary and polar satellite are shown.
(a) The angle between the equatorial plane and orbital plane of a polar satellite is 90°.
(b) The angle between equatorial plane and orbital plane of a geostationary satellite is 0°.

Q28. Mean solar day is the time interval between two successive noon when sun passes through zenith point (meridian).
Sidereal day is the time interval between two successive transit of a distant star through the zenith point (meridian).
By drawing appropriate diagram showing the earth’s spin and orbital motion, show that mean solar day is 4 minutes longer than the sidereal day. In other words, distant stars would rise 4 minutes early every successive day.
Sol: According to the diagram alongside, when the earth revolves about its polar axis in one sidereal day, it also moves from E to E’ around the sun due to translational motion and the point P’ is at P’’.

When the Earth still rotates through an angle about its axis to complete one solar day until the point P’ is at P”, again facing the sun.
Earth advances in its orbit by approximately 1° every day, i.e. in 24 hours. Then, it will have to rotate by 361° (which we define as 1 day) to have the sun at zenith point again,

Q29. Two identical heavy spheres are separated by a distance 10 times their radius. Will an object placed at the mid-point of the line joining their centres be in stable equilibrium or unstable equilibrium? Give reason for your answer.
Sol: We have to displace the object through a small distance, to determine the nature of equilibrium from the middle point and then force will be calculated in displaced position.
Let the mass and radius of each identical heavy sphere be M and R respectively. P is the midpoint of AB. An object of mass m be placed at the mid-point P of the line joining their centres.
The magnitude of force applied by each sphere on the object mass m is given by

F₁ = F₂ =GMm/(5R)²

The direction of forces are opposite, therefore resultant force acting on the object is zero. And the mass m is in stable condition.
If mass m is displaced towards right by small distance r. Now, force acting towards sphere, A due to object B is

Q30. Show the nature of the following graph for a satellite orbiting the earth.
(a) KE versus orbital radius R
(b) PE versus orbital radius R
(c) TE versus orbital radius R

Q31. Shown are several curves [Fig. (a), (b), (c), (d), (e) and (f)]. Explain with reason, which ones amongst them can be possible trajectories traced by a projectile (neglect air friction).

Sol. The trajectory or the path followed by a projectile under gravitational force of the earth instead of parabolic will be a conic section (for motion outside the earth) with the centre of the earth as a focus. Only the diagram in option (c) will fulfill the requirements.
Important point: In projectile we have taken the value of gravitational acceleration ‘g’ as a constant because up to some distance its variation will be neglected. The trajectory of the particle depends upon the velocity of projection. Depending upon the magnitude and direction of velocity it may be parabolic or elliptical.

Q32. An object of mass m is raised from the surface of the earth to a height equal to the radius of the earth, that is, taken from a distance R to 2R from the centre of the earth. What is the gain in its potential energy?

Sol:According to the diagram shown below, where an object of mass m is raised from the surface of the earth to a distance (height) equal to the radius of the earth (R).
Initial potential energy of the object when it is at the surface of the earth

U_i =  GMm/R

R where, Mis the mass of earth and R is the radius of earth or distance of object from centre of earth.
Final potential energy of the object when it is at a height equal to the radius

  U_f =  – GMm/ 2 R

Q33. A mass m is placed at P a distance h along the normal through the centre O of a thin circular ring of mass M and radius r (figure).
If the mass is moved further away such that OP becomes 2h, by what factor the force of gravitation will decrease, if h = r?

Sol: According to the problem, let us first consider the diagram, in which a system consisting of a ring and a point mass is shown.
Gravitational force acting on an object of mass m, placed at point P at a distance h along the normal through the centre of a circular ring of mass M and radius r.
To find that gravitation force due to ring, we have to consider a small element of the ring of mass dM.
Let the distance between elementary mass dM and m is x.

Long Answer Type Questions
Q34. A star like the sun has several bodies moving around it at different distances. Consider that all of them are moving in circular orbits. Let r be the distance of the body from the centre of the star and let its linear velocity be v, angular velocity to, kinetic energy K, gravitational potential energy U, total energy E and angular momentum /. As the radius r of the orbit increases, determine which of the above quantities increase and which ones decrease.
Sol: In equilibrium, the gravitation pull provides the necessary centripetal force. The situation is shown in the diagram, where a body of mass m is revolving around a star of mass M.

Q35. Six point masses of mass m each are at the vertices of a regular hexagon of side l. Calculate the force on any of the masses.
Sol: Resultant force will be equal to sum of individual forces by each point mass (m).
According to the diagram below, in which six point masses are placed at six vertices A, B, C, D, E and F.

Q36. Earth’s orbits an ellipse with eccentricity 0.0167. Thus, the earth’s distance from the sun and speed as it moves around the sun varies from day-to-day. This means that the length of the solar day is not constant through the year. Assume that the earth’s spin axis is normal to its orbital plane and find out the length of the shortest and the longer, day. A day should be taken from noon to noon. Does this explain variation of length of the day during the year?
Sol: From the geometry of the ellipse of eccentricity e and semi major axis a, the aphelion and perihelion distances are:

Q37. If mean angular velocity co corresponds to 1° per day, then _p = 1.034° per day and _a = 0.967° per day.
Since, 361° = 24 mean solar day we get (360 + 1.034) which corresponds to 24 h, 8.14″ (8.1″ longer) and 360.967°, corresponds to 23 h 59 min 52″ (7.9″ smaller).
This does not explain the actual variation of the length of the day during the year.
[G = 6.67 x 10^-11 SI unit and M= 6 x 10²⁴ kg]

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NCERT Exemplar Class 11 Physics Chapter 8 Mechanical Properties of Solids are part of NCERT Exemplar Class 11 Physics. Here we have given NCERT Exemplar Class 11 Physics Chapter 8 Mechanical Properties of Solids.

NCERT Exemplar Class 11 Physics Chapter 8 Mechanical Properties of Solids

Single Correct Answer Type
Mechanical Properties Of Solids Exemplar NCERT Class 11

Q1. Modulus of rigidity of ideal liquids is
(a) infinity
(b) zero
(c) unity
(d) some finite small non-zero constant value
Sol: (b) Key concept: Modulus of Rigidity:
Within limits of proportionality, the ratio of tangential stress to the shearing strain is called modulus of rigidity of the material of the body and is denoted by η

In this case the shape of a body changes but its volume remains unchanged.
Consider a cube of material fixed at its lower face and acted upon by a tangential force F at its upper surface having area A.
Only solids can exhibit a shearing as these have definite shape.

In liquids,η = 0
So, the frictional (viscous) force cannot exist in case of ideal fluid and since they cannot sustain shearing stress or tangential forces are zero, so there is no stress developed.

Mechanical Properties Of Solids Ncert Exemplar Solutions Class 11

Q2. The maximum load a wire can withstand without breaking, when its length is reduced to half of its original length, will
(a) be double (b) be half
(c) be four times (d) remain same
Sol: (d)
When the wire is loaded beyond the elastic limit, then strain increases much more rapidly. The maximum stress corresponding to B (see stress—strain curve) after which the wire begin to flow and breaks, is called breaking stress or tensile strength and the force by application of which the wire breaks is called the breaking force.
(i) Breaking force depends upon the area of cross-section of the wire, i.e. Breaking force ∝ A
Breaking force = P x A
Here P is a constant of proportionality and known as breaking stress.
(ii) Breaking stress is a constant for a given material and it does not depend upon the dimension (length or thickness) of wire.
(iii) If a wire of length L is cut into two or more parts, then again its each part can hold the same weight as breaking force is independent of the length of wire.

Mechanical Properties Of Solids Ncert Exemplar Class 11

Q3. The temperature of a wire is doubled. The Young’s modulus of elasticity
(a) will also double
(b) will become four times
(c) will remain same
(d) will decrease
Sol:(d)
Key concept: Youngs modulus (Y)
It is defined as the ratio of normal stress to longitudinal strain within limit of proportionality.

As Y ∝ 1/ ∆T

When temperature increases ∆T increases, hence Y decreases.

Q4. A spring is stretched by applying a load to its free end. The strain produced in the spring is
(a) volumetric
(b) shear
(c) longitudinal and shear
(d) longitudinal

Sol: (c) According to the diagram where a spring is suspended with a fixed rigid support. Now a load is attached with the lower end of that spring. So, it is stretched by applying a load to its free end. Clearly the length and shape of the spring changes and the weight of the load behaves as a deforming force.
The change in length corresponds to longitudinal strain and change in shape corresponds to shearing strain.

Q5. A rigid bar of mass M is supported symmetrically by three wires each of length /. Those at each end are of copper and the middle one is of iron. The ratio of their diameters, if each is to have the same tension, is equal to

Sol: (b) As the bar is supported symmetrically by the three wires, therefore extension in each wire is same.
Let T be the tension in each wire and diameter of the wire is D, then Young’s modulus is

Q6. A mild steel wire of length 2L and cross-sectional area A is stretched, well within elastic limit, horizontally between two pillars (figure ). A mass m is suspended from the mid-point of the wire. Strain in the wire is

Q7. A rectangular frame is to be suspended symmetrically by two strings of equal length on two supports (figure). It can be done in one of the following three ways:

The tension in the strings will be
(a) the same in all cases
(b) least in (i)
(c) least in (ii)
(d) least in (iii)
Sol:(c) According to the FBD diagram of the rectangular frame. Let M be the mass of rectangular frame and 0be the angle which the tension T in the string makes with the horizontal



Q8. Consider two cylindrical rods of identical dimensions, one of rubber and the other of steel. Both the rods are fixed rigidly at one end to the roof. A mass M is attached to each of the free ends at the centre of the rods.
(a) Both the rods will elongate but there shall be no perceptible change in shape.
(b) The steel rod will elongate and change shape but the rubber rod will only elongate.
(c) The steel rod will elongate without any perceptible change in shape, but the rubber rod will elongate and the shape of the bottom edge will change to an ellipse.
(d) The steel rod will elongate, without any perceptible change in shape, but the rubber rod will elongate with the shape of the bottom edge tapered to a tip at the centre.
Sol: (d)

According to the diagram shown below in which a mass AT is attached at the centre of each rod, then both rods will be elongated. But due to different elastic properties of material the steel rod will elongate without making any perceptible change in shape, but the rubber rod will elongate with the shape of the bottom edge tapered to a tip at the centre.

More Than One Correct Answer Type
Q9. The stress-strain graphs for two materials are shown in figure(assume same scale).

(a) Material (ii) is more elastic than material (i) and hence material (ii) is more brittle
(b) Materials (i) and (ii) have the same elasticity and the same brittleness.
(c) Material (ii) is elastic over a larger region of strain as compared to (i).
(d) Material (ii) is more brittle than material (i).
Sol: (c, d)
Key concept: Representation of different points on Stress-Strain graph:

When the strain is small (< 2%) (i.e., in region OP) stress is proportional to strain. This is the region where the so called Hooke’s law is obeyed. The point P is called limit of proportionality and slope of line OP gives the Young’s modulus Y of the material of the wire. If 6is the angle of OP from strain axis, then Y= tan θ.
If the strain is increased a little bit, i.e. in the region PE, the stress is not proportional to strain. However, the wire still regains its original length after the removal of stretching force.
This behaviour is shown up to point E known as elastic limit or yield- point. The region OPE represents the elastic behaviour of the material of wire. ‘
” If the wire is stretched beyond the elastic limit E, i.e. between EA, the strain increases much more rapidly and if the stretching force is removed the wire does not come back to its natural length. Some permanent increase in length takes place.
If the stress is increased further by a very small increase in it a very large increase in strain is produced (region AB) and after reaching point B, the strain increases even if the wire is unloaded and ruptures at C. In the region BC the wire literally flows. The maximum stress corresponding to B after which the wire begins to flow and breaks is called breaking or tensile strength. The region EABC represents the plastic behaviour of the material of wire.
Material having more ultimate tensile strength will be elastic over larger region.
So, from graph (i) linear limit vanishes soon and for small stress there is large strain as compared to graph (ii).
Hence, material (ii) is more elastic over a large region of strain as compared to (i). So, the ultimate tensile strength for material (ii) is greater than (i).
A material is said to be more brittle if its fracture point is more closer to ultimate strength point.
Hence material (ii) is more brittle than material (i).

Q10. A wire is suspended from the ceiling and stretched under the action of a weight F suspended from its other end. The force exerted by the ceiling on it is equal and opposite to the weight.
(a) Tensile stress at any cross-section A of the wire is F/A.
(b) Tensile stress at any cross-section is zero.
(c) Tensile stress at any cross-section A of the wire is 2F/A.
(d) Tension at any cross-section A of the wire is F.
Sol. (a, d)

According to the diagram a wire is stretched under the j
action of a weight F suspended from its other end.
Clearly, forces at each cross-section is F.
Now applying formula,

Tensile stress = Force applied/Area of cross-section = F/A
Tension = Applied force = F

 

Q11. A rod of length l and negligible  mass is suspended at its two ends by two wires of steel (wire A) and  aluminium (wire B) of equal lengths (figure).The cross-sectional areas of wires A and B are 1.0 mm² and 2.0 mm², respectively.(Y_Al= 70 x 10⁹ Nm^-2 and T_steel = 200 x 10⁹Nm^-2)

• Mass m should be suspended close to wire A to have equal stresses in both the wires.
• Mass m should be suspended close to B to have equal stresses in both the wires.
• Mass m should be suspended at the middle of the wires to have equal stresses in both the wires.
• Mass m should be suspended close to wire A to have equal strain in both wires.

Sol: (b, d) According to the diagram a massless rod is suspended at its two ends by two wires of steel (wire A) and aluminum (wire B) of equal lengths.

Q12. For an ideal liquid,
(a) the bulk modulus is infinite
(b) the bulk modulus is zero
(c) the shear modulus is infinite
(d) the shear modulus is zero

Q13. A copper and a steel wire of the same diameter are connected end to end. A deforming force F is applied to this composite wire which causes a total elongation of 1 cm. The two wires will have
(a) the same stress
(b) different stress
(c) the same strain
(d) different strain

Very Short Answer Type Questions
Q14. The Young’s modulus for steel is much more than that for rubber. For the same longitudinal strain, which one will have greater tensile stress?

Q15. Is stress a vector quantity?
Sol: Stress = Magnitude of internal reaction force / Area of cross-section
Hence, stress is not a scalar quantity not a vector quantity, it is a tensor quantity.

Q16. Identical springs of steel and copper are equally stretched. On which, more work will have to be done?
Sol: Key concept: Work Done in stretching a Wire or Spring:
In stretching a wire work is done against internal restoring forces. This work is stored in the wire as elastic potential energy or strain energy.
If a force F acts along the length L of the wire of cross-section A and stretches it by x, then
Q17. What is the Young’s modulus for a perfect rigid body?
Sol: According to Hooke’s law

Q18. What is the Bulk modulus for a perfect rigid body?
Sol: Bulk modulus is given by

For perfectly rigid body, change in volume AV = 0, therefore volumetric strain is zero.

Hence, bulk modulus for a perfectly rigid body is infinity

Short Answer Type Questions
Q19. A wire of length L and radius r is clamped rigidly at one end. When the other end of the wire is pulled by a force f its length increases by l. Another wire of the same material of length 2L and radius 2r, is pulled by a force 2 f Find the increase in length of this wire.
Sol: We have to apply Hooke’s law to compare the extension in each wire. According to the diagram which shows the situation.
Now, Young’s modulus

Solu:

Q21. To what depth must a rubber ball be taken in deep sea so that its volume is decreased by 0.1%. (The bulk modulus of rubber is 9.8 x 10⁸ N/m²; and the density of sea water is 10³ kg/m³)
Sol: According to the problem, Bulk modulus of rubber (B) = 9.8 x 10⁸ N/m² Density of sea water (p) = 10³ kg/m³ Percentage decrease in volume

Q22. A truck is pulling a car out of a ditch by means of a steel cable that is 9.1 m long and has a radius of 5 mm. When the car just begins to move, the tension in the cable is 800 N. How much has the cable stretched? (Young’s modulus for steel is 2 x 10¹¹ N/m²)

Sol. According to the problem,
Length of steel cable l = 9.1 m
Radius r = 5 mm = 5 x 10~³ m
Tension in the cable F = 800 N
Young’s modulus for steel Y= 2 x 10¹¹ N/m²
Change in length ∆l = ?

Q23. Two identical solid balls, one of ivory and the other of wet-clay, are dropped from the same height on the floor. Which one will rise to a greater height after striking the floor and why?
Sol: Since, ivory ball is more elastic than wet-clay ball, therefore, it tends to regain its original shape quickly. Due to this reason, more energy and momentum is transferred to the ivory ball in comparison to the wet-clay ball and hence, ivory ball will rise higher after striking the floor even though both are dropped from the same height.

Long Answer Type Questions
Q24. Consider a long steel bar under a tensile stress due to force F acting at the edges along the length of the bar (figure). Consider a plane making an angle 8 with the length. What are the tensile and shearing stresses on this plane?
(a) For what angle is the tensile stress a maximum?
(b) For what angle is the shearing stress a maximum?

Sol: According to the problem force F is applied along horizontal, so we resolve it in two perpendicular components—one is parallel to the inclined plane and other one is perpendicular to the inclined plane as shown in the diagram. Now, we can easily calculate the tensile and shearing stress. Here,

Important point: Here we are not applying direct formula for stress. Because to analyze different types of stresses, we have to divide the force in components. In normal stress, the force is applied normal to the surface. But in shear stress deforming force is applied tangential to one of the faces

Q25. A steel rod of length 21, cross-sectional area A and mass M is set rotating in a horizontal plane about an axis passing through the centre. If Y is the Young’s modulus for steel, find the extension in the length of the rod. (Assume the rod is uniform)

Q26.An equilateral triangle ABC is formed by two Cu rods AB and BC and one At It is heated in such a way that temperature of each rod increases by ∆T. Find change in the angle ABC. [Coefficient of linear expansion for Cu is α₁ coefficient of linear expansion for Al is α₂]

Sol: As the temperature of the rods increases length of each side will change, hence the angle corresponding to any vertex also changes as shown in the diagram.

Before heating, AB = BC = CA = l

After heating temperature of each rod is changed by ∆T and sides of ∆ABC is changed.

Let AB = l₁,BC = l₃,CA = l₂ Using cosine formula,


Q27. In nature, the failure of structural members usually result from large torque because of twisting or bending rather than due to tensile or compressive strains. This process of structural breakdown is called buckling and in cases of tall cylindrical structures like trees, the torque is caused by its own weight bending the structure. Thus, the vertical through the centre of gravity does not fall within the base. The elastic torque caused because of this bending about the central axis of the tree is given by YπR⁴/  4R. Y is the Young’s modulus, r is the radius of the trunk and is the radius of curvature of the bent surface along the height of the tree containing the centre of gravity (the neutral surface). Estimate the critical height of a tree for a given radius of the trunk.
Sol: According to the problem, the elastic torque or the bending torque is given and we have to find the torque caused by the weight due to bending.
The diagram of the given situation is as shown.
The bending torque on the trunk of radius r of tree = YπR⁴/  4R
where R the radius of curvature of the bent surface.



Q28. A stone of mass m is tied to an elastic string of negligible mass and spring constant k. The unstretched length of the string is L and has negligible mass. The other end of the string is fixed to a nail at a point P. Initially the stone is at the same level as the point P. The stone is dropped vertically from point P.
(a) Find the distance y from the top when the mass comes to rest for an instant, for the first time.
(b) What is the maximum velocity attained by the stone in this drop?
(c) What shall be the nature of the motion after the stone has reached its lowest point?
Sol:In this problem, the given string is elastic. Consider the diagram the stone is dropped from point P.
(a) When the stone is dropped, then it covers distance Y before coming to rest, for the first instant.

Y=L + (Y-L)
First it covers the distance L equal to length of string distance in free fall and a further distance (Y – L) due to extension in the string. So it covers a total distance Y until it instantaneously comes to rest at Q.
We have to find y, so by applying energy conservation principle,
Loss in potential energy of stone = Gain in elastic potential energy in string

(b) In SHM, the maximum velocity is attained when the body passes, through the “equilibrium, position”, i.e., when the instantaneous acceleration is zero. That is mg -kx = 0, where .r is the extension from L.

(c) When stone is at the lowest point Q, i.e. at instantaneous distance Y from P from where the stone is dropped, then equation of motion of the stone is
mass x acceleration = net force on the stone

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