Category: Class 11 Physics

NCERT Exemplar Class 11 Physics Chapter 9 Mechanical Properties of Fluids are part of NCERT Exemplar Class 11 Physics. Here we have given NCERT Exemplar Class 11 Physics Chapter 9 Mechanical Properties of Fluids.

NCERT Exemplar Class 11 Physics Chapter 9 Mechanical Properties of Fluids

Q1. A tall cylinder is filled with viscous oil. A round pebble is dropped from the top with zero initial velocity. From the plot shown in figure, indicate the one that represents the velocity (v) of the pebble as a function of time (t).

Sol: (c) In fluids, when the pebble is dropped from the top of a tall cylinder filled with viscous oil, a variable force called viscous force will act which increases with increase in speed. And at equilibrium this velocity becomes constant, that constant velocity is called terminal velocity.
When the pebble is falling through the viscous oil, the viscous force is F= 6πηrv
where r is the radius of the pebble, v is instantaneous speed, η is coefficient of viscosity. As the force is variable, hence acceleration is also variable so v-t graph will not be a straight line. First velocity increases and then becomes constant known as terminal velocity.

Problems On Mechanical Properties Of Fluids NCERT Class 11

Q2. Which of the following diagrams does not represent a streamline flow?

Sol. (d)
Streamline flow: Streamline flow of a liquid is that flow in which each element of the liquid passing through a point travels along the same path and with the same velocity as the preceding element passes through that point.
A streamline may be defined as the path, straight or curved, the tangent to which at any point gives the direction of the flow of liquid at that point.
The two streamlines cannot cross each other and the greater is the crowding of streamlines at a place, the greater is the velocity of liquid particles at that place. If we consider a cross¬sectional area, then a point on the area cannot have different velocities at the same time
Path ABC is streamline as shown in the figure and v1, v2 and v3 are the velocities of the liquid particles at A, B and C point respectively.

Q3. Along a streamline,
(a) the velocity of a fluid particle remains constant
(b) the velocity of all fluid particles crossing a given position is constant
(c) the velocity of all fluid particles at a given instant is constant
(d) ‘ the speed of a fluid particle remains constant,
Sol:(b) As discussed above for a streamline flow of a liquid velocity of each particle at a particular cross-section is constant, because Av = constant (law of continuity) between two cross-section of a tube of flow. So we can say that along a streamline, the velocity of every fluid particle while crossing a given position is the same.

Mechanical Properties Of Fluids Ncert Exemplar Class 11

Q4. An ideal fluid flows through a pipe of circular cross-section made of two sections with diameters 2.5 cm and 3.75 cm. The ratio of the velocities in the two pipes is
(a) 9:4 (b) 3:2 (c)√3: √2  (d)√2:√3

Sol: (a) The situation is shown in the diagram below in which an ideal fluid is flowing through a pipe of circular cross sections.

Numericals On Viscosity Class 11 NCERT

Q5. The angle of contact at the interface of water-glass is 0°, ethylalcohol-glass is 0°, mercury-glass is 140° and methyliodide-glass is 30°. A glass capillary is put in a trough containing one of these four liquids. It is observed that the meniscus is convex. The liquid in the trough is
(a) water
(b) ethylalcohol
(c) mercury
(d) methyliodide

According to the question, the observed meniscus of liquid in a capillary tube is of convex upward which is only possible when angle of contact is obtuse. It is so when one end of glass capillary tube is immersed in a trough of mercury. Hence, the combination will be of mercury-glass (140°) as shown in the figure.

More Than One Correct Answer Type
Q6. For a surface molecule,
(a) the net force on it is zero
(b) there is a net downward force
(c) the potential energy is less than that of a molecule inside
(d) the potential energy is more than that of a molecule inside
Sol: (b, d)

Key concept: To understand the concept of tension acting on the free
surface of a liquid, let us consider four liquid molecules like A, B, C and D. Their sphere of influence are shown in the figure.
(1) Molecule A is well within the liquid, so it is attracted equally in all directions. Hence the net force on this molecule is zero and it moves freely inside the liquid.
(2) Molecule B is little below the free surface of the liquid and it is atso attracted equally in all directions. Hence the resultant force acts on it is also zero
(3) Molecule C is just below the upper surface of the liquid film and the part of its sphere of influence is outside the free liquid surface. So the number of molecules in the upper half (attracting the molecules upward) is less than the number of molecule in the lower half (attracting the molecule downward). Thus the molecule C experiences a net downward force.
(4) Molecule D is just on the free surface of the liquid. The upper half of the sphere of influence has no liquid molecule. Hence the molecule D experiences a maximum downward force.
Thus all molecules lying on surface film experiences a net downward force. Therefore, free surface of the liquid behaves like a stretched membrane.
From the key concept, it is clear from point (4) that molecules on the surface experiences a net downward force. As shown in figure above, molecule D experiences a net downward force. Because on the above side of this molecule there is no liquid molecule. So, the potential energy is more than that of a molecule inside. Hence option (b) and (d) are correct.
Q7. Pressure is a scalar quantity, because
(a) it is the ratio of force to area and both force and area are vectors.
(b) it is the ratio of the magnitude of the force to area.
(c) it is the ratio of the component of the force normal to the area.
(d) ‘it does not depend on the size of the area chosen.
Sol: (b, c) Pressure is defined as the ratio of magnitude of component of the force normal to the area and the area under consideration.

i.e P = F/A
Pressure is a scalar quantity. Pressure acts normal to a surface and it is always compressive in nature, therefore, only its magnitude is required for its complete description.

Q8. A wooden block with a coin placed on its top, floats in water as shown in figure.
The distance / and h are shown in the figure. After sometime, the coin falls into the water. Then,
(a) l decreases
(b) h decreases
(c) l increases
(d) h increases
Sol: (a, b)
Key concept: When a body of density ρ and volume V is immersed in a liquid of density σ, the forces acting on the body are:
1. Weight of body W = mg = Vpg, acting vertically downwards through centre of gravity of the body

When coin is in water, volume of water displaced by coin is equal to the volume of coin V₁(say).
When the coin falls into the water, weight of the (block + coin) system decreases, which was balanced by the upthrust force earlier. As weight of the system decreases, block moves up. Hence ldecreases.
When coin is at the top of wooden block, it displaces a volume of water V₂, which is more than V₁ Because,
Weight of coin = Weight of volume of water displaced by coin (when coin is at .the top of wooden block)

Hence, upthrust force will also decrease. As volume of water displaced by the block decreases, hence h decreases.

Q9. With increase in temperature, the viscosity of
(a) gases decreases
(b) liquids increases
(c) gases increases
(d) liquids decreases
Sol: (c, d) The viscosity of gases increases with increase of temperature, because on increasing temperature the rate of diffusion increases.
The viscosity of liquid decreases with increase of temperature, because the cohesive force between the liquid molecules decreases with increase of temperature.
Relation between coefficient of viscosity and temperature (Andrade formula)

where T = Absolute temperature of liquid, p = density of liquid, A and C are constants.
Important point: With increase in temperature, the coefficient of viscosity of liquids decreases but that of gases increases. The reason is that as temperature rises, the atoms of the liquid become more mobile, whereas in case of a gas, the collision frequency of atoms increases as their motion becomes more random.

Q10. Streamline flow is more likely for liquids with
(a) high density (b) high viscosity
(c) low density (d) low viscosity
Sol: (b, c) Streamline flow is more likely for liquids having low density. We know that greater the coefficient of viscosity of a liquid more will be the velocity gradient, hence each line of flow can be easily differentiated. Streamline flow is related with critical velocity. The critical velocity is that velocity of liquid flow up to which its flow is streamlined and above which its flow becomes turbulent.
As the critical velocity is related to viscosity ( η) and density (ρ) of the liquid as:

(V_c) α η/ρ

Hence if the density will be low and viscosity will be high, the value of critical velocity will be more. So, option (b) and (c) are correct.

Very Short Answer Type Questions
Q11. Is viscosity a vector?
Sol: Viscosity is not a vector quantity. It is a scalar quantity because viscosity is a property of liquid as it does not have any direction.

Q12. Is surface tension a vector?
Sol: Surface tension of a liquid is measured by the force acting per unit length on either side of an imaginary line drawn on the free surface of liquid, the direction of this force being perpendicular to the line and tangential to the free surface of liquid. So if F is the force acting on one side of imaginary line of length L, then T= (F/L)
It depends only on the nature of liquid and is independent of the area of surface or length of line considered.
It is a scalar quantity as it has a unique direction which is not to be specified

Q14. A vessel filled with water is kept on a weighing pan and the scale adjusted to zero. A block of mass Mand density p is suspended by a massless spring of spring constants k. This block is submerged inside into the water in the vessel. What is the reading of the scale?
Sol: As shown in the diagram, A block of mass M and density p is suspended by a massless spring of spring constants k. This block is submerged into the water in the vessel.
The scale is adjusted to zero, therefore, when the block suspended to a spring is immersed in water, then the reading of the scale will be equal to the upthrust on the block due to water.

Important point: If the scale is not adjusted to zero as said earlier, reading on the scale will be different. Then weight of vessel and weight of water is also added in the reading.

Q15. A cubical block of density ρ is floating on the surface of water. Out of its height L, fraction x is submerged in water. The vessel is in an elevator accelerating upward with acceleration a. What is the fraction immersed?
Sol: Key concept: When a fluid is subjected to constant vertical acceleration, its free surface remains horizontal as the net effective gravity is acting vertically. But the magnitude of pressure at a point in the fluid increases or decreases depending upon the direction of acceleration.

Short Answer Type Questions

Q16. The sap in trees, which consists mainly of water in summer, rises in a system of capillaries of radius r = 2.5 x 10^-5 m. The surface-tension of sap is T = 7.28 x 10^-2 Nm^-1 and the angle of contact is 0°. Does surface tension alone account for the supply of water to the top of all trees?
Sol: According to the problem, radius (r) = 2.5 x 10^-5 m Surface tension (S) = 7.28 x 10^-2 N/m
Angle of contact (θ) = 0°

This is the maximum height to which the sap can rise due to surface tension. Many trees have heights much greater than 0.6 m, so only this action is not sufficient for supply of water to the top of such long tree.

Q17. The free surface of oil in a tanker, at rest, is horizontal. If the tanker starts accelerating the free surface will be titled by an angle θ. If the acceleration is a ms ^-2, what will be the slope of the free surface?
Sol: Key concept: The behaviour of a liquid contained in a horizontally accelerated vessel can be understood by understanding the behaviour of a pendulum suspended from the ceiling of a horizontally accelerated trolley.

Every fluid element attains an equilibrium position under the action of gravity and pseudo force. The free surface of the liquid orients itself perpendicular to the direction of net effective gravity.
tan θ = a/g
Suppose tanker accelerates along x-axis with acceleration a, free surface of the tanker will not be horizontal because pseudo force acts as shown in the diagram.

Consider an elementary particle of the oil of mass m.
The acting forces on the particle with respect to the tanker are shown in the figure alongside.
Now, balancing forces (as the particle is in equilibrium) along the inclined direction of surface.
ma = pseudo force
mg = weight of small part of oil.
Along free surface,
Net force = 0

=> ma cos θ = mg sin θ
=> a = g tan θ
=>  θ = tan^-1(a/g)

Q18.Two mercury droplets of radii 0.1 cm and 0.2 cm collapse into one single drop. What amount of energy is released? The surface tension of mercury T=5 x 10^-3 Nm^_1
Sol: When two drops form a bigger drop, volume remains conserved.
According to the problem, there is two mercury droplets of different radii collapse into one single drop.
Radius of smaller drop = r, = 0.1 cm = 10^-3 m,
Radius of bigger drop = r₂ = 0.2 cm = 2 x 10^-3 m
Surface tension (7) = 435.5 x 10 ^-3 N/m
Let V₁ and V₂ be the volumes of these two mercury droplets and volume of big drop formed by collapsing is V.
Volume of big drop = Volume of small droplets

Therefore, 3.22 x K^-6 J energy will be absorbed. So, the surface area of the water decreases means surface area of bigger drop is less than the sum of surface area of two smaller drops.

Q19. If a drop of liquid breaks into smaller droplets, it results in lowering of temperature of the droplets. Let a drop of radius R, breaks into N small droplets each of radius r. Estimate the drop in temperature.
Sol: The volume remains conserved, when a big drop, breaks into N small droplets.
Volume of liquid drop of radius R = (Volume of liquid droplet of radius r)xN

Due to releasing of this energy, the temperature is lowered.
If c is specific heat of liquid and its temperature is lowered by ΔT, then Energy released, ΔU – me ΔT

Q20. The surface tension and vapour pressure of water at 20°C is 7.28 x 10^-2 Nm^-1 and 2.33 x 10³ Pa, respectively. What is the radius of the smallest spherical water droplet which can form without evaporating at 20°C?
Sol:According to the problem, surface tension of water, T = 7.28 x 10^-2 N m^-1 Vapour pressure P = 2.33 x 10³ Pa Let r = radius of drop, which formed without evaporating.
The excess pressure (2T/r) should be greater than the vapour pressure. Then, the drop will evaporate.
Vapour pressure = Excess pressure in drop

Long Answer Type Questions

Q22. Surface tension is exhibited by liquids due to force of attraction between molecules of the liquid. The surface tension decreases with increase in temperature and vanishes at boiling point. Given that the latent heat of vaporisation for water L_v = 540 kcal kg^-1, the mechanical equivalent of heat J = 4.2 Jeal^-1, density of water p_w = 10³ kg^-1, Avogadro’s number
N_a = 6.0 xlO²⁶ kmole‘¹ and the molecular weight of water M_A = 10 kg for 1 kmole.

(a) Estimate the energy required for one molecule of water to evaporate

(c) 1 g of water in the vapour state at 1 atm occupies 1601 cm³. Estimate the inter-molecular distance at boiling point, in the vapour state.
(d) During vaporisation a molecule overcomes a force F, assumed constant, to go from an inter-molecular distance d to Estimate the value of F.
(e) Calculate Fid, which is a measure of the surface tension.

Q23. A hot air balloon is a sphere of radius 8 m. The air inside is at a temperature of 60°C. How large a mass can the balloon lift when the outside temperature is20°C? Assume air in an ideal gas, R = 8.314 J mole^-1 K ‘, 1 atm = 1.013 x 10^s Pa, the membrane tension is 5 Nm ^-1.

Sol: Pressure inside the curved surface Will be greater than of outside pressure

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NCERT Exemplar  Class 11 Physics Chapter 10 Thermal Properties of Matter are part of NCERT Exemplar Class 11 Physics. Here we have given NCERT Exemplar Class 11 Physics Chapter 10 Thermal Properties of Matter.

NCERT Exemplar Class 11 Physics Chapter 10 Thermal Properties of Matter

Q1. A bimetallic strip is made of aluminium and steel (aAI > astee|) On heating, the strip will
(a) remain straight (bj get twisted
(c) will bend with aluminium on concave side.
(d) will bend with steel on concave side
Sol: (d)
Key concept: Bi-metallic strip-. Two strips of equal lengths but of different materials (different coefficient of linear expansion) when join together, it is called “bi-metallic strip”, and can be used in thermostat to break or make electrical contact. This strip has the characteristic property of bending on heating due to unequal linear expansion of the two metal. The strip will bend with metal of greater a on outer side, i.e. convex side.

On heating, the metallic strip with higher coefficient of linear expansion (∝_Al) will expand more.
According to the question, ∝_AI > ∝_steel, so aluminum will expand more. So, it should have larger radius of curvature. Hence, aluminium will be on convex side. The metal of smaller ∝ (i.e., steel) bends on inner side, i.e., concave side.

Q2. A uniform metallic rod rotates about its perpendicular bisector with constant angular speed. If it is heated uniformly to raise its temperature slightly
(a) its speed of rotation increases
(b) its speed of rotation decreases
(c) its speed of rotation remains same
(d) its speed increases because its moment of inertia increases
Sol: (b) When the rod is heated uniformly to raise its temperature slightly, it expands. So, moment of inertia of the rod will increase.
Moment of inertia of a uniform rod about its perpendicular bisector

If the temperature increases, moment of inertia will increase.
No external torque is acting on the system, so angular momentum should be conserved.

Q3. The graph between two temperature scales A and B is shown in figure between upper fixed point and lower fixed point there are 150 equal division on scale A and 100 on scale B. The relationship for conversion between the two scales is given by

Sol: Key concept: Temperature on one scale can be converted into other scale by using the following identity.
Reading on any scale – LFP /UFP – LFP = Constant for all scales
where, LFP —> Lower fixed point
UFP —>Upper fixed point
From the graph it is clear that the lowest point for scale A is 30° and highest point for the scale A is 180°.
Lowest point for scale B is 0° and highest point for scale B is 100°. Hence, the relation between the two scales A and B is given by

Q4. An aluminium sphere is dipped into water.
Which of the following is true?
(a) Buoyancy will be less in water at 0°C than that in water at 4°C.
(b) Buoyancy will be more in water at 0°C than that in water at 4°C.
(c) Buoyancy in water at 0°C will be same as that in water at 4°C.
(d) Buoyancy may be more or less in water at 4°C depending on the radius of the sphere.
Sol: (a)
Key concept: Liquids generally increase in volume with increasing temperature but in case of water, it expands on heating if its temperature is greater than 4°C. The density of water reaches a maximum value of 1.000 g/cm3 at 4°C.
This behaviour of water in the range from 0°C to 4°C is called anomalous expansion.

Q5. As the temperature is increased, the period of a pendulum
(a) increases as its effective length increases even though its centre of mass still remains at the centre of the bob
(b) decreases as its effective length increases even though its centre of mass ‘ still remains at the centre of the bob
(c) increases as its effective length increases due to shifting to centre of mass below the centre of the bob
(d) decreases as its effective length remains same but the centre of mass shifts above the centre of the bob

Sol: (a) A pendulum clock keeps proper time at temperature θ₀. If temperature is increased to θ (>θ ₀), then due to linear expansion, length of pendulum increases and hence its time period will increase

So, as the temperature increases, length of pendulum increases and hence time period of pendulum increases. Due to increment in its time period, a pendulum clock becomes slow in summer and will lose time.

Q6. Heat is associated with
(a) kinetic energy of random motion of molecules
(b) kinetic energy of orderly motion of molecules
(c) total kinetic energy of random and orderly motion of molecules
(d) kinetic energy of random motion in some cases and kinetic energy of orderly motion in other
Sol: (a) When a body is heated its temperature rises and in liquids and gases vibration of molecules about their mean position increases, hence kinetic energy associated with random motion of molecules increases.
So, thermal energy or heat associated with the random and translatory motions of molecules.

Q7. The radius of a metal sphere at room temperature Tis Rand the coefficient of linear expansion of the metal is .The sphere heated a little by a temperature ∆Tso that its new temperature is T + ∆T.The increase in the volume of the sphere is approximately.


ρ = mass / volume

So, the volume of all object will also be same.
Here the cooling will be done in the form of radiations that is according to Stefan’s law. Since, emissive power is directly proportional to the surface. Here, for given volume, sphere has least surface area and circular plate of greatest surface area.

As thickness of the plate is least, hence surface area of the plate is maximum. According to Stefan’s law, heat loss (cooling) is directly proportional to the surface area.

H_sphere : H_cube: H_plate  = A_sphere : A_cube: A_plate
As A_plate is maximum, hence the plate will cool fastest.
As the sphere is having minimum surface area, hence the sphere cools slowest.

More Than One Correct Answer Type
Q9. Mark the correct options.
(a) A system X is in thermal equilibrium with Y but not with Z. The systems Y and Z may be in thermal equilibrium with each other.
(b) A system X is in thermal equilibrium with Y but not with Z. The systems Y and Z are not in thermal equilibrium with each other.
(c) A system X is neither in thermal equilibrium with Y nor with Z. The systems Y and Z must be in thermal equilibrium with each other.
(d) A system X is neither in thermal equilibrium with Y nor with Z. The systems Y-and Z may be in thermal equilibrium with each other.
Sol: (b, d)
Key concept:Two bodies are said to be in thermal equilibrium with each other, when no heat flows from one body to the other. That is when both the bodies are at the same temperature.

Q10. Gulab jamuns (assumed to be spherical) are to be heated in an oven. They are available in two sizes, one twice bigger (in radius) than the other. Pizzas (assumed to be discs) are also to be heated in oven. They are also in two sizes, one twice bigger (in radius) than the other. All four are put together to be heated to oven temperature. Choose the correct option from the following.
(a) Both size gulab jamuns will get heated in the same time
(b) Smaller gulab jamuns are heated before bigger ones
(c) Smaller pizzas are heated before bigger ones
(d) Bigger pizzas are heated before smaller
Sol: (b, c) Between these four which has the least surface area will be heated first because of less heat radiation. So, smaller gulab jamuns are having least surface area, hence they will be heated first.
Similarly, smaller pizzas are heated before bigger ones because they are of small surface areas.

Q11. Refer to the plot of temperature versus time (figure) showing the changes in the state if ice on heating (not to scale). Which of the following is correct?
(a) The region AB represents ice and water in thermal equilibrium
(b) At B water starts boiling
(c) At C all the water gets converted into steam
(d) C to D represents water and steam in equilibrium at boiling point

Sol: (a, d) During phase change process, temperature of the system remains constant.
(a) In region AB, a phase change takes place, heat is supplied and ice melts but temperature of the system is 0°C. it remains constant during process. The heat supplied is used to break bonding between molecules.
(b) In region CD, again a phase change takes place from a liquid to a vapour state during which temperature remains constant. It shows water and steam are in equilibrium at boiling point.

Q12. A glass full of hot milk is poured on the table. It begins to cool gradually. Which of the following is correct?
(a) The rate of cooling is constant till milk attains the temperature of the surrounding.
(b) The temperature of milk falls off exponentially with time.
(c) While cooling, there is a flow of heat from milk to the surrounding as well as from surrounding to the milk but the net flow of heat is from milk to the surrounding and that is why it cools.
(d) All three phenomenon, conduction, convection and radiation are responsible for the loss of heat from milk to the surroundings.
Sol: (b, c, d) When hot milk spread on the table heat is transferred to the surroundings by conduction, convection and radiation. Because the surface area of poured milk on a table is more than the surface area of milk filled in a glass. Hence, its temperature falls off exponentially according to Newton’s law of cooling. Heat also will be transferred from surroundings to the milk but will be lesser than that of transferred from milk to the surroundings. So, option (b), (c) and (d) are correct.

Very Short Answer Type Questions
Q13. Is the bulb of a thermometer made of diathermic or adiabatic wall?
Sol: The bulb of a thermometer is made up of diathermic wall because diathermic walls allow exchange of heat energy between two systems but adiabatic walls do not. So it receives heat from the body to measure the temperature of body.

Q14. A student records the initial length l , change in temperature ∆ T and change in length ∆ l of a rod as follows:

S. No	l(m)	∆T(°C)	∆l(m)
1.	2	10	4 x 10-4
2.	1	10	4 x 10-4
3.	2	20	2 x 10-4
4.	3	10	6x 10-4

If the first observation is correct, what can you say about observations 2, 3 and 4.

Sol: If the first observation is correct, hence from the 1st observation we get the coefficient of linear expansion

For 4th observation, ∆l = ∝l∆T

= 2 x 10^-5 x 3 x 10 = 6 x 10^-4 m [i.e., observed value is correct]

Q15. Why does a metal bar appear hotter than a wooden bar at the same temperature? Equivalently it also appears cooler than wooden bar if they are both colder than room temperature.

Q16. Calculate the temperature which has numeral value on Celsius and Fahrenheit scale.
Sol: To construct a scale of temperature, two fixed points are taken. First fixed point is the freezing point of water, it is called lower fixed point. The second fixed point is the boiling point of water, it is called upper fixed point. Temperature on one scale can be converted into other scale by using the following identity.

Q17. These days people use steel utensils with copper bottom. This is supposed to be good for uniform heating of food. Explain this effect using the fact that copper is the better conductor. Junction

Sol: The copper bottom of the steel utensil gets heated quickly.
Because of the reason that copper is a good conductor of heat as compared to steel. But steel does not conduct as quickly, thereby allowing food inside to get heated uniformly.

Short Answer Type Questions
Q18. Find out the increase in moment of inertia I of a uniform rod (coefficient of linear expansion a) about its perpendicular bisector when its temperature is slightly increased by ∆T.
Sol: Moment of inertia of a uniform rod of mass M and length l about its perpendicular bisector

Q19. During summers in India, one Of the common practice to keep cool is to make ice balls of crushed ice, dip it in flavored sugar syrup and sip it. For this a stick is inserted into crushed ice and is squeezed in the palm to make it into the ball. Equivalently in winter in those areas where it snows, people make snow balls and throw around. Explain the formation of ball out of crushed ice or snow in the light of p-T diagram of water.
Sol : Given diagram shows the variation of pressure with temperature for water. When the pressure is increased in solid state (at 0°, 1 atm), ice changes into liquid state while decreasing pressure in liquid state (at 0°, 1 atm), water changes to ice.

When crushed ice is squeezed, some of it melts, filling up the gap between ice flakes upon releasing pressure. This water freezes, binding all ice flakes and making the ball more stable.

Q20. 100 g of water is super cooled to -10°C. At this point, due to some disturbance mechanised or otherwise some of it suddenly freezes to ice. What will be the temperature of the resultant mixture and how much mass would freeze?

 

Q21. One day in the morning Ramesh filled up 1/3 bucket of hot water from geyser, to take bath. Remaining 2/3 was to be filled by cold water (at room temperature) to bring mixture to a comfortable temperature. Suddenly Ramesh had to attend to something which would take some times, say 5-10 min before he could take bath. Now, he had two options (i) fill the remaining bucket completely by cold water and then attend to the work, (ii) first attend to the work and fill the remaining bucket just before taking bath. Which option do you think would have kept water warmer? Explain

Sol:According to the Newton’s law o‘f cooling, the rate of loss of heat is directly proportional to the difference of temperature. Or we can say which gives a consequence about rate of fall of temperature of a body with respect to the difference of temperature of body and surroundings.
The first option would have kept water warmer because by adding hot water to cold water, the temperature of the mixture decreases. Due to this temperature difference between the mixed water in the bucket and the surrounding decreases, thereby the decrease in the rate of loss of the heat by the water.
In second option, the hot water in the bucket will lose heat quickly. So if he first attend to the work and fill the remaining bucket with cold water which already lose much heat in 5-10 minutes then the water become more colder as comparison with first case.

Long Answer Type Questions
Q22. We would like to prepare a scale whose length does not change with temperature. It is proposed to prepare a unit scale of this type whose length remains, say 10 cm. We can use a bimetallic strip made of brass and iron each of different length whose length (both components) would change in such a way that difference between their length B remain constant. If α_iron = 1.2 x 10 ^-5/K and α_brass = 1.8xl0^-5/K, what should we take as length of each strip?

Sol: According to the problem, L₁-L_b = 10 cm where,
L₁ = length of iron scale
L_b = Length of brass scale
This condition is possible if change in length both the rods is remain same at all temperatures.

Q23. We would like to make a vessel whose volume does not change with temperature (take a hint from the problem above). We can use brass and iron (β_vbrass ⁼ 6 x 10 ⁵ / K and β_viron = 3.55 x 10^-5/K) to create a volume of 100 cc. How do you think you can achieve this?
Sol:Here we are making a vessel whose Brass volume does not change with temperature.
To make the desired vessel, we should have an iron vessel with a brass rod inside as shown in the diagram.

Therefore, an iron vessel with a volume of 249.9 cm³ fitted with a brass rod of volume 144.9 cm³ will serve as a vessel of volume 100 cm³, which will not change with temperature.
Important points:

• Solids can expand in one dimension (linear expansion), two dimensions (superficial expansion) and three dimensions (volume expansion) while liquids and gases usually suffers change in volume only.
• Thermal expansion is minimum in case of solids but maximum in case of gases because intermoleeular force is maximum in solids but minimum in gases.

Q24. Calculate the stress developed inside a tooth cavity filled with copper when hot tea at temperature of 57°C is drunk. You can take body (tooth) temperature to be 37°C and a = 1.7x 10^-5/°C bulk modulus for copper = 140x 10⁹N/m².

This is about 10³ times of atmospheric pressure

Q25. A rail track made of steel having length 10 m is clamped on a railway line at its two ends (figure). On a summer day due to rise in temperature by 20°C. It is deformed as shown in figure. Find x (displacement of the centre) if27

Q26. A thin rod. having length L₀ at 0°C and coefficient of linear expansion α has its two ends maintained at temperatures θ₁, and θ₂, Find its new length.
Sol. When temperature of a rod varies linearly, then average temperature of the middle point of the rod can be taken as mean of temperatures at the two ends. According to the diagram,

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We hope the NCERT Exemplar Class 11 Physics Chapter 10 Thermal Properties of Matter help you. If you have any query regarding NCERT Exemplar Class 11 Physics Chapter 10 Thermal Properties of Matter, drop a comment below and we will get back to you at the earliest.

NCERT Exemplar Class 11 Physics Chapter 13 Oscillations are part of NCERT Exemplar Class 11 Physics. Here we have given NCERT Exemplar Class 11 Physics Chapter 13 Oscillations.

NCERT Exemplar Class 11 Physics Chapter 13 Oscillations

Multiple Choice Questions
Single Correct Answer Type

Q1. The displacement of a particle is represented by the equation

. The motion of the particle is
(a) simple harmonic with period 2KUO
(b) simple hannonic with period nia)
(c) periodic but not simple harmonic
(d) non-periodic
Sol: (b)
Key concept:

Q2. The displacement of a particle is represented by the equation y= sin³ ωtThe motion is
(a) non-periodic
(b) periodic but not simple harmonic
(c) simple harmonic with period 2π/ω
(d) simple harmonic with period π/ω
Sol: (b)
Key concept: There are certain motions that are repeated at equal intervals of time. Let the the interval of time in which motion is repeated. Then x(t) =x(t + T), where T is the minimum change in time. The function that repeats itself is known as a periodic function.

Q3. The relation between acceleration and displacement of four particles are given below:
(a) a_x = +2x            (b) a_x = +2x²            (c) a_x = -2x²               (d) a_x = -2x
Which one of the particle is exempting simple harmonic motion?

Sol: (d)
Key Concept: In case of simple harmonic motion, the acceleration is always directed towards the mean position and so is always opposite to displacement, i.e. a α = -x or a = -ω²x
In option (d) a_x = -2x or a α -x, the acceleration of the particle is proportional to negative of displacement. Hence it represents S.H.M.

Q4. Motion of an oscillating liquid column in a U-tube is
(a) periodic but not simple harmonic
(b) non-periodic
(c) simple harmonic and time period is independent of the density of the  liquid
(d) simple harmonic and time period is directly proportional to the density of the liquid

Sol: (c)
Key Concept: If the liquid in U-tube is filled to a height h and cross¬section of the tube is uniform and the liquid is incompressible and non- viscous. Initially the level of liquid in the two limbs will be at the same height equal to h. If the liquid is-pressed by y in one limb, it will rise by y along the length of the tube in the other limb, so the restoring force will be developed by hydrostatic pressure difference

Q5. A particle is acted simultaneously by mutually perpendicular simple harmonic motion x = a cos ωt and y = a sin ωt. The trajectory of motion of the particle will be
(a) an ellipse (b) a parabola (c) a circle (d) a straight line
Sol: (c)
Key concept: If two S.H.M’s act in perpendicular directions, then their resultant motion is in the form of a straight line or a circle or a parabola etc. depending on the frequency ratio of the two S.H.Ms and initial phase difference. These figures are called Lissajous figures.
Let the equations of two mutually perpendicular S.H.M’s of same frequency be

Q6. The displacement of a particle varies with time according to the relation
y = a sint + b cost
(a) The motion is oscillatory but not SHM
(b) The motion is SHM with amplitude a + b
(c) The motion is SHM with amplitude a² + b²
(d) The motion is SHM with amplitude √a² + b<sup>2


Q7. Four pendulums A, B, C and D are suspended from the same elastic support as shown in figure. A and C are of the same length, while B is smaller than A and D is larger than A. If A is given a transverse displacement,
(a) D will vibrate with maximum amplitude
(b) C will vibrate with maximum amplitude
(c) B will vibrate with maximum amplitude
(d) All the four will oscillate with equal amplitude

Sol: (b) Here A is given a transverse displacement. Through the elastic support the disturbance is transferred to all the pendulums.
A and C are having same length, hence they will be in resonance, because of their time period of oscillation. Since length of pendulums A and C is same  and T =2 π√L/g , hence their time period is same and they will have
frequency of vibration. Due to it, a resonance will take place and the pendulum C will vibrate with maximum amplitude.</sup>

Q8. Figure shows the circular motion of a particle. The radius of the circle, the period, sense of revolution and the initial position are indicated on the figure. The simple harmonic motion of the x-projection of the radius vector of the rotating particle P is


Sol: (a)
Key concept: Suppose a particle P is moving uniformly on a circle of radius A with angular speed. Q and R are the two feets of the perpendicular drawn from P on two diameters one along .Y-axis and the other along Y-axis.

Suppose the particle P is on the X-axis at t = 0. Radius OP makes an angle with the X-axis at time t, then
x = A cosωt and y = A sinωt
Here, x and v are the displacements of Q and R from the origin at time t, which are the displacement equations of SHM. It implies that although P is under uniform circular motion, Q and R are performing SHM about O with the same angular speed as that of P.

Q9. The equation of motion of a particle is x = a cos(∝t)² . The motion
(a) periodic but not oscillatory
(b) periodic and oscillatory
(c) oscillatory but not periodic
(d) neither periodic nor oscillatory
Sol: (c) The equation of motion of a particle is
x = a cos(∝t)²
is a cosine function and x varies between -a and +a, the motion is oscillatory. Now checking for periodic motion, putting t+T in place of t. T is supposed as period of the function ω(t).

Q10. A particle executing SHM has a maximum speed of 30 cm/s and a maximum acceleration of 60 cm/s². The period of oscillation is

(a) sec
(b) /2 sec
(c) 2 sec
(d) /t

=

Q11. When a mass in is connected individually to two springs S₁ and S₂, the oscillation frequencies are V₁ and V₂. If the same mass is attached to the two springs as shown in figure, the oscillation frequency would be

More Than One Correct Answer Type
Q12. The rotation of earth about its axis is
(a) periodic motion
(b) simple harmonic motion
(c) periodic but not simple harmonic motion
(d) non-periodic motion
Sol: (a, c) Rotation of earth about its axis repeats its motion after a fixed interval of lime, so its motion is periodic.
The rotation of earth is obviously not a to and fro type of motion about a fixed point, hence its motion is not an oscillation. Also this motion does not follow S.H.M equation, a ∝ -x.
Hence, this motion is not a S.H.M.

Q13. Motion of a ball bearing inside a smooth curved bowl, when released from a point slightly above the lower point is
(a) simple harmonic motion
(b) non-periodic motion
(c) periodic motion
(d) periodic but not SHM
Sol: (a, c) For small angular displacement, the situation is shown in the figure. Only one restoring force creates .
motion in ball inside bowl.

Q14. Displacement versus time curve for a particle executing SHM is shown in figure. Choose the correct statements.
(a) Phase of the oscillator is same at t = 0s and t = 2 s
(b) Phase of the oscillator is same at t = 2s and t = 6 s
(c) Phase of the oscillator is same at t = 1s and t = 7 s
(d) Phase of the oscillator is same at t = 1s and t = 5 s

Sol: (b,d)
Key concept:
Phase: The physical quantity which represents the state of motion of particle (e.g. its position and direction of motion at any instant).
The time varying quantity (t + ) is called the phase of the motion, and the constant is called the phase constant (or phase angle). Phase determines the status of the particle at t = 0.
Suppose we choose t= 0, at an instant when the particle is passing through its mean position and is going towards the positive direction. The phase (t + )becomes zero.
=> = 0 and x= Asint and v = Acost
If we choose t = 0, at an instant when the particle is at its position extreme position, then is π/2 at that instant.
Thus t + = π/2 at t = 0 ⇒ = π/2, or x = A sin (t + π/2) – A cost

In option (a) at t =0 s and t = 2 s, the displacements are in opposite directions, hence phase of the oscillator is not same which makes option incorrect.

In option (b) it is clear from the curve that points corresponding to t = 2 s and t = 6 s are separated by a distance belonging to one time period. Hence, these points must be in same phase.
In option (c) t = 1 s and t – 7 s though the displacement is zero but the particle moves in opposite directions, downwards at t = 1 s and upwards at t — 7 s. Hence phase of the oscillator is not same which makes option incorrect.
In option (d) points belong to t = 1 s and t = 5 s are at separation of one time period, hence must be in phase.

Q15. Which of the following statements is/are true for a simple hannonic oscillator?
(a) Force acting is directly proportional to displacement from the mean position and opposite to it
(b) Motion is periodic
(c) Acceleration of the oscillator is constant
(d) The velocity is periodic
Sol: (a, b, d)
Key concept: The simple harmonic motion is a type of periodic motion or oscillation motion where the restoring force is directly proportional to the displacement and acts in the direction opposite to that of displacement. When the system is displaced from its equilibrium position, a restoring force that obeys Hooke’s law tends to restore the system to equilibrium. As a result, it accelerates and starts going back to the equilibrium position. An oscillation follows simple harmonic motion if it fulfils the following two rules:
1. Acceleration is always in the opposite direction to the displacement from the equilibrium position.
2. Acceleration is proportional to the displacement from the equilibrium position.

Q16. The displacement-time graph of a particle executing SHM is shown in figure. Which of the following statement is/are true? 

Displacement is maximum, i.e corresponds to extreme position, it means PE is maximum and KE is zero.

Q17. A body is performing SHM, then its
(a) average total energy per cycle is equal to its maximum kinetic energy
(b) average kinetic energy per cycle is equal to half of its maximum kinetic energy                                                                         
(c) mean velocity over a complete cycle is equal to 2/π times of its maximumvelocity                                          
(d) root mean square velocity is 1/√2 times of its maximum velocity

Q18. A particle is in linear simple harmonic motion between two points A and B, 10 cm apart (figure). Take the direction from A to B as the positive direction and choose the correct statements. __ _
AO = OB = 5 cm
BC= 8 cm

(a) The sign of velocity, acceleration and force on the particle when it is 3 cm away from A going towards B are positive
(b) The sign of velocity of the particle at C going towards B is negative
(c) The sign of velocity, acceleration and force on the particle when it is 4 cm away from B going towards A are negative
(d) The sign of acceleration and force on the particle when it is at point B is negative

In option (a): When the particle is 3 cm away from A going towards B. So, velocity is towards AB, i.e. positive. In SHM, acceleration is always towards mean position (O) it means both force and acceleration act towards O, have positive sign.
Hence option (a) is correct.
In option (b): When the particle is at C, velocity is towards B hence positive. Hence option (b) is not correct.
In option (c): When the particle is 4 cm away from B going towards A velocity is negative and acceleration is towards mean position (O), hence negative. Hence option (c) is correct.
In option (d): Acceleration is always towards mean position (O). When the particle is at B, acceleration and force are towards BA that is negative. Hence option (d) is correct.

Very Short Answer Type Questions
Q19. Displacement versus time curve for a particle executing SHM is shown in figure. Identify the points marked at which (i) velocity of the oscillator is zero, (ii) speed of the oscillator is maximum.
Sol: Key concept: In displacement-time graph of SHM, zero displacement values correspond to mean position; where velocity of the oscillator is maximum. Whereas the crest and troughs represent amplitude positions, where displacement is maximum and velocity of the oscillator is zero.
(i) The points A, C, E, G lie at extreme positions (maximum displacement, y = A). Hence the velocity of the oscillator is zero.
(ii) The points B, D, F, H lie at mean position (zero displacement, y = 0). We know the speed is maximum at mean position.

Q20. Two identical springs of spring constant k are attached to a block of mass m and to fixed supports as shown in figure. When the mass is displaced from equilibrium position by a distance x towards right, find the restoring force.

When mass is displaced from equilibrium position by a distance x towards right, the right spring gets compressed by x developing a restoring force kx towards left on the block. The left spring is stretched by an amount x developing a restoring force kx left on the block.

Q21. What are the two basic characteristics of a simple harmonic motion?
Sol: The two basic characteristics of a simple harmonic motion
(i) Acceleration is directly proportional to displacement.
(ii) The direction of acceleration is always towards the mean position, that is opposite to displacement.

Q22. When will the motion of a simple pendulum be simplehannonic?
Sol: Simple pendulum perform angular S.H.M. Consider the bob of simple pendulum is displaced through an angle θ shown. Q
The restoring torque about the fixed point O is

τ = mgl sinθ
If θ is small angle in radians, then sin θ =  0
=> mglθ
In vector form τ ∝ θ
Hence, motion of a simple pendulum is SHM for small angle of oscillations.

Q23. What is the ratio of maximum acceleration to the maximum velocity of a simple harmonic oscillator?

Q24. What is the ratio between the distance travelled by the oscillator in one time period and amplitude?
Sol: In the diagram shown a particle is executing SHM between P and Q. The particle starts from mean position ‘O’ moves to amplitude position ‘P’, then particle turn back and moves from ‘P’ to iQ\ Finally the particle turns back again and return to mean position ‘O’. In this way the particle completes one oscillation in one time period.

Total distance travelled while it goes from O → P  →  O  →  Q  →  O
= OP + PO + OQ+QO = A+A+A+A=4A
Amplitude = OP = A
Hence, ratio of distance and amplitude = 4A/A = 4

Q25. In figure, what will be the sign of the velocity of the point P’, which is the projection of the velocity of the reference particle P. P is moving in a circle of radius R in anti-clockwise direction.

Sol. As the particle on reference circle moves in anti-clockwise direction. The projection will move from P’ to O towards left, i.e. from right to left, hence sign is negative.

Q26. Show that for a particle executing SHM, velocity and displacement have a phase difference of π/2 .

Q27. Draw a graph to show the variation of PE, KE and total energy of a simple harmonic oscillator with displacement.

Important point: From the graph we note that potential energy or kinetic energy completes two vibrations in a time during which S.H.M. completes one vibration. Thus the frequency of potential energy or kinetic energy is double than that of S.H.M.

Q28. The length of a second’s pendulum on the surface of earth is 1 m. What will be the length of a second’s pendulum on the moon?

Short Answer Type Questions
Q29. Find the time period of mass M when displaced from its equilibrium position and then released for the system as shown in figure.

Sol: Key concept: For observing oscillation, we have to displace the block slightly beyond equilibrium position and find the acceleration due to the restoring force.
Let in the equilibrium position, the spring has  extended by an amount x₀.
Tension in the spring = kx₀
For equilibrium of the mass M, Mg = 2kx₀

Let the mass be pulled through a distance y and then released. But, string is inextensible, hence the spring alone will contribute the total extension y + y = 2y, to lower the mass down by y from initial equilibrium mean position x₀. So, net extension in the spring (x₀ + 2y). From F.B.D of the block,

Q30.Show that the motion of a particle represented by y = sin ax – cos cot is simple harmonic with a period of 2π/ω
Sol: The given equation is in the form of combination of two harmonic functions. We can write this equation in the form of a single harmonic (sine or cosine) function.

We have displacement function: y = sin ωt – cos ωt

Hence the function represents SHM with a period T = 2π/ω

Q31. Find the displacement of a simple harmonic oscillator at which its PE is half of the maximum energy of the oscillator.
Sol: Let us assume that the required displacement where PE is half of the maximum energy of the oscillator be x.
The potential energy of the oscillator at this position,

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We hope the NCERT Exemplar Class 11 Physics Chapter 13 Oscillations help you. If you have any query regarding NCERT Exemplar Class 11 Physics Chapter 13 Oscillations, drop a comment below and we will get back to you at the earliest.

NCERT Exemplar Class 11 Physics Chapter 14 Waves are part of NCERT Exemplar Class 11 Physics. Here we have given NCERT Exemplar Class 11 Physics Chapter 14 Waves.

NCERT Exemplar Class 11 Physics Chapter 14 Waves

Multiple Choice Questions
Single Correct Answer Type

Q1. Water waves produced by a motorboat sailing in water are
(a) neither longitudinal nor transverse
(b) both longitudinal and transverse
(c) only longitudinal
(d) only transverse
Sol: (b) Water waves produced by a motorboat sailing on the surface of deep water are both longitudinal and transverse because the waves, produce transverse as well as lateral vibrations in the particles of the medium. The water molecules at the surface move up and down; and back and forth simultaneously describing nearly circular paths as shown in Figure.

As the wave passes, water molecules at the crests move in the direction of the wave while those at the troughs move in the opposite direction.

Q3. Speed of sound wave in air
(a) is independent of temperature
(b) increases with pressure
(c) increases with increase in humidity
(d) decreases with increase in humidity

Q4. Change in temperature of the medium changes
(a) frequency of sound waves
(b) amplitude of sound waves
(c) wavelength of sound waves
(d) loudness of sound waves

Q5. With propagation of longitudinal waves through a medium, the quantity transmitted is •
(a) matter
(b) energy
(c) energy and matter
(d) energy, matter and momentum
Sol: (b) A wave is a disturbance which propagates energy and momentum from one place to the other without the transport of matter. In propagation of longitudinal waves through a medium leads to transmission of energy through the medium without matter being transmitted. There is no movement of matter (mass) and hence momentum.

Important point:
Characteristics of wave motion: ;
• It is a sort of disturbance which travels through a medium. ,
• Material medium is essential for the propagation of mechanical waves. .
• When a wave motion passes through a medium, particles of the medium only vibrate simple harmonically about their mean position. They do leave their position and move with the disturbance.
• There is a continuous phase difference amongst successive particles of the medium, i.e. particle 2 starts vibrating slightly later than particle 1 and so on.

•The velocity of the particle during there vibration is differnet at different positions different positions.
• The velocity of wave motion througha particular medium is constant.
It depends only on die nature of medium not op the frequency, wavelength or intensity,
•Energy is, propagated along with thewave motion without any net transport of themedium.

Q6. Which of the following statements are true for wave motion?
(a) Mechanical transverse waves can propagate through all mediums.
(b) Longitudinal waves can propagate through solids only.
(c) Mechanical transverse waves can propagate through solids only.
(d) Longitudinal waves can propagate through vacuum.
Sol:(c) In case of mechanical transverse wave propagates through a medium, the medium particles oscillate right angles to the direction of wave motion or energy propagation. It travels in the form of crests and troughs.
When mechanical transverse wave propagates through a medium element of the medium is subjected to shearing stress. Solids and strings have shear modulus, that is why, sustain shearing stress. Fluids have no shape of their own, they yield to shearing stress. Transverse waves can be transmitted through solids, they can be setup on the surface of liquids. But they cannot be transmitted into liquids and gases.

Q7. A sound wave is passing through air column in the form of compression and rarefaction. In consecutive compressions and rarefactions,
(a) density remains constant
(b) Boyle’s law is obeyed
(c) bulk modulus of air oscillates
(d) there is no transfer of heat
Sol: (d)
Key concept:
• When sound wave travels through a medium, say air, the particles of medium disturb in the same fashion, i.e. compression and rarefaction (depression). When air particles come closer it is called compression. On the other hand, when particles go farther than their normal position it is called rarefaction. In the condition of compression, molecules of medium come closer to each other and in the condition of rarefaction, molecules of the medium go farther from each other; compared to their normal positions.
• When compression takes place in the medium, the density and pressure of the medium increase. When rarefaction takes place in the medium, density and pressure of the medium decrease. This increase and decrease in density and pressure are temporary. Thus, compression is called the region of high density and pressure. Rarefaction is called the region of low density and pressure.

When sound wave travels through a medium:
• Due to compression and rarefaction, density of the medium (air) changes. As density is changing, so Boyle’s law is not obeyed.
• The Bulk modulus of medium remains unchanged.
• The time of compression and rarefaction is too small, i.e. we can assume adiabatic process and hence no transfer of heat.

Sol: (b)
Key concept:
Reflection of Mechanical Waves:

Medium	Longi­ tudinal wave	Trans­ verse wave	Change in direc­tion	Phase change	Time change	Path change
Reflec­tion from rigid end/ denser medium	Compres­sion as rarefac­tion and vice- versa	Crest as crest and Trough as trough	Reversed		T/2	λ/2
Reflec­tion from free end/ rarer medium	Compres­sion as compres­sion and rarefac­tion as rarefac­tion	Crest as trough and Trough as crest	No change	Zero	Zero	Zero

Q9. A string of mass 2.5 kg is under tension of 200 N. The length of the stretched string is 20.0 m. If the transverse jerk is struck at one end of the string, the disturbance will reach the other end in
(a) 1 s
(b) 0.5 s
(c) 2 s
(d) data given is insufficient

Q10. A train whistling at constant frequency is moving towards a station at a constant speed v. The train goes past a stationary observer on the station. The frequency n of the sound as heard by the observer is plotted as a function of time t (figure). Identify the expected curve.

Key concept: General expression for Apparent Frequency. Suppose observer (O) and source (S) are moving in the same direction along a line with velocities v₀ and v_s respectively. Velocity of sound is v and velocity of medium is v_m, then apparent frequency observed by observer is given

Sign convention for different situations:

• The direction of v is always taken from source to observer.
• All the velocities in the direction of v are taken positive.
• All the velocities in the opposite direction of v are taken negative.

More Than One Correct Answer Type
Q11. A transverse harmonic wave on a string is described by

where x and y are in cm and t is in sec. The positive direction of x is from left to right.
(a) the wave is travelling from right to left
(b) the speed of the wave is 20 m/s
(c) frequency of the wave is 5.7 Hz
(d) the least distance between two successive crests in the wave is 2.5 cm
Sol: (a, b, c)
Key concept: The general equation of a plane progressive wave with initial phase is

(a) It represents a progressive wave of frequency 60 Hz.
(b) It represents a stationary wave of frequency 60 Hz.
(c) It is the result of superposition of two waves of wavelength 3 m, frequency 60 Hz each travelling with a speed of 180 m/s in opposite direction.
(d) Amplitude of this wave is constant.
Sol: (b, c)
Key concept: Standing Waves or Stationary Waves:
When two sets of progressive wave trains of same type (both longitudinal or both transverse) having the same amplitude and same time period/ frequency/wavelength travelling with same speed along the same straight line in opposite directions superimpose, a new set of waves are formed. These are called stationary waves or standing waves.

Q13. Speed of sound waves in a fluid depends upon

(a) directly on density of the medium
(b) square of Bulk modulus of the medium
(c) inversely on the square root of density
(d) directly on the square root of bulk modulus of the medium

Q14. During propagation of a plane progressive mechanical wave,
(a) all the particles are vibrating in the same phase
(b) amplitude of all the particles is equal
(c) particles of the medium executes SHM
(d) wave velocity depends upon the nature of the medium
Sol: (b, c, d)
Key concept:
Characteristics of wave motion:
• When a wave motion passes through a medium, particles of the medium only vibrate simple harmonically about their mean position. They do not move with the disturbance.
• Medium particles oscillate with same frequency and also the amplitude of oscillation of all the particles is equal. All the particles marked as 1, 2, 3,4 and 5 oscillate with the same frequency.
• There is a continuous phase difference amongst successive particles of the medium, i.e. particle 2 starts vibrating slightly later than particle 1 and so on.
• The velocity of the particle during their vibration is different at different positions.
• The velocity of wave motion through a particular medium is constant. It depends only on the nature of medium not on the frequency, wavelength or intensity.

Option (a): Clearly, the particles 1, 2 and 3 are having different phase.
Option (b) and (c): Particles of the wave shown in the figure executes SHM with same amplitude.
Option (d): The wave velocity of mechanical wave depends only on elastic and inertia property of medium for a progressive wave propagating in a fluid. Hence wave velocity depends upon the nature of the medium.

Q15. The transverse displacement of a string (clamped at its both ends) is given by

All the points on the string between two consecutive nodes vibrate with
(a) same frequency
(b) same phase
(c) same energy
(d) different amplitude

Sol: (a, b, d)
Key concept:
• The points for which amplitude is minimum are called nodes in a stationary wave, nodes are equally spaced at a distance λ/2.
• The points for which amplitude is maximum are called antinodes. Like nodes, antinodes are also equally spaced with spacing (λ/2) Furthermore, nodes and antinodes are alternate with spacing (λ/4).
• The nodes divide the medium into segments (or loops). All the particles in a segment vibrate in same phase, but in opposite phase with the particles in the adjacent segment. Twice in one period all the particles pass through their mean position simultaneously with maximum velocity (A at), the direction of motion being reversed after each half cycle.

Q16. A train, standing in a station yard, blows a whistle of frequency 400 Hz in still air. The wind starts blowing in the direction from the yard to the station with a speed of 10 m/s. Given that the speed of sound in still air is 340 m/s. Then.
(a) the frequency of sound as heard by an observer standing on the platform is 400 Hz
(b) the speed of sound for the observer standing on the platform is 350 m/s
(c) the frequency of sound as heard by the observer standing on the platform will increase
(d) the frequency of sound as heard by the observer standing on the platform will decrease
Sol: (a, b) When the wind is blowing in the same direction as that of sound wave, then net speed of the wave is sum of speed of sound wave and speed of the wind.
Given, f₀ = 400 Hz, v = 340 m/s Speed of wind v_w = 10 m/s
Option (a): As there is no relative motion between the source and observer, hence frequency observed will be the same as natural frequency f₀ = 400 Hz
Option (b): When the wind is blowing in the same direction as that of sound wave, then net speed of the wave is sum of speed of sound wave and speed of the wind. The speed of sound v = v + v_w
= 340+ 10 = 350 m/s

Option (b) and (c): There will be no change in frequency because there is no relative motion between source and observer. Hence (c) and (d) are incorrect.

Q17. Which of the following statement are true for a stationary waves?
(a) Every particle has a fixed amplitude which is different from the amplitude of its nearest particle.
(b) All the particles cross their mean position at the same time.
(c) All the particles are oscillating with same amplitude.
(d) There is no net transfer of energy across any plane.
(e) There are some particles which are always at rest.
Sol: (a, b, d, e) Consider the equation of a stationary wave
y(x, t) = 2a sin (kx) cos (wt)
Option (a): In stationary wave any particle at a given position have amplitude 2a sin kx.
Option (b): The time period of oscillation of all the particles is same, hence all the particles cross their mean position at the same time.
Option (c): Amplitude of all the particles are 2a sin kx which is different for different particles at different values of x.
Options (d) and (e): Nodes are the points which is always at rest hence no transfer of energy across the nodes. It means the energy is a stationary wave is confined between two nodes.

Q18. A sonometer wire is vibrating in resonance with a tuning fork. Keeping the tension applied same, the length of the wire is doubled. Under what conditions would the tuning fork still be in resonance with the wire?
Sol: Wire of sonometer is twice the length which it vibrates in its second harmonic. Thus, if the tuning fork resonates at L, it will resonate at 2L. This can be explained as below:
The frequency of sonometer is given by

Hence, when the wire is doubled the number of loops also get doubled to produce the resonance. That is it resonates in second harmonic.

Q19. An organ pipe of length L open at both ends is found to vibrate in its first harmonic when sounded with a tuning fork of 480 Hz. What should be the length of a pipe closed at one end, so that it also vibrates in its first harmonic with the same tuning fork?

Q20. A tuning fork A, marked 512 Hz, produces 5 beats per second, when sounded with another unmarked tuning fork B. If B is loaded with wax the number of beats is again 5 per second. What is the frequency of the tuning fork B when not loaded?
Sol: When the prong of B is loaded with wax, its frequency becomes less than the original frequency.
If we assume that the original frequency of B is 507, then on loading its frequency will be less than 507. The beats between A and B will be more than 5.
If we assume that the original frequency of B is 517, then on loading its frequency will be less than 517. The beats between A and B may be equal to 5. Hence the frequency of the tuning fork B when not loaded should be 517.

Q22. The displacement of an elastic wave is given by the function, y = 3 sin ωx + 4 cos ωt, where y is in cm and t is in second. Calculate the resultant amplitude.

Sol: Given, displacement of the wave
y = 3 sin ωx + 4 cos ωx
Let us assume, 3 = A cos θ                                   .                                                .. .(i)
3=Acos θ                                                                                …(ii)
On dividing Eq. (ii) by Eq. (i)
tan θ = 4/3 => ϕ= tan^-1(4/3)
Squaring and adding equations (i) and (ii),
A² cos² θ + A² sin² θ = 3² + 4²
=>              A² (cos² θ + sin² θ) = 25
A² = 25 => A = 5. Hence, amplitude = 5 cm

Q23. At what temperatures (in °C) will the speed of sound in air be 3 times its value at 0°C?

Q24. When two waves of almost equal frequencies n_l and n₂ reach at a point simultaneously, what is the time interval between successive maxima?

Short Answer Type Questions

Q25. A steel wire has a length of 12 m and a mass of 2.10 kg. What will be the speed of a transverse wave on this wire when a tension of 2.06 x 10⁴N is applied?

Q26. A pipe 20 cm long is closed at one end. Which harmonic mode of the pipe is resonantly excited by a source of 1237.5 Hz? (Sound velocity in air = 330 ms^-1)

Q27. A train standing at the outer signal of a railway station blows a whistle of frequency 400 Hz in still air. The train begins tomove with a speed of 10 ms’¹ towards the platform. What is the frequency of the sound for an observer standing on the  platform? (Sound velocity in air = 330 ms^-1)

Q28. The wave pattern on a stretched string is shown in figure. Interpret what kind of wave this is and find its
wavelength.

Sol. If we observe the graph there are some points on the graph which are always at rest. The points on positions x = 10,20,30,40 never move, always at mean position with respect to time. These are forming nodes which characterize a stationary wave.
We know the distance between two successive nodes is equal to λ/2
=>      λ = 2 x (node to node distance)
= 2 x (20 – 10) = 20 cm

Q29. The pattern of standing waves formed on a stretched string at two instants of time are shown in figure. The velocity of two waves superimposing to form stationary waves is 360 ms^-1 and their frequencies are 256 Hz.

(a) In second plot, the displacement of each particle is zero. It means all the points are crossing mean position. At first plot point at A¹ is at amplitude position. Time taken to move from amplitude position to mean position to is equal to one-fourth of the time period.

Q30. A tuning fork vibrating with a frequency of 512 Hz is kept close to the open end of a tube filled with water (figure). The water level in the tube is gradually lowered. When the water level is 17 cm below the open end, maximum intensity of sound is heard.
If the room temperature is 20°C, calculate
(a) speed of sound in air at room temperature.
(b) speed of sound in air at 0°C.
(c) if the water in the tube is replaced with mercury, will there be any difference in your observations?

Sol: If a pipe partially filled with water whose upper surface of the water acts as a reflecting surface of a closed organ pipe. If the length of the air column is varied until its natural frequency equals the frequency of the fork, then the column resonates and emits a loud note.
The frequency of tuning fork,f= 512 Hz.
For observation of first maxima of intensity,

(c) The resonance will still be observed for 17 cm length of air column above mercury. However, due to more complete
reflection of sound waves at mercury surface, the intensity of reflected sound increases.

Q31. Show that when a string fixed at its two ends vibrates in 1 loop, 2 loops, 3 loops and 4 loops, the frequencies are in the ratio 1 : 2 : 3 : 4.

Sol: Key concept: For a string fixed at its two ends vibrate in a normal mode if it vibrates according to the equation

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