CBSE Sample Papers for Class 11 Physics Set 1 with Solutions

Students must start practicing the questions from CBSE Sample Papers for Class 11 Physics with Solutions Set 1 are designed as per the revised syllabus.

CBSE Sample Papers for Class 11 Physics Set 1 with Solutions

Time Allowed : 3 hours
Maximum Marks : 70

General Instructions:

1. All questions are compulsory. There are 40 questions.
2. This Question paper has four sections : Section A, Section B, Section C, Section D
3. Section A contains twenty five questions of one mark each, Section B contains five questions of two marks each, Section C contains seven questions of three marks each, Section D contains three question of five marks each.
4. There is no overall choice. However, internal choices have been provided in seven questions of one mark, two questions of two marks, two questions of three marks and two questions of five marks weightage. You have to attempt only one of the choices in such questions.
5. You may use the following values of physical constants wherever necessary:
   c = 3 × 10⁸ m/s,
   h = 6.63 × 10^-34 Js
   e = 1.6 × 10^-19 C,
   Radius of Earth, R_e = 6.4 × 10⁶ m
   Universal Gravitational constant. G = 6.67 × 10^-11 Nm2kg-2
   mass of electron, m_e = 9.1 × 10^-31 kg,
   mass of neutron, m_n = 1.675 × 10^-27 kg
   mass of proton, m_p = 1.673 × 10^-27 kg
   Avogadro’s number = 6.023 × 10²³ atom per gram
   Boltzmann constant = 1.38 × 10^-23 JK^-1

Section – A (25 Marks)

Question numbers 1 to 25 carry 1 mark each.

Question 1.
Find number of significant figure in 16500. [1]
(A) 3
(B) 4
(C) 5
(D) none of the above
Answer:
Option (A) is correct.

Question 2.
A person moves 30 m north, then 20 m east then 30\(\sqrt{2}\) m south west. The displacement from the original position is: [1]
(A) 15 m east
(B) 28 m south
(C) 10 m west
(D) 15 m south west
Answer:
Option (B) is correct. 1

Explanation:
Angle formed at B = 45°

= -10i
So, final displacement is – 10 m towards west.

Question 3.
An athlete runs some distance before taking a long jump because : [1]
(A) it helps him to gain energy
(B) it helps to apply large force
(C) it given him larger amount of inertia
(D) to increase reaction forces
OR
A car moving on a horizontal road may be thrown out of the road in taking a turn [1]
(A) by the gravitational force
(B) due to lack of proper centripetal force
(C) due to the rolling frictional force between the car and the road
(D) due to the reaction of the road.
Answer:
Option (C) is correct.

Explanation:
An athelete runs some distance before taking along jump because of inertia of motion, due to this motion he jump higher and larger distance.

OR
Option (B) is correct.

Question 4.
Number of joule contained in 1 kg metre is : [1]
(A) 9.8
(B) 980
(C) 1000
(D) 105
Answer:
Option (A) is correct.

Explanation:
Since 1 kg-m = 9.80665 joule.
≈ 9.8 joule

Question 5.
Moment of inertia of a ring of mass M and radius R about its diameter is : [1]
(A) \(\frac{1}{4}\) MR²
(B) \(\frac{1}{2}\) MR²
(C) MR²
(D) 2MR²
Answer:
Option (B) is correct.

Explanation:
Let us consider a ring of mass M and radius R lies in the xy plane, with its centre at the origin.
∴ According to perpendicular axis theorem for calculating moment of inertia.
I_z ≈ I_x + I_y
where I_z, I_x, I_y are moment of inertia of ring about z, x, and y-axis respectively.
∵ By symmetry I_x = I_y
and I_z = MR²
I_x = \(\frac{1}{2}\) MR²

Question 6.
Artificial satellite once set in motion continues to move around the earth by receiving energy from : [1]
(A) the sun
(B) burning of chemical fuel
(C) none
(D) storage cell
OR
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 for 0.001 s, the force that the batsman had to apply to hold the bat firmly at its place would be : [1]
(A) 10.5 N
(B) 21.0 N
(C) 1.05 × 10⁴ N
(D) 2.1 × 10⁴ N
Answer:
Option (C) is correct.

Explanation:
Any artificial satellite once set in motion around their planet stays in their orbit using the force generated by centripetal acceleration.

OR

Option (C) is correct.
Explanation:
F = \(\frac{\Delta p}{t}=\frac{m v-(-m v)}{t}\)
= \(\frac{2 m v}{t}\)
= \(\frac{2\times 150 \times 10^{-3} \mathrm{~kg} \times 35 \mathrm{~m} / \mathrm{s}}{0.001 \mathrm{~s}}\)
= 10500N
= 1.05 × 10⁴ N

Question 7.
Shear strain is represented by : [1]
(A) angle of twist
(B) angle of shear
(C) increase in area
(D) decrease in volume
Answer:
Option (B) is correct.

Explanation: Since shear strain is defined as the ratio of shearing stress that laterally deforms the substance to the shearing strain.

Question 8.
Specific heat of water is [1]
(A) Zero
(B) one
(C) infinity
(D) unknown
Answer:
Option (B) is correct.

Explanation:
The specific heat of a substance is the amount of heat energy needed to raise a unit mass of the material by 1°C.
∵ for water at 20°C the specific heat is :
1 Calorie/g/°C = 4.18 J/g/°C

Question 9.
Which of the following processes is reversible ? [1]
(A) transfer of heat by conduction
(B) transfer of heat by radiation
(C) isothermal compression
(D) electrical heating of microhm wire
OR
The perfect gas equation for 4 gram of Hydrogen gas is : [1]
(A) PV = RT
(B) PV = 2RT
(C) PV = \(\frac{1}{2}\)RT
(D) PV = 4RT
Answer:
Option (C) is correct.

Explanation:
All quasi-static process is revisable where in all changes take place infinitely slow. Since, isothermal process is very slow which is quasi-static. Hence, it will be reversible.

OR

Option (B) is correct.
Explanation:
Perfect gas equation or ideal gas equation:
PV = nRT
∵ Weight of Hydrogen = 4 gm
So, n = number of moles = \(\frac{\text { given weight }}{\text { molecular weight }}\)
= \(\frac{4 \mathrm{~g}}{2 \mathrm{~g}}\) = 2
∴ Required equation ⇒ PV = 2 RT

Question 10.
Unit of frequency is : [1]
(A) maxwell
(B) Newton
(C) Hertz
(D) Henry
Answer:
Option (C) is correct.

Explanation:
S.I. unit of frequency is Hertz.

Question 11.
Which is the most accurate clock ? [1]
Answer:
Cesium atomic clock.

Question 12.
Indicate the accelerated portion in the graph shown below : [1]

Answer:
Only portion PQ indicates accelerated motion.

Commonly Made Error
Students may indicate the portion RS as the accelerated portion.

Answering Tip
Before answering these type of questions, students should ensure about the axes on the graph is plotted and then apply the knowledge of quantities to find the correct answer.

Question 13.
Can there be motion in two dimensions with an acceleration in only one dimension ? [1]
Answer:
Yes, projectile motion.

Question 14.
What is the dimensional formula of work ? [1]
Answer:
[ML²T^-2]

Question 15.
Give an example of rigid body motion in which the centre of mass of the body remains at rest. [1]
Answer:
Motion of a point cylinder capable of rotating without friction about its own axis is the required example.

Question 16.
An artificial satellite is at a height of 36,500 km above earth’s surface. What is the work by earth’s gravitational force in keeping it in this orbit ? [1]
OR
Is it possible for a body to have inertia but no weight?
Answer:
No, energy is required by a satellite to keep it orbiting because the work done by the centripetal force is zero.

Commonly Made Error
Students could not calculate the work done by Earth’s gravitational force.

Answering Tip
Work done = Fs cos θ,where the angle between centripetal force and displacement (i.e. tangent to circle) is 90 degrees. So work done is zero.
OR
Yes, because weight (mg) of a body can be zero as it depend upon acceleration due to gravity but every body will always have inertia (i.e. mass). For example: At centre of earth, g = 0. So, weight is zero, but mass is not zero. So, inertial will not be zero.

Question 17.
Which is more elastic water or air ? [1]
OR
Name the material in whose capillary, water will descent instead of rising.
Answer:
Elasticity is reciprocal of compressibility. Air is more compressive than water. Hence water is more elastic than air.

Commonly Made Error
Students could not compare the elasticity of air and water.

Answering Tip
Students should understand that the bulk modulus of elasticity is reciprocal of compressibility.

OR

Paraffin wax.

Question 18.
Give the relation between Celsius, Fahrenheit and Reaumur scale temperatures. [1]
OR
Why a gas is cooled when expanded ?
Answer:
Celsius, Fahrenheit and Reaumur scales are used to express temperature. They are related as :
\(\frac{\mathrm{C}-0}{100-0}\) = \(\frac{\mathrm{F}-32}{212-32}\) = \(\frac{\mathrm{R}-0}{80-0}\)

Commonly Made Error
Students could not relate Celsius, Fahrenheit and reaumur scale of temperatures.

Answering Tip
Students should be familiarized with the inter-conversion formulas for different scale of temperatures.
OR
Gas is cooled when expanded due to decrease in internal energy.

Question 19.
Draw a graph to show the variation of EE., K.E. and total energy‘of a simple harmonic oscillator with displacement. [1]
Answer:

Question 20.
Can a motion be periodic but not oscillatory? If yes, give an example. If not, explain why ? [1]
Read the following text and answer any 4 of the following questions based on the same:
Flywheel and sewing machine: There is a difference between inertia and moment of inertia of a body. Inertia depends on the mass of the body but the moment of inertia about an axis depends on the mass of the body and the distribution of its mass about the axis.

In the following figure, the masses of the two wheels are exactly equal but in the wheel (A) the mass is uniformly distributed and in the wheel (B) most of the mass is situated at the rim. Both the wheels rotate about axis passing through the centre. It is noticed that while the two wheels are set in rotation and left, wheel (B) continues rotating for a longer time.

This means that the moment of inertia of wheel (B) is greater than the wheel (A). Also greater is the part of the mass of the body away from the axis of rotation, greater the moment of inertia of the body about the axis. Such a wheel is known as flywheel.
Consider a foot operated sewing machine. It has two wheels – one big and the other small. The wheels are connected by a rope. The bigger wheel acts as flywheel.

The rope transfers the motion from this flywheel to the smaller wheel. Smaller wheel works as a pulley and also as a small fly wheel. We see even we stop supply of driving force to the bigger wheel it still continues to run for a short time because of its moment of inertia.
So, flywheel acts as an energy reservoir by storing and supplying mechanical energy when required. The kinetic energy stored in a flywheel is
E = 1/2Iω²
where, I = moment of inertia and ω = angular velocity.
Answer:
Yes, uniform circular motion is the example of it.

Question 21.
Which of the following statement is true? [1]
(A) Moment of inertia depends on the total mass and the distribution of mass from the axis of rotation
(B) Inertia depends on the total mass and the distribution of mass from the axis of rotation
(C) If the masses of two objects are equal then their moment of inertia are also equal
(D) As mass of an object is distributed away from the axis of rotation, the moment of inertia decreases
Answer:
Option (A) is correct.

Explanation:
Inertia depends on the mass of the body but the moment of inertia about an axis depends on the mass of the body and the distribution of its mass about the axis.
Moment of inertia = Σmr²

Question 22.
How many flywheels are there in foot operated sewing machine ? [1]
(A) One
(B) Two
(C) Three
(D) Zero
Answer:
Option (B) is correct.

Explanation:
In foot operated sewing machine there are two wheels – one big and the other small. The wheels are connected by a rope. The bigger wheel acts as flywheel. The rope transfers the motion from this flywheel to the smaller wheel. Smaller wheel works as a pulley and also as a small flywheel.

Question 23.
We see even we stop supply of driving force to the bigger wheel of foot operated sewing machine, it still continues to run for a short time. If the rim of this wheel is made thicker then: [1]
(A) It will run for the same period when the driving force is stopped
(B) It will run for shorter period when the driving force is stopped
(C) It will run for longer period when the driving force is stopped
(D) It will stop immediately when the driving force is stopped
Answer:
Option (C) is correct.

Explanation: If the rim of the bigger wheel of foot operated sewing machine is made thicker then it will run for longer period when the driving force is stopped. This is due to the increase of its moment of inertia.

Question 24.
Energy stored in flywheel is: [1]
(A) \(\frac{1}{2}\) mv²
(B) \(\frac{1}{2}\) mω²
(C) \(\frac{1}{2}\) Iv²
(D) \(\frac{1}{2}\) Iω²
Answer:
Option (D) is correct.

Explanation:
Kinetic energy of an object moving in a straight line is E = \(\frac{1}{2}\) mv².
The kinetic energy of a spinning object is
E = \(\frac{1}{2}\) Iω².

Question 25.
Which one of the following wheel (having same mass) will have highest moment of inertia about axis passing through the centre? [1]

Answer:
Option (D) is correct.

Explanation:
Mass of all the wheels are same. In wheel D, the mass is distributed furthest from the axis. Since moment of inertia = Σmr², hence wheel D will have highest moment of inertia.

Section – B (10 Marks)

Question numbers 26 to 30 carry 2 marks each.

Question 26.
Does the nature of a vector change when it is multiplied by a scalar ? [2]
OR
Find the direction for an umbrella when rain falls vertically with speed 20 ms-1 and wind blows from east to west with a speed of 15 ms^-1. [2]
Answer:
The nature of a vector may or may not change when it is multiplied by a scalar.
For example, when a vector is multiplied by a pure number like 1, 2, 3, …. etc., then the nature of the vector does not change.
On the other hand, when a vector is multiplied by a scalar physical quantity, then the nature of the vector changes. For example, when acceleration (vector) is multiplied by mass (scalar) of a body, then it indicates force (a vector quantity) whose nature is different than acceleration.

Commonly Made Error
Students may get confused in the nature and direction of a vector quantity and answer the same affirmatively only.

Answering Tip
Students should remember the basic rules of multiplication of a vector and their corresponding examples before answering this question.

OR

From the figure

tan θ = \(\frac{v_\omega}{v_r}=\frac{15}{20}=\frac{3}{4}\)
tan θ = 0.75
or θ ≈ 37° east from vertical.

Question 27.
What is the pressure on a swimmer 10 m below the surface of a lake ? [2]
Answer:
P = P_a + Pgh
= 1.01 × 10⁵ + 10³ × 9.8 × 10
= 1.99 × 10⁵ Pa ≈ 2 atm.

Commonly Made Error
Students fails to include the atmospheric pressure on the swimmer.

Answering Tip
Students should understand that pressure at a point is given by the relation
P = P_a + hPg
where,
P_a is the atmospheric pressure
hρg is the column pressure.

Question 28.
A metallic sphere is heated. Out of its radius, surface area, volume and mass which will undergo the maximum percentage increase ? [2]
OR
Why is it impossible for a ship to use the internal energy of sea water to operate its engine ? [2]
Answer:
Mass remains same in this case. Radius of the sphere will increase. Surface area of sphere ∝ (radius)² and Volume ∝ (Radius)³. So the volume will undergo maximum percentage increase.

OR

For using the internal energy of sea water to operate the engine of a ship, the internal energy of the sea water has to be converted into mechanical energy. Since whole of the internal energy cannot be converted into mechanical energy, a part has to be rejected to a colder body (sink). Since, no such body is available, so the internal energy of the sea water cannot be used to operate the engine of the ship.

Commonly Made Error
Students could not reason out for the impossibility to convert the whole internal energy of sea water to work done.

Answering Tip
Second law of thermodynamics should be understood completely.

Question 29.
Will a pendulum gain or lose time when taken to the top of a hill ? [2]
Answer:
Value of acceleration due to gravity decreases at the top of the hill. Time period of a simple pendulum is given by :
T = 2π \(\sqrt{\frac{l}{g}}\)
Decrease in g means, increase in T, i.e., the pendulum takes more time to complete vibration, 1 it implies that it will lose time.

Question 30.
Liquids and gases cannot propagate transverse waves. Why ? [2]
Answer:
Liquids and gases cannot sustain shearing stress. Therefore, transverse waves in the form of crests and troughs (involving change of shape) are not possible in fluids. Rather, the fluid possess volume elasticity. Therefore, compressions and rarefactions (involving changes in volume) can be propagated through fluids.

Section – C (21 Marks)

Question numbers 31 to 37 carry 3 marks each.

Question 31.
Find the magnitudes and directions of the vectors \(\hat{i}\) + \(\hat{j}\) and \(\hat{i}\) – \(\hat{j}\). [3]
Answer:
(i) Magnitude of vectors \(\hat{i}\) + \(\hat{j}\)
= |\(\hat{i}\) + \(\hat{j}\)|
= \(\sqrt{1^2+1^2}\) = \(\sqrt{2}\)
tan θ = \(\frac{|\hat{j}|}{|\hat{i}|}=\frac{1}{1}\) = 1
or θ = tan^-1 1 = 45° with X-axis

(ii) Magnitude of vectors \(\hat{i}\) – \(\hat{j}\)
= |\(\hat{i}\) – \(\hat{j}\)|
= \(\left|\sqrt{1^2+(-1)^2}\right|\) = \(\sqrt{2}\)

tan θ = \(\frac{|\hat{-j}|}{|\hat{i}|}=\frac{-1}{1}\) = 1
or θ = tan^-1 (-1)
= – 45° with X-axis

Question 32.
A force of 100 N gives a mass m₁, an acceleration of 10 ms^-2 and of 20 ms 2 to a mass m^-2. What acceleration must be given to it if both the masses are tied together ? [3]
Answer:
Suppose, a = acceleration produced if m₁ and m₂ are tied together.
F = 100 N.
Let a₁ and a₂ be the acceleration produced in m₁ and m₂ respectively.
∴ a₁ = 10 ms^-2, a₂ = 20 ms^-2 (given)
Again
m₁ = \(\frac{\mathrm{F}}{a_1}\) and m₂ = \(\frac{\mathrm{F}}{a_2}\)
or m₁ = \(\frac{100}{10}\) = 10 kg
and m₂ = \(\frac{100}{20}\) = 5 kg
∴ m₁ + m₁ = 10 + 5 = 15 kg
So, a = \(\frac{\mathrm{F}}{m_1+m_2}\) = \(\frac{100}{15}\) = \(\frac{20}{3}\)
= 6.67 ms^-2.

Question 33.
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. [3]
Answer:
The horizontal speed of the ball is same as that of the car, therefore ball as well as car travels equal horizontal distance. u = car speed of car, v = vertical speed of ball

For a ground-based observer, the ball is a projectile with speed v₀ and the angle of projection θ with horizontal in as shown above.

Question 34.
(n -1) equal point masses each of mass m are placed at the vertices of a regular tt-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. [3]
OR
A comet orbits the sun in a highly elliptical orbit. Does the comet have a constant : [3]
(A) Linear speed,
(B) Angular speed,
(C) Angular momentum,
(D) Kinetic energy,
(E) Potential energy,
(F) Total energy throughout it orbit ?
Neglect mass loss of the comet when it comes very close to the sun.
Answer:
The centre of mass of a regular n-polygon lies at its geometric centre.
Let \(\vec{b}\) be the position vector of the centre of mass of regular n-polygon.
From question, (n – 1) equal point masses each of mass m are placed at the(n – 1) vertices of a regular n-polygon,
Then, r_cm = \(\frac{(n-1) m b+m a}{(n-1) m+m}\)
Now, mass m is placed at nth remaining vertex, then r_cm = \(\vec{0}\)
\(\frac{(n-1) m b+m a}{(n-1) m+m}\) = 0
or \(\vec{b}\) = \(\frac{-m \vec{a}}{(n-1) m}=\frac{-\vec{a}}{(n-1)}\)
Negative sign indicates that c.m. lies other side from n^th vertex geometrical centre of n-polygon.

OR

(a) The linear speed (v = ωR) changes because the distance, i.e., (R) of the comet from the sun changes due to it elliptical orbit around the sun.
(b) The angular speed of the comet also changes because it covers different angles in equal intervals of time.
(c) The angular momentum of the comet is same throughout due to the conservation of angular momentum in the absence of any torque.
(d) Kinetic energy changes because linear speed is different at different points.
(e) The potential energy at different points is different because the comet is not at the same distance from the sun (the orbit is not circular).
(f) The total energy of comet remains, the same throughout the motion.

Commonly Made Error
Confusion in identifying the physical quantities which remain constant in motion of the body in an elliptical orbit.

Answering Tip
Student should study Kepler’s II law in detail.

Question 35.
AA 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 × 10¹¹ Nm^-2.) [3]
Answer:
Given:
Steel cable’s length, l = 9.1 m
Radius, r = 5mm = 5 × 10^-3 m.
Tension in cable, F = 800 N
Young’s modulus = 2 × 10¹¹ N/m^-2
Young’s modulus ,
Y = \(\frac{F}{A} \times \frac{l}{\Delta l}\)
or ∆l = \(\frac{F}{\pi r^2} \times \frac{l}{\mathrm{Y}}\)
= \(\frac{800 \times 9.1}{3.14 \times\left(5 \times 10^{-3}\right)^2 \times 2 \times 10^{11}}\)
= 4.64 × 10^-4 m

Question 36.
For Brownian motion of particles of suspensions in liquid answer the following : [3]
(A) What should be the typical size of suspended particles ?
(B) Bombardments of the suspended particles by molecules of the liquid are random. We should then expect equal no. of molecules hitting a suspended particle from all directions. Why the net impact is not zero ?
(C) Can the assembly of suspended particles be considered a ‘gas’ of ‘heavy molecules’ ? If so, what is the temperature of this gas if temperature of the liquid is T ?
OR
Molar volume is the volume occupied by 1 mol of any (ideal) gas at standard temperature and pressure (STP 1 atm pressure, 0°C). Show that it is 22.4 litres. [3]
Answer:
(a) 10^-6 m.
(b) The average number of molecules hit in different directions so their net impact is not zero.
(c) Yes, since the heavy gas is in thermal equilibrium with the liquid, its temperature is equal to the temperature of liquid, i.e., T.

OR

Here n = 1 mol, T = 273 K, R = 8.31 J/mole/K.
P = 1 atm = 0.76 m of Hg
= 0.76 × 9.8 × 13.6 × 10³ N/m²
Using the relation
PV = nRT,
or, V = \(\frac{n \mathrm{RT}}{\mathrm{P}}\)
= \(\frac{1 \times 8.31 \times 273}{0.76 \times 9.8 \times\left(13.6 \times 10^3\right)}\)
= 22.4 × 10^-3 m³
= 22.4 lit.

Question 37.
The periodic time of a mass suspended by a spring (force constant k) is T. If the spring is cut in three equal pieces, what will be the force constant of each part ? If the same mass be suspended from one piece, what will be the periodic time ? [3]
Answer:
Consider the spring be made of combination of three springs in series each of spring constant k. The effective spring constant K is given by :
\(\frac{1}{\mathrm{~K}}=\frac{1}{k}+\frac{1}{k}+\frac{1}{k}=\frac{3}{k}\)
or K = \(\frac{\mathrm{k}}{\mathrm{3}}\)
or, k = 3K
∴ Time period of vibration of a body attached to the end of this spring,
T = 2π\(\sqrt{\frac{m}{K}}\)
= 2π\(\sqrt{\frac{m}{(k / 3)}}\)
= 2π\(\sqrt{\frac{3 m}{k}}\) …………….. (i)
When the spring is cut into three pieces, the spring constant – k.
Time period of vibration of a body attached to the end of this spring.
T₁ = 2π\(\sqrt{\frac{m}{K}}\) ……………. (ii)
From equation (i) and (ii),
\(\frac{\mathrm{T}_1}{\mathrm{~T}}=\frac{1}{\sqrt{3}}\)
or, T₁ = \(\frac{\mathrm{~T}}{\sqrt{3}}\)

Section – D (15 Marks)

Question numbers 38 to 40 carry 5 marks each.

Question 38.
The velocity time graph of a particle is given by [5]

(i) Calculate distance and displacement of particle from given v-t graph.
(ii) Specify the time for which particle undergone acceleration, retardation and moves with constant velocity.
(iii) Calculate acceleration, retardation from given v-t graph.
(iv) Draw acceleration-time graph of given v-t graph.
Answer:
(i) Distance = area of ∆OAB + area of trapezium BCDE
= 12 + 28 = 40 m
Displacement = area of ∆OAB – area of trapezium BCDE
= 12 – 28 = -16 m.

(ii) Time for acceleration is 0 ≤ t ≤ 4
Time for retardation is 4 ≤ t ≤ 8
Time for constant velocity is 8 ≤ t ≤ 12

(iii) Acceleration = \(\frac{v}{t}=\frac{4}{4}\) = 1 m/s²
Retardation = \(=\frac{-4}{2}\) = -2 m/s²

(iv)

Commonly Made Error
Students may commit mistake while taking out the values of different variables from the graph.

Answering Tip
Students should read the graph carefully and take out the corresponding values of the variables to be used.

Question 39.
A rocket is fired ‘ vertically’ from the surface of Mars with a speed of 2 km/s. If 20% of its initial energy is lost due to motion through atmospheric resistance, how far will the rocket go from the surface of Mars before returning to it ? Mass of Mars = 6.4 × 10²³ kg, Radius of Mars = 3395 km, G = 6.67 × 10^-11 Nm²kg^-2. [5]
OR
Read each statement below carefully, and state, with reasons, if it is true or false : [5]
(A) During rolling, the force of friction acts in the same direction as the direction of motion of the CM of the body.
(B) The instantaneous speed of the point of contact during rolling is zero.
(C) The instantaneous acceleration of the point of contact during rolling is zero.
(D) For perfect rolling motion, work done against friction is zero.
(E) A wheel moving down a perfectly frictionless inclined plane will undergo slipping (not rolling) motion.
Answer:
Initial K.E = \(\frac{1}{2}\) mv² Initial P.E = \(\frac{\mathrm{-GMm}}{\mathrm{R}}\)
where, m = mass of the rocket, M = mass of the Mars,
R = radius of Mars
∴ Total initial energy = \(\frac{1}{2}\) mv² – \(\frac{\mathrm{GMm}}{\mathrm{R}}\)
Since 20% of K.E is lost, only 80% remains to reach the height.
∴ 80% Total initial energy available
= \(\frac{4}{5} \times \frac{1}{2} m v^2-\frac{\mathrm{GMm}}{\mathrm{R}}\)
= 0.4 mv² – \(\frac{\mathrm{GMm}}{\mathrm{R}}\)
when the rocket reaches the highest point, at a height h above the surface, its K.E. = 0 and

OR

(a) True
Frictional force acts opposite to the direction of motion of the centre of mass of a body. In the case of rolling, the direction of motion of the centre of mass is forward. Hence, frictional force acts in the forward direction.

(b) True
Rolling can be considered as the rotation of a body about an axis passing through the point of contact of the body with the ground. Hence, its instantaneous speed is zero.

(c) False
When a body is rolling, its instantaneous acceleration is not equal to zero. It has some value.

(d) True
When perfect rolling begins, the frictional force acting at the lowermost point becomes zero. Hence, the work done against friction is also zero.

(e) True
The rolling of a body occurs when a frictional force acts between the body and the surface. This frictional force provides the torque necessary for rolling. In the absence of a frictional force, the body slips from the inclined plane under the effect of its own weight.

Question 40.
An equilateral triangle ABC is formed by two Cu rods AB and BC and one A1 rod. It is heated in such a way that temperature of each rod increases by AT. Find change in the angle ABC. [Coeff. of linear expansion for Cu is α₁, coeff. of linear expansion for A1 is α₂] . [5]
OR
A body cools from 80°C to 50°C in 5 minutes. Calculate the time it takes to cool from 60°C to 30°C. The temperature of the surroundings is 20°C. [5]
Answer:
Due to temperature increase, length of each side changes, so, the angle corresponding to any vertex also changes. Let us consider the following diagram.

Let l₁ = AB, l₂ = AC, l₃ = BC
cos θ = \(\frac{l_3^2+l_1^2-l_2^2}{2 l_3 l_1}\)
or 2l₃l₁cos θ = l₃² + l₁² – l₂²
Differentiating the above equation
2 (l₃dl₁ + l₁dl₃) cos θ – 2l₁l₃ sin θdθ
= 2l₃dl₃ + 2l₁dl₁ – 2l₂dl₂
Now,
dl₁ = l₁α₁∆t
dl₂ = l₂α₁∆t [∵ ∆t = change in temperature]
dl₃ = l₃α₂∆t
and l₁ = l₂ = l₃ = l
(l² α₁∆t + l² α₁ ∆t)cos θ + l² sin θ dθ
= l² α₁ ∆t + l² α₁ ∆t – l² α₂ ∆t
sin θ d θ = 2α₁∆t(1 – cos θ) – α₂ ∆t
Putting θ = 60°
dθ × sin 60° = 2 α₁∆t (1 – cos 60°) – α₂∆t
dθ × \(\frac{\sqrt{3}}{2}\) = 2 α₁∆t × \(\frac{1}{2}\) – α₂∆t
dθ × \(\frac{\sqrt{3}}{2}\) = (α₁ – α₂)∆t
change in angle, dθ = \(\frac{\left(\alpha_1-\alpha_2\right) \Delta t}{\frac{\sqrt{3}}{2}}\)
= \(\frac{2\left(\alpha_1-\alpha_2\right) \Delta t}{\sqrt{3}}\)

OR

We know that, as per Newton’s law of cooling,
\(\frac{\mathrm{dT}}{\mathrm{dt}}\) = – K(T – T_s)
\(\frac{d \mathrm{~T}}{\mathrm{~T}-\mathrm{T}_{\mathrm{s}}}\) = -K dt
\(\int_{\mathrm{T}_1}^{\mathrm{T}_2} \frac{d \mathrm{~T}}{\mathrm{~T}-\mathrm{T}_{\mathrm{s}}}\) = – \(\int_{t_1}^{t_2} \mathrm{~K} d t\)
log_e \(\left[\mathrm{T}-\mathrm{T}_s\right]_{\mathrm{T}_1}^{\mathrm{T}_2}\) = -K(t; – t₁)
As it is given in the first part of this question
T_s = 20°C,
T₁ = 80°C,
T₂ = 50°C,
t₂ – t₁ = 5 minutes = 300 s.
∴ log_e \(\left(\frac{50-20}{80-20}\right)\) = – K(300)
log_e \(\left(\frac{30}{60}\right)\) =-300K
-0.6931 = -300K
or K = 0.002310°C s^-1 ……………. (i)
Using this value of K for the second part of the question,
log_e \(\left(\frac{30-20}{60-20}\right)\) = -(0.002310) × t
– 0.002310t = – 1.38629
or t = 600 s = 10 min

Commonly Mode Error
Students could not evaluate the time required by the body to cool.

Answering Tip
Students should be familiarized with the concept of Newton’s law of cooling.