Students can access the CBSE Sample Papers for Class 11 Physics with Solutions and marking scheme Set 3 will help students in understanding the difficulty level of the exam.
CBSE Sample Papers for Class 11 Physics Set 3 with Solutions
Time Allowed : 3 hours
Maximum Marks : 70
General Instructions:
- There are 33 questions in all. All questions are compulsory.
- This question paper has five sections: Section A, Section B, Section C, Section D and Section E.
- All the sections are compulsory.
- Section A contains sixteen questions, twelve MCQ and four Assertion Reasoning based of 1 mark each, Section B contains five questions of two marks each, Section C contains seven questions of three marks each, Section D contains two case study based questions of four marks each and Section E contains three long answer questions of five marks each.
- There is no overall choice. However, an internal choice has been provided in one question in Section B, one question in Section C, one question in each CBQ in Section D and all three questions in Section E. You have to attempt only one of the choices in such questions
- Use of calculators is not allowed.
- You may use the following values of physical constants where ever necessary
- c = 3 × 108 m/s
- me = 9.1 × 10-31 kg
- µ0 = 4π × 10-7 TmA-1
- ε0 = 8.854 × 10-12 × C2 N-1 m-2
- Avogadro’s number = 6.023 × 1023 per gram mole
Section – A
Question 1.
One astronomical unit is a distance equal to
(A) 9.46 × 1015 m
(B) 1.496 × 1011 m
(C) 3 × 108 m
(D) 3.08 × 1016 m
Answer:
(B) 1.496 × 1011 m
Question 2.
The number of significant zeroes present in 0.010020, is
(A) Five
(B) Two
(C) One
(D) Three
Answer:
(D) Three
Question 3.
What is the ratio of the average acceleration during the intervals OA and AB in the velocity-time graph as shown below?
(A) \(\frac{1}{2}\)
(B) \(\frac{1}{3}\)
(C) 1
(D) 3
Answer:
(B) \(\frac{1}{3}\)
Explanation:
Average acceleration during OA = tan 30°
Average acceleration during AB = – tan 60°
The required ratio = \(\frac{\tan 30^{\circ}}{\tan 60^{\circ}}=\frac{1}{3}\)
Question 4.
A car travels \(\frac{2}{5}\)th of the total distance with a speed 10 ni/s and rest with a speed 20 m/s. The average speed is
(A) \(\frac{100}{7}\) m/s
(B) \(\frac{200}{7}\) m/s
(C) \(\frac{133}{7}\) m/s
(D) \(\frac{5}{7}\) m/s
Answer:
(A) \(\frac{100}{7}\) m/s
Explanation:
Total distance = d meter
Time taken to cover \(\frac{2 d}{5}\) distance = \(\frac{2 d}{50}\) s
Time taken to cover rest \(\frac{3 d}{5}\) s distance = \(\frac{3 d}{100}\) s
Total time = \(\frac{2 d}{50}+\frac{3 d}{100}=\frac{7 d}{100}\) s
Average speed = \(\frac{d}{\frac{7 d}{100}}=\frac{100}{7}\) m/s
Question 5.
If a particle moves from point P (2, 3, 5) to point Q (3, 4, 5). Its displacement vector is
(A) \(\hat{i}+\hat{j}+10 \hat{k}\)
(B) \(\hat{i}+\hat{j}+5 \hat{k}\)
(C) \(\hat{i}+\hat{j}\)
(D) \(2 \hat{i}+\hat{j}+6 \hat{k}\)
Answer:
(C) \(\hat{i}+\hat{j}\)
Explanation:
\(\overrightarrow{P Q}=(3-2) \hat{i}+(4-3) \hat{j}+(5-5) \hat{k}\)
= \(\hat{i}+\hat{j}\)
Question 6.
According to first law of thermodynamics
(A) ∆Q = ∆U + P∆V
(B) ∆Q = ∆U – P∆W
(C) ∆Q = (∆U + ∆V) P
(D) ∆Q = P∆U + ∆V
Answer:
(A) ∆Q = ∆U + P∆V
Explanation:
∆Q = ∆U + ∆W
= ∆U + P∆V
Question 7.
Stress vs. strain graphs for two materials A and B are shown below. The ration of the Young Modulus of the two materials is
(A) 3
(B) \(\frac{1}{3}\)
(C) √3
(D) \(\frac{1}{\sqrt{3}}\)
Answer:
(A) 3
Explanation:
γA = tan 60°
γB = tan 30°
\(\frac{Y_{\mathrm{A}}}{Y_{\mathrm{B}}}=\frac{\tan 60^{\circ}}{\tan 30^{\circ}}\) = 3
Question 8.
The spherical shape of rain drop is due to
(A) Atmospheric pressure
(B) Viscosity
(C) Density
(D) Surface Tension
Answer:
(D) Surface Tension
Explanation:
To minimize the surface energy at equilibrium, raindrops takes the shape of sphere. Since surface tension equals to the surface energy, surface tension is responsible due to this phenomenon.
Question 9.
A body is initially at rest. It undergoes one-dimensional motion with constant acceleration. The powerdelivered to it at time fis proportional to:
(A) t\(\frac{1}{2}\)
(B) t
(C) t\(\frac{3}{2}\)
(D) t2
Answer:
(B) t
Explanation:
Power delivered = P = Fv
Or, P = F (u + at)
Or, P = F × a × t (since u = 0)
Or, P = ma × a × t
∴ P ∝ t (since m and a are constants)
Question 10.
To make the frequency double of a spring oscillator
(A) Mass is to be halved
(B) Mass is to be doubled
(C) Mass is to be quadrupled
(D) Mass is to be reduced to one-fourth
Answer:
(D) Mass is to be reduced to one-fourth
Explanation:
υ ∝ \(\frac{1}{\sqrt{m}}\)
∴ \(\frac{v_1}{v_2}=\sqrt{\frac{m_2}{m_1}}\)
Or, \(\frac{1}{2}=\sqrt{\frac{m_2}{m_1}}\)
Or, \(\frac{1}{4}=\frac{m_2}{m_1}\)
∴ m2 = \(\frac{m_1}{4}\)
Question 11.
Two adjacent piano keys are struck simultaneously. The frequencies of sound emitted by υ1 and υ2. The number of beats produced
(A) \(\frac{1}{2}\) (υ1 + υ2)
(B) \(\frac{1}{2}\) (υ1 – υ2)
(C) (υ1 + υ2)
(D) (υ1 ~ υ2)
Answer:
(D) (υ1 ~ υ2)
Question 12.
Two superimposed sound waves produce beats when they have
(A) Slightly different amplitudes
(B) Slightly different frequencies
(C) Slightly different phases
(D) Slightly different velocities
Answer:
(B) Slightly different frequencies
For Questions 13 to 16 two statements are given – one labelled Assertion (A) and other labelled Reason (R). Select the correct answer to these questions from the options as given below.
(A) If both Assertion and Reason are true and Reason is correct explanation of Assertion.
(B) If both Assertion and Reason are true and Reason is not the correct explanation of Assertion.
(C) If Assertion is true but Reason is false.
(D) If both Assertion and Reason are false.
Question 13.
Assertion (A) : Number of significant figures in 0.005 is one and that in 0.500 is three.
Reason (R) : This is because zeros are not significant.
Answer:
(C) If Assertion is true but Reason is false.
Explanation:
In a number less then one, zero between the decimal point and first non zero digit are not significant. But zeros to the right of last non zero digit are significant.
Question 14.
Assertion (A) : Vector i + j + k is perpendicular to vector i – 2j + k .
Reason (R) : Two non-zero vectors are perpendicular if their dot product is equal to zero.
Answer:
(A) If both Assertion and Reason are true and Reason is correct explanation of Assertion.
Explanation:
Angle between two vectors = cos-1 \(\frac{(\hat{i}+\hat{j}+\hat{k})(\hat{i}-2 \hat{j}+\hat{k})}{|(\hat{i}+\hat{j}+\hat{k})(\hat{i}-2 \hat{j}+\hat{k})|}\)
= cos-1 \(\frac{1-2+1}{\sqrt{3 \sqrt{6}}}\)
= 90°
So, the assertion is true.
\(\vec{A} \cdot \vec{B}=|\vec{A}||\vec{B}|\)
if the vectors are non-zero, then the dot product will be 0 when cos Φ = 0, i.e. Φ = 90°, i.e., the vectors are perpendicular. So, the reason is also true and it explains the assertion.
Question 15.
Assertion (A) : Mass and energy are not conserved separately, but they are conserved as a single entity called mass- energy.
Reason (R) : Mass and energy are inter-convertible in accordance with Einstein’s relation E = mxc2
Answer:
(A) If both Assertion and Reason are true and Reason is correct explanation of Assertion.
Explanation:
Mass and energy are inter-convertible according to Einstein’s relation E = mc2. Hence, law of conservation of mass and law of conservation of energy can be stated as law of conservation of mass- energy which is more versatile.
Question 16.
Assertion (A) : A person standing on a rotating platform suddenly stretched his arms, the platform slows down.
Reason (R) : A person by stretching his arms increases the moment of inertia and decreases angular velocity.
Answer:
(A) If both Assertion and Reason are true and Reason is correct explanation of Assertion.
Explanation:
When the person suddenly outstretches his arms, the moment of inertia of the person increases. This is because the distribution of mass of the person changes due to the change in the position of the arms. But the angular momentum is conserved.
Since L = Iω, as I increases, L remains constant, the angular velocity (ω) will decrease. So the assertion and reason both are true and the reason explains the assertion.
Section – B
Question 17.
Give conventional rules for the rounding off of uncertain digits.
Answer:
The conventional rules are:
(i) If the insignificant digit to be dropped is more than 5, the preceding digit is increased by 1, but if it is less than five, then preceding digits is not changed.
e.g., 1.748 is rounded off to 3 significant figures as 1.75 and 1.742 as 1.74.
(ii) If the insignificant digit to be dropped is 5, then this digit is simply dropped if the preceding digit is even but if odd, then the preceding digit is increased by 1.
e.g., the number 1.845 rounded off to three significant digits is 1.84 but for number 1.875 it is 1.88.
Question 18.
What shall be the effect on the length of the day if the polar ice caps of earth melt ?
Answer:
Melting of polar ice caps will increase the radius of gyration and hence, M.I. also increase. In order ti conserve angular momentum, the angular velocity ω shall decrease.
So, the length of the day (T = \(\frac{2 \pi}{\omega}\)) shall increase.
Question 19.
What is phase in wave equation y = r sin (2π \(\frac{t}{T}\) + 2π \(\frac{x}{\lambda}\))
Answer:
Rewriting y = r sin 2π \(\left(\frac{t}{T}+\frac{x}{\lambda}\right)\).
Here, 2π \(\left(\frac{t}{T}+\frac{x}{\lambda}\right)\) is phase of the wave.
Question 20.
(a) What are harmonics ?
Answer:
Harmonics are the notes of frequencies which are integral multiple of the fundamental frequency.
(b) Is superposition principle applicable to electromagnetic waves ?
Answer:
Yes, it is applicable to electromagnetic waves.
Question 21.
A man standing on the edge of a cliff throws a stone straight up with initial speed u and then throw another stone straight down with same initial speed u and from the same position. Find ratios of speeds, two stones would attain, when they hit ground at the base of the cliff.
Answer:
The stone thrown upward reaches back to the thrower with speed u.
Thus both the stones fall under the influence of gravity with same initial velocity u so the two stones will hit the ground with same speed. Hence, ratio of their speed when they hit the ground is 1.
OR
Two satellites A and B go around a planet in circular orbits having radius 4R and R respectively. If the speed of the satellite A is 3v, find the speed of the satellite B.
Answer:
As, υ0 = \(\sqrt{\frac{G M}{R}}\)
so, For A,
3υ = \(\sqrt{\frac{G M}{4 R}}\)
and For B
υ’ = \(\sqrt{\frac{G M}{R}}\)
Dividing (2) by (1) we get,
∴ \(\frac{v^{\prime}}{3 v}\) = 2
or υ’ = 6υ.
Section – C
Question 22.
What are the conditions under which a rocket fired from the Earth, launches an artificial satellite of Earth ?
Answer:
Following are the basic conditions:
- The rocket must take the satellite to a suitable height above the surface of Earth.
- From the desired height, the satellite must be projected with a suitable velocity, called the orbital velocity.
- In the orbital path of satellite, the air resistance should be negligible so that its velocity does not decrease and it does not burn due to the heat produced.
Question 23.
(a) Why does an iron needle float on clean water but sink when some detergent is added to this water ?
Answer:
Due to surface tension, the free surface of liquid at rest behaves like a stretched membrane. When an iron needle floats on the surface of clean water, its weight is supported by the stretched membrane. When some detergent is added to this water, its surface tension decreases.
As a result of it, the stretched membrane on the surface of water is weakened and is not be able to support the weight of needle. Hence, needle sinks in such water.
(b) What is the effect of temperature on surface tension ?
Answer:
Surface tension of liquids decreases with increase in temperature.
Question 24.
(a) Write the relation among α, β and γ.
Answer:
β = 2α
and γ = 3α
α : β : γ = 1 : 2 : 3
where,
α → Coefficient of linear expansion,
β → Coefficient of superficial expansion,
γ → Coefficient of cubical or volume expansion.
(b) Why is a gap left between the ends of two railway lines in a railway track ?
Answer:
It is done to accommodate the linear expansion of railway line during summer. If the gap is not left, in summer, the lines will bend causing a threat of derailment.
Question 25.
(a) Two identical rectangular strips one of copper and the other of steel are riveted to form a bimetallic strip. What will happen on heating ?
Answer:
Coefficient of linear expansion of copper is more than steel. On heating, the expansion in copper strip is more than the steel strip.
The bimetallic strip will bend with steel strip on inner.
(b) What is latent heat energy ?
Answer:
Latent heat energy is the heat energy required to change the state of 1 kg of substance from one to another at constant temperature.
Question 26.
(a) Two rods A and B are of equal lengths. Each rod has its ends at temperature T1 and T2. What is the condition that will ensure equal rates of flow of heat through the rods A and B ?
Answer:
Since, Q = KA (θ1 – θ2)t/x
QA/t = KAAA (θ1 – θ2)/x
QB/t = KBAB (θ1 – θ2)/x
For equal rates of flow
QA/t = QB/t
or KAAA = KBAB.
(b) Why are the bottoms of cooking vessels painted black ?
Answer:
Black paint absorbs more heat. The black painted bottoms absorbs more heat and help in cooking meals faster.
Question 27.
Discuss three important characteristics of wave motion.
Answer:
The important characteristics of wave motion:
- A material medium is essential for the propagation of mechanical waves. The medium must posses three properties, viz., elasticity, inertia and minimum friction amongst the particles of the medium.
- Energy is propagated along with the wave motion without any net transport of the medium.
- There is a continuous phase difference amongst successive particles of the medium.
Question 28.
List the S.I. base quantities and find their units with symbols.
Answer:
Base quantity | S.I. unit | Symbol |
(i) Length | metre | m |
(ii) Mass | kilogram | kg |
(iii) Time | second | s |
(iv) Electric current | ampere | A |
(v) Temperature | kelvin | K |
(vi) Amount of substance | mole | mol |
(vii) Luminous intensity | candela | cd |
OR
(i) What can be represented by the graph given below, where d is height and v is velocity ?
Answer:
This graph can be for a ball dropped vertically from a height d.
It hits the ground with some downward velocity and bounces up to height d/2 where its upward velocity becomes zero.
(ii) How many dimensions of motion does the following have ?
(a) Train moving fast on its track.
(b) A lizard moving on a wall in a room.
(c) Kite flying in the sky.
(d) Bee flying in a closed room.
Answer:
(a) One dimensional motion
(b) Two dimensional motion
(c) Three dimensional motion
(d) Three dimensional motion.
Section – D
Question 29.
Read the following text and answer the following questions on the basis of the same:
A black hole is a place in space where gravity pulls so much that even light can not get out. The gravity is so strong because matter has been squeezed into a tiny space.
From Newton’s universal law of gravitation, F = \(\frac{G M m}{R^2}\)
The photons have no mass. From Einstein’s mass-energy equivalence relation E = mc2
According to quantum theory of radiation,
E = hυ
∴ m = \(\frac{h}{c \lambda}\)
Putting in the expression of Newton’s law of gravitation,
Force acting on the light particles = G \(\frac{M h}{c \lambda R^2}\)
Because no light can get out, people can’t see black holes. They are invisible. Space telescopes with special tools can help find black holes.
A beam of light just grazing the edge of the sun also bend a little. This may be observed during total solar eclipse. The position of a star will appears closer to the sun.
Black holes can be big or small. Scientists think the smallest black holes are as small as just one atom.
These black holes are very tiny but have the extremely high density. Another kind of black hole is called “stellar.” Its mass can be up to 20 times more than the mass of the sun. There may be many, many stellar mass black holes in Milky Way. The largest black holes are called “super massive.” These black holes have masses those are more than 1 million suns together. Scientists have found proof that every large galaxy contains a super massive black hole at its centre. The super massive black hole at the centre of the Milky Way galaxy is called Sagittarius A. It has a mass equal to about 4 million suns and would fit inside a very large ball that could hold a few million Earth. A black hole cannot be seen because strong gravity pulls all of the light into the middle of the black hole. But scientists can see how the strong gravity affects the stars and gas around the black hole. If a black hole of the same mass as the sun replaces the sun, Earth and the other planets will orbit the black hole as they orbit the sun now.
(i) The gravity of black hole is so strong because
(A) Its density is too low
(B) Its density is too high
(C) Its volume is too small
(D) Both (A) and (C) Ail
Answer:
(B) Its density is too high
Explanation:
The gravity of black hole is so strong because matter has been squeezed into a tiny space i.e., density is too high.
(ii) The super massive black hole at the center of the Milky Way galaxy is
(A) Sagittarius A
(B) Sagittarius B2
(C) Cygnus X – 1
(D) Cygnus X – 3
Answer:
(A) Sagittarius A
Explanation:
The largest black holes are called “super massive.” These black holes have masses more than 1 million Sun’s together. Scientists have found proof that every large galaxy contains a super massive black hole at its centre. The super massive black hole at the centre of the Milky Way galaxy is called Sagittarius A. It has a mass equal to about 4 million suns.
(iii) What will happen to the earth and other planets if a black hole of same mass as the sun replaces the sun?
(A) Earth and other planets will fall into the black hole
(B) Earth and other planets will orbit the black hole as they orbit the Sun now
(C) The orbits of earth and other planets will be much shorter.
(D) Earth and other planet will stop orbiting.
Answer:
(B) Earth and other planets will orbit the black hole as they orbit the Sun now
Explanation:
If a black hole of same mass as the sun replace the sun, Earth and the other planets will orbit the black hole as they orbit the sun now.
OR
The attractive force acting on a light particle by a black hole is
(A) G \(\frac{M h}{c \lambda R^2}\)
(B) G \(\frac{M c h}{h R^2}\)
(C) G \(\frac{M c}{h \lambda R^2}\)
(D) G \(\frac{M \lambda}{c h R^2}\)
Answer:
(A) G \(\frac{M h}{c \lambda R^2}\)
Explanation:
From Newton’s universal law of gravitation,
F = \(\frac{G M m}{R^2}\)
The photons have no mass. From Einstein’s mass-energy equivalence relation
E = mc2
According to quantum theory of radiation,
E = hn
∴ m = \(\frac{h}{c \lambda}\)
Putting in the expression of Newton’s law of gravitation,
Force acting on the light particles = \(\frac{G M h}{c \lambda R^2}\)
(iv) A beam of light just grazing the edge of the sun also bend a little. It is observable during
(A) Total solar eclipse
(B) Partial solar eclipse
(C) Full lunar eclipse
(D) Partial lunar eclipse
Answer:
(A) Total solar eclipse
Explanation:
How to observe a beam of light just grazing the edge of the sun bends? This can be observed during a total solar eclipse. Carefully measuring the location of a star, it appears closer to the sun. In 1919, teams of astronomers observed the solar eclipse from Brazil and from an island in the Atlantic. They intercepted the path of the total solar eclipse, and carefully photographed the positions of nearby stars. Several months later, they announced that Einstein was right! The rays really bend!
Question 30.
Read the following text and answer any 4 of the following questions on the basis of the same:
An object is dropped from the top of a tower, it is accelerated downward under the influence of the force of gravity. The acceleration developed in the body is equal to the acceleration due to gravity. If air resistance is neglected, the object is said to be in free fall and in such situation time of fall is irrespective of the dimension of the body, its mass and density. Free fall is thus a case of motion with uniform acceleration.
At the time of drop, the velocity of the body is 0. As the body comes down, the velocity increases and it becomes maximum just before touching the earth.
(i) What is the acceleration of a freely falling body?
(A) 9.8 m/s2
(B) Less than 9.8 m/s2
(C) More than 9.8 m/s2
(D) Depends on the mass of the body
Answer:
(A) 9.8 m/s2
Explanation:
For a freely falling body, the acceleration developed in the body is equal to the acceleration due to gravity of the earth.
(ii) Under which consideration, the acceleration developed in an object is taken to be constant?
(A) Height through which the object falls is greater than earth’s radius.
(B) Height through which the object falls is smaller than earth’s radius.
(C) The mass of the object is negligible.
(D) Both (A) and (D) Ail
Answer:
(B) Height through which the object falls is smaller than earth’s radius.
Explanation:
If the height through which the object falls is small compared to the earth’s radius, the acceleration developed in the object i.e., g can be taken to be constant.
(iii) Which of the following equation of motion is correct for free fall?
(A) Velocity = v = 9.8t
(B) Height travelled = h = 9.8t
(C) Velocity = v = \(\sqrt{(19.6 h)}\)
(D) Both (A) and (C)
Answer:
(D) Both (A) and (C)
Explanation:
In the equation of motion v = u + at, if we put u = 0 and a = g = 9.8 for free fall, then v = 9.8t
In the equation of motion, v2 = u2 + 2 as, if we put u = 0 and a = g = 9.8 for free fall, then
v = \(\sqrt{19.6h}\)
(iv) For a free fall which one of the following represents the v-t graph properly?
(A)
(B)
(C)
(D)
Answer:
(A)
Explanation:
Equation of motion:
v = u + at
putting u = 0, a = g = 9.8
v = 9.8t
So, it is an equation of a straight time passing through origin.
OR
Which of the following statements is true?
(A) During free fall air resistance is neglected.
(B) Free fall is a motion with uniform acceleration.
(C) Acceleration remains constant throughout the motion.
(D) All of the above
Answer:
(D) All of the above
Section – E
Question 31.
Derive the relations:
(i) v = u + at
(ii) s = ut + \(\frac{1}{2}\) at2
(iii) v2 = u2 + 2as
Answer:
(i) Derivation of v = u + at
Let, u = Initial velocity
v = Final velocity
a = Acceleration
a = \(\frac{(v-u)}{t}\) (By definition of acceleration)
∴ v = u + at ……………..(i)
(ii) Derivation of s = ut + \(\frac{1}{2}\) at2
Let, u = Initial velocity
y = Final velocity
a = Acceleration
Average velocity = \(\frac{v+u}{2}\)
Distance covered = s = average velocity × time
or, s = \(\frac{v+u}{2}\) × t
or, s = \(\frac{u^2+a t+u}{2}\) × t (using equation (i))
or, s = \(\frac{2 u+a t}{2}\) × t
∴ s = ut + \(\frac{1}{2}\) at2 ………………(ii)
(iii) Derivation of v2 = u2 + 2as
v = u + at
or, v2 = (u + at)2
or, v2 = u2 + 2uat + a2t2
or, v2 = u2 + 2a (ut + \(\frac{1}{2}\) at2)
∴ v2 = u2 + 2as
(using equation (ii))
OR
The velocity – time graph of a particle is given by
(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 – 28s = – 16 m.
(ii) Time for acceleration : 0 ≤ t ≤ 4 and – 4 ≤ t ≤ 0
Time for retardation : 4 ≤ t ≤ 8
Time for constant velocity : 8 ≤ t ≤ 12
(iii) Acceleration (0 ∈ t ∈ 4) = \(\frac{\text { Difference in velocity }}{\text { Difference in time }}\)
= \(\frac{4}{4}\)
= 1 m/s2
Acceleration (- 4 ≤ t ≤ 0) = \(\frac{[0-(-4)]}{4}\) = 1 m/s2
Retardation = \(\frac{-4}{2}\) = – 2 m/s2
(iv)
Question 32.
(a) Derive an expression for the gravitational potential energy above the surface of Earth.
Answer:
Let the body of mass n, be taken at height h above the surface of Earth. At any instant of time t it reaches at a distance x from the centre of Earth.
The work done in raising through dx is.
dW = \(\frac{G M m}{x^2}\) . dx
= ∆ (P.E.)
Hence, the work done in taking the body from surface of the Earth (x = R) to a height h (x = R + h) is
P.E. = W
= \(\int_R^{R+h} \frac{G M m}{x^2}\) dx
= GMm \(\int_{\mathrm{R}}^{\mathrm{R}+h} \frac{1}{x^2}\) dx
PE. above the surface of Earth = mgh
(b) Will 1 kg sugar be more at poles or at the equator ?
Answer:
The value of g is larger at the poles than at the equator. if the sugar is weighed in a common balance then there will be no difference. If it is weighed by a spring balance, calibrated at the equator, then 1 kg of sugar will have a lesser amount at poles.
OR
(a) Two identical spheres are separated by a distance 10 times of their radius. A third sphere is placed at the mid point of the line joining their centres. Will the third sphere be in stable equilibrium? Explain your answer.
Answer:
Given:
m1 = m2 = M, r = 10R
Let mass m is placed at mid-point A (line joining the centres of P & Q sphere)
Now, = |F1| = \(\frac{G M m}{(5 R)^2}\)
|F1| = |F2| = \(\frac{G M m}{25 (R)^2}\)
F1 & F2 are equal and opposite forces are acting on m at A.
Net force F1 = F2
or F1 + F2 = 0
So, mass m is in equilibrium.
If m is slightly displaced x from A towards P then
F1 = \(\frac{G M m}{(5 R-x)^2}\)
F2 = \(\frac{G M m}{(5 R+x)^2}\)
or F1 > F2
(b) Why are space rockets usually launched from west to east in the equatorial line ?
Answer:
We know that Earth revolves from west to east about its polar axis. Therefore, all the particles on the Earth have velocity from the west to east. This velocity is maximum in the equatorial line, as u = Rω, where R is the radius of Earth and ω is the angular velocity of revolution of Earth about its polar axis.
When a rocket is launched from west to east in equatorial plane, the maximum linear velocity is added to the launching velocity of the rocket, due to which launching becomes easier.
Question 33.
Derive the expression for heat flow through a compound wall.
Answer:
Consider a compound wall (or a slab) made of two materials A and B of thickness d1 and d2.
Let K1 and K2 be the coefficients of thermal conductivity and θ1 and θ2 are the temperatures of the end faces (θ1 > θ2) and θ is the temperature of the surface in contact.
For material A,
Q1 = \(\frac{K_1 A_1\left(\theta_1-\theta\right)}{d_1}\) ……………(i)
For the material B,
Q2 = \(\frac{K_2 A_2\left(\theta-\theta_2\right)}{d_2}\) ……………..(ii)
From eqn. (i) and (ii),
Q1 = Q2
\(\frac{K_1 A_1\left(\theta_1-\theta\right)}{d_1}=\frac{K_2 A_2\left(\theta-\theta_2\right)}{d_2}\)
θ = \(\frac{\frac{K_1 \theta_1}{d_1}+\frac{K_2 \theta_2}{d_2}}{\frac{K_1}{d_1}+\frac{K_2}{d_2}}\)
Considering
A1 = A2
Substituting the value θ in eqn. (i),
Q1 = \(\frac{A\left(\theta_1-\theta_2\right)}{\frac{d_1}{K_1}+\frac{d_2}{K_2}}\)
In general for any no. of walls
Q = \(\frac{A\left(\theta_1-\theta_2\right)}{\Sigma\left(\frac{d}{K}\right)}\)
OR
(a) Derive the equation of adiabatic changes.
Answer:
For 1 mole ideal gas,
PV = RT
Differentiating,
PdV + VdP = RdT
or PdV + VdP = (CP – CV) dT
or CvdT + PdV = CpdT – VdP …………….(1)
From the 1st equation of thermodynamics,
dQ = dU + dW
puttting dU = CvdT and dW = PdV
dQ = CvdT + PdV = CpdT – VdP
(from equation 1) ………..(2)
In adiabatic process, dQ = 0
∴ 0 = CpdT – VdP (Putting in equation 2)
∴ CpdT = VdP …………..(3)
Also
CvdT + PdV = 0
(Putting in equation 2)
∴ CvdT = – PdV ……………..(4)
Dividing equation 3 by equation 4,
\(\frac{C_p}{C_v}=-\frac{V}{P} \frac{d P}{d V}\)
or γ = – \(-\frac{V}{P} \frac{d P}{d V}\)
or \(\frac{d P}{P}+\gamma \frac{d V}{V}\) = 0
Integrating,
In P + γ ln V = ln c
(in C is an integration constant)
or, In PVγ = In c = constant
∴ PVγ = constant
(b) By applying the first law of thermodynamics to isobaric process, obtain relation between two specific heats of a gas.
Answer:
In an isobaric process, pressure remains constant.
If an amount of heat dQ is supplied to one mole of a gas at constant pressure and its temperature increases by r, then
dQ = Cp dT
Here, C is molar specific heat of the gas at constant pressure. Therefore, for an isobaric process, the first law of thermodynamics becomes:
Cpdt = dU + PdV ……………..(i)
From perfect gas equation it follows that
PdV = RdT
In the eqn. (i), substituting PdV and dU, we have
CpdT = CVdT + RdT
Cp = CV + R