Students can access the CBSE Sample Papers for Class 11 Physics with Solutions and marking scheme Set 2 will help students in understanding the difficulty level of the exam.
CBSE Sample Papers for Class 11 Physics Set 2 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.
Increase in length is ∆l of a wire of length L by longitudinal stress. The stress is proportional to
(A) \(\frac{L}{\Delta l}\)
(B) \(\frac{\Delta l}{L}\)
(C) (∆l)2
(D) \(\left(\frac{\Delta I}{L}\right)^2\)
Answer:
(B) \(\frac{\Delta l}{L}\)
Explanation:
Stree ∝ strain
Strain = \(\frac{\Delta l}{L}\)
So, stress ∝ \(\frac{\Delta l}{L}\)
Question 2.
Wavelength of a radiation is 0.00008 m. It is equal to
(A) 80 micron
(B) 8 micron
(C) 0.8 micron
(D) 0.08 micron
Answer:
(A) 80 micron
Explanation:
0.000008 m = 8 × 10-5 m
= 80 × 10-6 m
= 80 micron
Question 3.
A person travels half of the distance of a straight road with velocity v1 and other half with a velocity v2. His average velocity is
(A) \(\frac{\left(v_1+v_2\right)}{2}\)
(B) \(\sqrt{v_1 v_2}\)
(C) \(\frac{2 v_1+v_2}{\left(v_1+v_2\right)}\)
(D) \(\sqrt{v_1^2+v_2^2}\)
Answer:
(C) \(\frac{2 v_1+v_2}{\left(v_1+v_2\right)}\)
Explanation:
Length of the road = d
\(\frac{d}{2}\) distance travelled with velocity v1.
Time taken = \(\frac{d}{2 v_2}\)
\(\frac{d}{2}\) distance travelled with velocity v2.
Time taken = \(\frac{d}{2 v_2}\)
Total time taken = \(\frac{d}{v}+\frac{d}{v}\)
= \(\frac{d v+d v}{v v}\)
Average velocity = \(\frac{\text { Total distance }}{\text { Total time }}\)
= \(\frac{\frac{d}{d v+d v}}{v v}\)
= \(\frac{2 v_1 v_2}{v_1+v_2}\)
Question 4.
The angle between \(\vec{A}=2 \hat{i}+3 \hat{j}+8 \hat{k}\) and \(\vec{B}=4 \hat{j}+4 \hat{j}+a \hat{k}\) is 90°. The value of a is
(A) \(\frac {1}{2}\)
(B) – \(\frac{1}{2}\)
(C) 2
(D) – 2
Answer:
(B) – \(\frac{1}{2}\)
Explanation:
Since the angle between the two vectors is 90°, hence their dot product is 0.
\(\overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{B}}\) = 0
or, \((\hat{i}+3 \hat{j}+8 \hat{k}) \cdot(-4 \hat{i}+4 \hat{j}+a \hat{k})\) = 0
or, – 8 + 12 + 8a = 0
∴ a = – \(\frac{1}{2}\)
Question 5.
A person sitting in a open car, moving with a constant velocity, throws a ball vertically upward. Which one of the following will happen?
(A) The ball will fall outside the car.
(B) The ball will fall in front of the man.
(C) The ball will fall behind the man
(D) The ball will fall in the hands on the man.
Answer:
(D) The ball will fall in the hands on the man
Explanation:
Both ball and car have the same horizontal velocities, both will cover same horizontal distances and hence the ball will fall in the hands of the man.
Question 6.
A cyclist turns around a curve at 20 km/hour. If he turns at double the speed the tendency to overturn will be
(A) Double
(B) Quadrupled
(C) Halved
(D) Unchanged
Answer:
(B) Quadrupled
Explanation:
Centripetal force = F = \(\frac{m v^2}{r}\)
So, F ∝ v2
As the speed becomes double, the force becomes quadrupled and hence the tendency to over turn will also be quadrupled.
Question 7.
Distance between two masses is doubled. The gravitational attraction between them will be.
(A) Doubled
(B) Quadrupled
(C) Halved
(D) None of these
Answer:
(D) None of these
Explanation:
F = \(\frac{m m}{r}\)
∴ F ∝ \(\frac{m}{r}\)
So, when the distance is doubled, the gravitational attraction between the masses will be one fourth.
Question 8.
The total mechanical energy of S.H.M. is
(A) Zero at extreme position
(B) Zero at the mean position
(C) Negative
(D) Constant
Answer:
(D) Constant
Explanation:
Total mechanical energy throughout the motion is – mω2A2.
Question 9.
Diagram shows the propagation of a wave. Which points are in same phase?
(A) 6 and 7
(B) 6 and 5
(C) 3 and 7
(D) 2 and 6
Answer:
(D) 2 and 6
Explanation:
Displacements of the points 2 and 6 are same is sign and magnitude.
Question 10.
The fundamental note produced by one end closed pipe is 900 Hz. If the closed end of the pipe is opened the fundamental note produced will be
(A) 900 Hz
(B) 450 Hz
(C) 1800 Hz
(D) 2700 Hz
Answer:
(B) 450 Hz
Explanation:
Fundamental note produced by both ends open pipe = 2 × fundamental note produced by one end closed pipe.
Question 11.
3 kcal heat is given to a system which performs 1000 J work. The change in internal energy is
(A) 10600 J
(B) 12600 J
(C) 13600 J
(D) None of these
Answer:
(D) None of these
Explanation:
∆Q = ∆U + ∆W
Or, 3000 × 4.2 = ∆U + 1000
Or, 12600 – 1000 = ∆U
∴ ∆U = 11600 J
Question 12.
According to kinetic theory of gas, the pressure exerted by a gas on the wall of the container is
(A) Change of momentum per unit volume
(B) Change of momentum per unit area
(C) Total change of momentum
(D) Rate of change of momentum per unit area
Answer:
(D) Rate of change of momentum per unit area
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) : Light year and wavelength of light both represent distance.
Reason (R) : Both have the dimension of time
Answer:
(C) If Assertion is true but Reason is false.
Explanation:
Light-year is the distance travelled by light in one year. Wavelength of light wave is the distance over which the wave shape repeats.
Hence,both represent distance. The assertion is true.
Both have the dimension is length. So, the reason is false.
Question 14.
Assertion (A) : In a projectile motion, the angle between the instantaneous velocity and acceleration at the highest point is 0°.
Reason (R) : At the highest point, the projectile has no horizontal component.
Answer:
(D) If both Assertion and Reason are false.
Explanation:
A projectile has only horizontal component of velocity at the highest point. Acceleration is acting downward. Hence, the angle between the instantaneous velocity and acceleration is 90°.
So, the assertion and reason both are false.
Question 15.
Assertion (A) : All oscillatory motions are periodic, but all periodic motions are not oscillatory.
Reason (R) : Simple pendulum executes oscillatory motion.
Answer:
(B) If both Assertion and Reason are true and Reason is not the correct explanation of Assertion.
Explanation:
In oscillatory motion, there is a concept of mean position through which the particle executes a periodic motion. The particle passes through the mean position in very \(\frac{T}{4}\) time.
There is non-oscillatory type of periodic motion also in which there is no concept of mean position. So, all periodic motions are not oscillatory. Hence the assertion is true.
Simple pendulum is an example of oscillatory motion. The reason is also true but it does not explain the assertion.
Question 16.
Assertion (A) : When a simple pendulum oscillates on the surface of moon its time period increases.
Reason (R) : On moon there is no atmosphere.
Answer:
(B) If both Assertion and Reason are true and Reason is not the correct explanation of Assertion.
Explanation:
Time period = T = 2π \(\sqrt{\frac{l}{g}}\)
In moon the value of g is less than that of earth. Hence the time period increases. The assertion is true.
In moon there is no atmosphere. The reason is also true, but it does not explain the assertion.
Section – B
Question 17.
Why an optical microscope is not used to measure the size of a molecule ?
Answer:
Size of molecule ranges from 10-8 m to 10-10 m. An optical microscope uses visible light of average wavelength 6000 Å, i.e., 6000 × 10-10 m to measure the sizes. Since, size of molecule is smaller than the wavelength of light used, so optical microscope cannot resolve a molecule.
Question 18.
Explain why waves on strings are always transverse ?
Answer:
A string is non-stretchable, i.e., compressions and rarefactions cannot be produced in strings. Therefore, longitudinal waves in strings are not possible. Strings do have elasticity of shape. Therefore, waves on strings are transverse.
Question 19.
A bullet weighing 10 g is fired with a velocity of 800 ms-1. After passing through a mud wall 1 m thick, its velocity decreases to 100 ms-1. Find the average resistance offered by the mud wall.
Answer:
Using F.s = \(\frac{1}{2}\) mv2 – \(\frac{1}{2}\) mu2
where m = 10 g = \(\frac{10}{1000}\) kg
= 0.01 kg
and v = 100 ms-1, u = 800 ms-1
and s = 1 m,
we get F = \(\frac{1}{2}\) × 0.01 × (1002 – 8002)
= – 3150 N.
Question 20.
What will happen if a satellite stops orbiting ?
Answer:
When a satellite stops orbiting, its kinetic energy becomes zero. This energy is converted into potential energy of the satellite. It then, starts falling towards the Earth while crossing the atmosphere with great speed and it may even burn.
Question 21.
The displacement of a body is proportional to t3, where t is time elapsed. What is the nature of acceleration-time graph of the body?
Answer:
s = kt3
or, v = \(\frac{d s}{d t}\)
or, v = 3kt2
or, a = \(\frac{d v}{d t}\)
∴ a = 6kt
Acceleration is directly proportional to time, so a – t graph is a straight line.
OR
If \(\vec{A}+\vec{B}=\vec{C}\) prove that C = (A2 + B2 + 2AB cos θ)1/2, where θ is the angle between A and B ?
Answer:
Section – C
Question 22.
(a) Two different gases have exactly the same temperature. Does this mean that their molecules move with the same r.m.s. speed?
Answer:
When the two gases have exactly the same temperature, the average kinetic energy per molecule for each gas is the same. But as the different gases may have molecules of different masses, the r.m.s. speed (c) of molecules of different gases shall be different.
(b) Mention the different ways of increasing the number of molecular collisions per unit time in a gas.
Answer:
The number of collisions per unit time can be increased by
(i) increasing the temperature of the gas.
(ii) increasing the number of molecules, and
(iii) decreasing the volume of the gas.
Question 23.
Give an analytical method to find the vector sum of three vectors \(\vec{P}, \vec{Q} \text { and } \vec{R}\).
Answer:
Let \(\vec{P}, \vec{Q} \text { and } \vec{R}\) be represented in component form, i.e.,
\(\vec{P}=P_x \hat{i}+P_y \hat{j}+P_z \hat{k}\)
\(\vec{Q}=Q_x \hat{i}+Q_y \hat{j}+Q_z \hat{k}\)
\(\vec{R}=R_x \hat{i}+R_y \hat{j}+R_z \hat{k}\)
Let \(\vec{S}\) be their summation vector, i.e.,
\(\vec{S}=\vec{P}+\vec{Q}+\vec{R}\)
= \(\left(P_x+Q_x+R_x\right) \hat{i}+\left(P_y+Q_y+R_y\right) \hat{j}+\left(P_z+Q_z+R_z\right) \hat{k}\)
Addition of vectors obey the commutative as well as associative laws
∴ Sx = Px + Qx + Rx,
Sy = Py + Qy + Ry, and
Sz = Pz + Qz + Rz
Question 24.
A raindrop of mass 1 g falling from a height of 1 km hits the ground with a speed of 50 ms-1. Calculate
(a) the loss of P.E. of the drop.
Answer:
Given mass of rain drop, (m) = 0.001 kg
= 1.0 × 10-3 kg
Height, h = 1 km = 1000 m
Speed, v = 50 m/s, u = 0.
Loss of PE = mgh
= 1 × 10-3 × 10 × 10-3 = 10 J
(b) the gain in K.E. of the drop.
Answer:
(b) Gain in KE = \(\frac{1}{2}\) mv2
= \(\frac{1}{2}\) × 10-3 × 2500
= 1.25 j
(c) Is the gain in K.E. equal to loss of P.E.? If not why.?
Answer:
No, because a part of PE is used up in doing work against the viscous drag of air.
Question 25.
(a) What is column pressure ?
Answer:
Pressure exerted by a liquid due to its height is called column pressure.
(b) What are absolute pressure and gauge pressure ?
Answer:
Pressure at a point is given by the relation
P = Pa + hρg
Where, Pa is the atmospheric pressure and hpg is the column pressure.
Here, P is the absolute pressure and (P – Pa) is the gauge pressure normally measured.
Question 26.
(a) Why does the free surface of a liquid behave like an elastic stretched membrane ?
Answer:
The liquid molecules on the surface experience a downward force. So, they have greater potential energy.
(b) When wax is rubbed on cloth, the cloth becomes water proof. Why ?
Answer:
In order to have minimum energy, the free surface tends to contract to minimum area and hence behaves like an elastic stretched membrane. The capillaries formed in threads disappear when wax is rubbed on cloth.
Question 27.
An air bubble of radius r rises steadily through a liquid of density p at the rate of v. Neglecting density of air, find the coefficient of viscosity of liquid.
Answer:
Buoyant force = weight of liquid displaced
= \(\frac{4}{3}\) πr3ρg
Viscous force = Stoke’s drag force
= 6πηrv
6πηrv = \(\frac{4}{3}\) πr3ρg
η = \(\frac{2}{9} \frac{r^2 \rho g}{v}\)
Question 28.
(a) What is periodic wave function ?
Answer:
A wave function (x, t) which satisfies the periodicity conditions of position and time is called a periodic wave function, i.e.,
(i) y (x + mλt) = y (x, t)
(ii) y (x, t + nT) = y (x, t)
Where λ = wavelength of wave,
T = period of the wave, n and in are integers.
(b) Is it possible to have transverse wave in a steel rod ?
Answer:
Yes, transverse waves are possible in a steel rod, because steel has elasticity of shape.
OR
(a) Why are longitudinal waves called pressure waves?
Answer:
This is because propagation of longitudinal waves through a medium involves changes in pressure and volume of au when compressions and rarefactions are formed.
(b) Write down the expression for speed of transverse waves in solids and in a stretched string.
Answer:
In a solid:
Speed of transverse wave, v = \(\sqrt{\frac{\eta}{\rho}}\)
Where, η is modulus of rigidity and ρ is density of material a soild.
In a stretched string:
Here, v = \(\sqrt{\frac{T}{m}}\)
Where T is tension in the string and in is mass per unit length of the string.
Section – D
Question 29.
Read the following text and answer the following questions on the basis of 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 – on6 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.
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 = 1/2 Iω2
where, I = moment of inertia
and ω = angular velocity.
(i) Which of the following statement is true?
(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:
(A) Moment of inertia depends on the total mass and the distribution of mass from the axis of rotation
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
(ii) How many flywheels are there in foot operated sewing machine ?
(A) One
(B) Two
(C) Three
(D) Zero
Answer:
(B) Two
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.
(iii) 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
(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:
(C) It will run for longer period when the driving force is stopped
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.
OR
Energy stored in flywheel is
(A) \(\frac {1}{2}\) mv2
(B) \(\frac {1}{2}\) mω2
(C) \(\frac {1}{2}\) Iv2
(D) \(\frac {1}{2}\) Iω2
Answer:
(D) \(\frac {1}{2}\) Iω2
Explanation:
Kinetic energy of an object moving in a straight line is
E = – mv2
The kinetic energy of a spinning object is
E = – Iω2
(iv) Which one of the following wheel (having same mass) will have highest moment of inertia about axis passing
(A)
(B)
(C)
(D)
Answer:
(A)
Explanation:
Mass of all the wheels are same. In wheel D, the mass is distributed furthest from the axis. Since moment of inertia = Σ mr2, hence wheel D will have highest moment of inertia.
Question 30.
Read the following text and answer any 4 of the following questions on the basis of the same:
Student “A” recorded 5 reading of measurement of length of a rod. The average value obtained was 2.6250 m. He rounded off it to 2.63 m. His teacher commented that the result was not correct. So, he repeated the experiment. This time he obtained the average value 2.6350 m. He rounded off it to 2.63 m. This time also his teacher commented that the result was not correct.
“A” had a peep into the readings of his friend “B”. “B” obtained an average value 2.6750 m and rounded it off to 2.68 m. Teacher remarked that B’s result was perfect.
(i) Mass of an object is 11.25 g and its volume is 2 cm3. What should be the most significant representation of its density?
(A) 5.625 g/cm3
(B) 5.63 g/cm3
(C) 5 g/cm3
(D) 5.62 g/cm3
Answer:
(C) 5 g/cm3
Explanation:
During multiplication and division the significant digit in the result should be equal to the lowest significant digit of the operands. In this problem, significant digit of volume is 1 and it is the lowest. So, the most significant result should have 1 significant digit.
(ii) What should be the correct result of student “A” after proper rounding off to two decimal places when he
(A) 2.62
(B) 2.63
(C) 2.60
(D) None of these
Answer:
(A) 2.62
Explanation:
If the digit from where the number is truncated is 5, the digit after it is 0 and the digit previous to it is even then the digit should remain unchanged.
In the problem, 2.6250 the number is to be truncated at 5. The digit after it is 0. The digit just before it is 2, which is an even digit. So, as per rule there should not be any change. The result after rounding off should be 2.62.
(iii) What should be the correct result of student “A” after proper rounding off to two decimal places when he obtained an average value 2.6350?
(A) 2.60
(B) 2.63
(C) 2.64
(D) None of these
Answer:
(C) 2.64
Explanation:
If the digit from where the number is truncated is 5, digit after it is O and the digit previous to it is odd then the digit is to be increased by 1.
In the problem, 2.6350 the number is to be truncated at 5. The digit after it is 0. The digit just before it is 3, which is an odd digit. So, as per rule the digit 3 is
to be increased by 1. The result after rounding off should be 2.64.
(iv) 3.6453 and 3.6554 are rounded off to two decimal places. The results will be
(A) 3.64, 3.65
(B) 3.65, 3.66
(C) 3.65, 3.65
(D) 3.6, 3.6
Answer:
(B) 3.65, 3.66
Explanation:
If the digit from where the number is truncated is 5. The digit after it is 3 in one number and 4 in other number i.e., both are non-zero digits. So, as per rule the digit before 5 is to be increased by 1.
OR
355.0 kg = 355000 g. Significant figures in the two numbers are
(A) 4, 6
(B) 3, 6
(C) 3, 3
(D) 4, 4
Answer:
(D) 4, 4
Explanation:
Number of significant figures in 355.0 kg is 4.
Decimal is removed and additional zeros have been put to convert it to gram.
In such case the significant figures do not increase due to zeros. It remains 4.
Section – E
Question 31.
(a) The velocity time graph for a particle is shown in figure. Draw acceleration time graph from it.
Answer:
Acceleration – Time Graph
(b) Can a body have zero velocity and still be accelerating? If yes, give any situation.
Answer:
Yes, at the highest point of vertical upward motion under gravity.
OR
(a) The two straight rays OA and OB on the same displacement-time graph make angle 30° and 60° with time axis respectively as shown in figure :
(i) Which ray represents greater velocity ?
(ii) What is the ratio of two velocity represented by OA and OB?
Answer:
(i) Since, the slope of the displacement-time graph of uniform motion in one dimension represents the velocity of the object, hence the line showing greater slope in graph corresponds to greater velocity of the object. Therefore, OB represents greater velocity.
(ii) Ratio of two velocities
\(\frac{v_{\mathrm{A}}}{v_{\mathrm{B}}}=\frac{\tan 30^{\circ}}{\tan 60^{\circ}}\)
= \(\frac{1 / \sqrt{3}}{\sqrt{3}}\)
= \(\frac{1}{3}\)
(b) What are the different methods to derive the equation for uniform accelerated motion?
Answer:
- From definitions of average velocity and acceleration.
- From velocity – time graphs.
- By using calculus.
Question 32.
(a) Earth is continuously pulling Moon towards its centre. Why does not Moon fall on to Earth ?
Answer:
It is so because the gravitational attraction of Earth provides the necessary centripetal force to the Moon for its orbital motion around the Earth. Due to which the Moon is revolving around the Earth.
(b) Gravitational force is a weak force but still it is considered the most important force. Why ?
Answer:
Gravitational force plays an important role for initiating the birth of stars, for controlling the entire structure of the universe and evolution of the universe. It helps to explain many natural phenomena.
(c) Compare variation in g at pole and equator due to rotation of Earth ?
Answer:
Value of g is maximum at pole whereas minimum at equator. Effect of rotation of Earth on g is almost nil on poles and maximum at equator.
OR
(a) State Kepler’s laws of planetary motion.
Answer:
(a) Kepler’s Laws of Planetary Motion-:
Kepler’s I Law (Law of Orbits) :
Each planet revolves around the Sun in an elliptical orbit. The Sun is situated at one foci of the ellipse.
Kepler’s II Law (Law of Areas)-:
The position vector of the planet from the Sun sweeps out equal area in equal interval of time. That is the areal velocity of the planet around the Sun is constant.
Kepler’s III Law (Law of Periods)-:
The square of the time period of any planet about the Sun is proportional to the cube of the semi-major axis of the elliptical orbit.
(b) Name the physical quantities which remain constant during the planetary motion, point without acceleration.
Answer:
- Mass
- Angular Momentum
- Total Energy
(c) According to Kepler’s second law, the radius vector to a planet from the Sun sweeps out equal areas in equal interval of time. The law is consequence of which conservation law ?
Answer:
It is consequence of law of conservation of angular momentum.
Question 33.
(a) Why do the hair of a shaving brush cling together when taking out of water ?
Answer:
When the brush is taken out of water, thin water film is formed at the tips of the hair. It contracts due to surface tension and so, the hair cling together.
(b) Why the tip of the nib of a pen is split?
Answer:
The tip of the nib of a pen is split in order to provide a capillary which helps the ink to rise to the end of the nib and enables it to write continuously.
(c) What will happen if the length of the capillary tube smaller than the height to which the liquid rises ?
Answer:
The liquid will not over flow in the capillary. It will be full upto the top of the capillary and at open end it will bulge out.
OR
(a) Find the coefficient of thermal conductivity.
Answer:
Consider a cube of side x and area of each face A. The opposite faces of the cube are maintained at temperature, θ1 and θ2 where θ1 > θ2. Heat gets conducted in the direction of the fall of temperature. The flow of heat depends upon the following factors:
(i) Q ∝ A …………..(i)
(ii) Q ∝ \(\frac{d \theta}{d x}\) ……………(ii)
(iii) Q ∝ t ………………(iii)
Combining equations (i), (ii) & (iii)
Q ∝ A\(\frac{d \theta}{d x}\) × t
or Q = KA\(\frac{d \theta}{d x}\) t
or Q = K \(\frac{A\left(\theta_1-\theta_2\right)}{x}\) t
Here, K is a constant called the coefficient of thermal conductivity of the material of the cube and t stands for time interval.
If A = 1 m2, x = 1 m, t = 1 s, ∆θ = 1°C then Q = K
Thus coefficient of thermal conductivity is defined as the amount of heat required to flow through a solid having unit area of cross section, unit length to rise 1°C in 1 s.
We can also write as
H = KA \(\left[\frac{\Delta \theta}{\Delta x}\right]\)
where H = Heat flow per second
\(\left[\frac{\Delta \theta}{\Delta x}\right]\) = Temperature gradient
T = θ
when, A = 1 m2
(θ1 – θ2) = 1°C
t = 1s
x = 1 cm
then H = K.
(b) Light, from the Moon, is found at the peak (or wavelength of maximum emission) at λ = 14 μm. Given that the Wien’s constant ‘b’ equals 2.898 × 10-3 mK, estimate the temperature of the Moon.
Answer:
Using Wien’s law (λmT = b), we get
T = \(\frac{b}{\lambda_m}\)
= \(\frac{2.898 \times 10^{-3} \mathrm{mK}}{14 \times 10^{-6} \mathrm{~m}}\)
Thus the surface temperature of Moon is about 207 K.