Category: Class 11 Physics Objective Questions

250+ TOP MCQs on Second Law of Thermodynamics | Class 11 Physics

Physics Multiple Choice Questions on “Second Law of Thermodynamics”.

1. A process is carried out between 2 systems. It is possible for entropy of one system to decrease. True or False?
a) True
b) False
Answer: a
Clarification: The net entropy change has to be positive. It is possible for the entropy of a system to decrease, but the entropy of the second system should increase enough to ensure net entropy increases.

2. According to the first law of thermodynamics, a book lying on a table can fly upwards on its own by using the internal energy of the table. True or False?
a) True
b) False
Answer: a
Clarification: In the above said process, energy is conserved and according to first law, this is valid. It only violates the second law. So, the given statement is true.

3. Which law of thermodynamics says that efficiency of a heat engine cannot be 1?
a) First
b) Zeroth
c) Second
d) Third
Answer: c
Clarification: The second law of thermodynamics fundamentally disallows many processes that may be consistent with the law of conservation of energy. It says that entropy change in an irreversible process has to be positive. This means that heat given to an engine can’t all be converted into work. Therefore, efficiency cannot be 1.

4. Which of the following is the Kelvin Planck statement?
a) Heat from a cold reservoir cannot be transferred to a hot reservoir without external work
b) Efficiency for a reversible cyclic process can be 1
c) Heat from a hot reservoir can be completely converted into work if process is reversible
d) Efficiency of a heat engine is less than 1
Answer: d
Clarification: Kelvin Planck statement is: No process is possible whose sole result is the absorption of heat from a reservoir and complete conversion of that heat into work. So, this implies that efficiency of a heat engine is less than 1.

5. The second law of thermodynamics states that?
a) Entropy of an isolated system can never decrease over time
b) Entropy of a pure crystalline substance at 0K is 0
c) Energy is conserved
d) Entropy of a system can never decrease
Answer: a
Clarification: The second law states that entropy of an isolated system can never decrease over time. There can be a system whose entropy decreases, but, at the same time in that process another system’s increase in entropy will more than compensate for that decrease.

250+ TOP MCQs on Transverse and Longitudinal Waves | Class 11 Physics

Physics Multiple Choice Questions on “Transverse and Longitudinal Waves”.

1. All waves require material for propagation. True or False?
a) True
b) False
Answer: b
Clarification: Only material waves,like the name suggests, require material for propagation. Electromagnetic waves on the other hand don’t require any medium. They can travel through vacuum in the form of electric and magnetic fields.

2. Transverse waves involve energy transfer by movement of medium particles, while longitudinal waves don’t. True or False?
a) True
b) False
Answer: b
Clarification: Transverse waves involve vibration, & not movement, of medium particles to transfer energy. Even in longitudinal waves, medium particles move up & down but don’t travel to transfer energy.

3. Electromagnetic waves are considered to be which of the following types?
a) Transverse
b) Longitudinal
c) Both Transverse & Longitudinal
d) Neither longitudinal nor transverse
Answer: a
Clarification: Electromagnetic waves are formed by transverse motion of electric & magnetic fields. Thus, they are said to be transverse waves.

250+ TOP MCQs on Motion in a Plane – Scalars and Vectors | Class 11 Physics

Physics Multiple Choice Questions on “Motion in a Plane – Scalars and Vectors”.

1. What is a scalar?
a) A quantity with only magnitude
b) A quantity with only direction
c) A quantity with both magnitude and direction
d) A quantity without magnitude
Answer: a
Clarification: A scalar quantity is one which only has a magnitude. Examples of scalar quantities are mass, volume, work, energy etc.

2. What is a vector quantity?
a) A quantity with only magnitude
b) A quantity with only direction
c) A quantity with both magnitude and direction
d) A quantity without direction
Answer: c
Clarification: A vector quantity is one which has both magnitude and direction. Unlike scalars, it can also tell the direction in which is the entity is acting. Examples of vector quantities are force, velocity, displacement etc.

3. Which one of the following operations is valid?
a) Vector multiplied by scalar
b) Vector added to scalar
c) Vector subtracted from scalar
d) Vector divided by vector
Answer: a
Clarification: Only operations valid for vector with scalars are multiplication and division. A vector can be multiplied or divided by a scalar. Other operations like addition and subtraction are not valid for vectors with scalars.

4. The vector obtained by addition of two vectors is termed as ______
a) New vector
b) Resultant vector
c) Derived vector
d) Sum vector
Answer: b
Clarification: The resultant vector is the vector which is obtained by the addition or subtraction of two vectors. The resultant vector can either be calculated by using the graphical method or the analytical method.

5. Mass is a ____
a) Scalar quantity
b) Vector quantity
c) Relative quantity
d) Dependent quantity
Answer: a
Clarification: Mass represents the amount of matter in a body. It does not have any direction but only mass. Hence, it is a scalar quantity.

6. The operation used to obtain a scalar from two vectors is ______
a) Cross product
b) Dot product
c) Simple product
d) Complex product
Answer: b
Clarification: Dot product of two vectors gives a scalar quantity as the output. Cross product gives a vector as the output. It is also known as scalar product.

7. The operation which does not give you a vector as an output from two vector inputs is ______
a) Dot product
b) Cross product
c) Vector addition
d) Vector subtraction
Answer: a
Clarification: Vector addition and subtraction give a vector as the output. Cross product also gives a vector as the output. Only Dot product gives a scalar after acting on two vectors.

250+ TOP MCQs on Introduction to Work, Energy and Power | Class 11 Physics

Physics Multiple Choice Questions on “Introduction to Work, Energy and Power”.

1. The energy possessed by a body by the virtue of its motion is called ______
a) Kinetic energy
b) Potential energy
c) Total energy
d) Motion energy
Answer: a
Clarification: The kinetic energy of a body emerges due to its motion. More specifically, the velocity of a body decides the amount of kinetic energy it has. Usually, the kinetic energy of a body with mass m is given as K.E. = (Big(frac{1}{2}Big))mv2, where v is the velocity of the body.

2. The energy possessed by a body by the virtue of its position is called ______
a) Kinetic energy
b) Potential energy
c) Total energy
d) Position energy
Answer: b
Clarification: The potential energy of a body emerges due to its position. More specifically, the displacement of a body from the reference position decides the amount of potential energy it has. Usually, the potential energy of a body with mass m is given as P.E.=mgh, where h is the height of the body from the ground plane and g is the acceleration due to gravity. Absolute potential of a body cannot be found. Only the relative value can be found out.

3. Energy is ________
a) Work
b) The ability to create work
c) Quantification of work
d) Force multiplied by displacement
Answer: b
Clarification: Energy is defined as the ability to create work. When a force is applied on a body to create displacement, work takes place. Work is quantified by the force applied by the displacement. Whereas the energy is the ability of the force to create work. In general, the total input energy = total output energy + work.

4. Power is ______
a) Rate of doing work
b) Ability to do work
c) Rate of energy creation
d) Equivalent to work
Answer: a
Clarification: Power is defined as the rate of doing work. The ability to do work is energy. Energy can neither be created nor be destroyed, hence, the rate of energy creation does not exist.

5. The unit of energy has been named after ______
a) James Prescott Joule
b) John Prescott Joule
c) Jammie Joule
d) Jessy Joule
Answer: a
Clarification: The SI unit of energy is Joule. It has been named after the famous scientist James Prescott Joule. He made a significant contribution to the word of science by discovering the relationship of energy with mechanical work.

6. In which category do potential and kinetic energy fall?
a) Mechanical energy
b) Electrical energy
c) magnetic energy
d) Usual energy
Answer: a
Clarification: Mechanical energy is composed of potential and kinetic energy. Potential energy is due to the position of the body. Kinetic energy is due to the motion of the body. Both when combined represent the total mechanical energy of the body.

7. How many Ergs are there in 1 Joule?
a) 10
b) 104
c) 107
d) 109
Answer: c
Clarification: Erg is the unit of energy in CGS system. One Joule = 107 ergs. This can be found out by putting in the CGS units in the expression for energy.

8. What is the correct expression for power?
a) P = dW/dt
b) P = F * d
c) P = E
d) P = dE/dt
Answer: a
Clarification: Power is defined as the rate of change of work. Hence, P = dW/dt. Instantaneous power = Force x instantaneous speed. Average power = total work / total time.

9. What is the correct expression for Work?
a) W = F * ds
b) W = P/t
c) W = E
d) W = E/t
Answer: a
Clarification: Work occurs when a force is applied on a body some displacement occurs as a result. Work is quantified as force multiplied by displacement. Hence, W = F * s. Instantaneous work is given as W = F * ds.

10. Energy involved in creating work _____
a) Gets used up
b) Gets transferred
c) Gets exhausted
d) Gets lost
Answer: b
Clarification: Energy involved in creating work gets transferred or converted into some other kind of energy. Energy can neither be created nor be destroyed. Hence, the energy involved will get converted into some other form of energy. Usually in mechanical work, the energy gets dissipated as heat energy.

250+ TOP MCQs on Vector Product of Two Vectors | Class 11 Physics

Physics Multiple Choice Questions on “Vector Product of Two Vectors”.

1. Regarding the velocity of a particle in uniform circular motion about a fixed axis, select the correct option. w & r angular velocity and radius vectors respectively. ‘X’ & ‘ . ’ represent cross & dot products respectively.
a) v = r X w
b) v = w X r
c) v = w.r
d) w = v.r
Answer: b
Clarification: For a particle rotating about a fixed axis, its angular velocity vector points along the axis. Velocity of a particle is a vector so it will be a cross product and not a dot product. Now if we keep our fingers along the direction of the angular velocity vector and curl them in the direction of the radius we get the direction of velocity for that radius vector. Therefore, v = w X r. Refer to the diagram.

2. Which of the following statements is false?
a) Cross product is commutative
b) Cross product is distributive over addition
c) Dot product of two vectors gives a scalar
d) Dot product is commutative
Answer: a
Clarification: Cross product a X b ≠ b X a, therefore it is not commutative. a X (b + c) = (a X b) + (a X c), therefore it is distributive over addition. Dot product is also known as scalar product & a.b = b.a, so it is commutative.

3. Find the vector product (a X b) of the two given vectors: a = 2i + 3j + 4k, b = 3i + 5j. Here, i, j & k are unit vectors along three mutually perpendicular axes.
a) -20i + 12j + k
b) 10i + 6j + 1/2k
c) 20i – 12j – k
d) 10i – 6j -1/2k
Answer: a
Clarification: The cross product of a & b = i(0-20) – j(0 – 12) + k(10 – 9)
= -20i + 12j + k.
Note that this is not the same as b X a.

4. There are 6 vectors: a, b, c, d, e, f. Simply the following expression: [ a(b . c) X (e . f)d ] X [ a – d ].‘X’ represents cross product while ’.‘ represents dot product. Vectors a & d are perpendicular.
a) 0
b) (b . c)(e . f) [ d + a ]
c) (b . c)(e . f) [ d – a ]
d) (b . c)(e . f) [ d X a ]
Answer: b
Clarification: [ a(b . c) X (e . f)d ] X [ a – d ]
= (b . c)(e . f) { [ a X d ] X [ a – d ] }
= (b . c)(e . f) { [ (a X d) X a ] – [ (a X d) X d ] }
Now, (a X d) is perpendicular to the plane of a & d. Refer to the diagram below.
We see that (a X d) X a = d & (a X d) X d = −a.
Therefore, our expression = (b . c)(e . f) [ d + a ].

250+ TOP MCQs on Gravitation – Energy of an Orbiting Satellite | Class 11 Physics

Physics Multiple Choice Questions on “Gravitation – Energy of an Orbiting Satellite”.

1. What is the magnitude of the ratio of KE and PE of a satellite revolving around a planet in a circular orbit of radius “R”?
a) 1:4
b) 1:2
c) 2:1
d) 3:2

Answer: b
Clarification:Let the mass of the satellite be “m” and mass of the planet be “M”.
Kinetic energy (KE) = 1/2 x m x v2
= (1/2) x (G x m x M)/R
Potential energy (PE) = – (G x m x M)/R
KE : PE = -1:2
Therefore, (the magnitude of KE:PE) = 1:2.

2. What is the total energy possessed by a satellite of mass “m” orbiting above the earth of mass “M” and radius “R” at a height “h”?
a) [(G*M*m)/2*(R+h)]
b) -[(G*M*m)/2*(R+h)]
c) [(G*M*m)/(R+h)]
d) -[(G*M*m)/(R+h)]

Answer: b
Clarification: Total energy (E) of a satellite is the sum of its potential energy (PE) and kinetic energy (KE).
PE = – (G x M x m)/(R+h)
KE = (G x M x m)/(2 x (R+h))
E = PE + KE
= -(G x M x m)/(R+h) + (G x M x m)/(2 x (R+h))
= -(G x M x m)/(2 x (R+h)).

3. A satellite of mass “m” is in a circular orbit of radius 1.5R around the earth of radius R. How much energy is required to move it to an orbit of radius 2R?
a) -(G*M*m)/12R
b) (G*M*m)/12R
c) -(G*M*m)/4R
d) (G*M*m)/4R

Answer: a
Clarification: Total energy at 1.5R (TE) = -[(G*M*m)/3R]
Total energy at 2R (TE’) = -[(G*M*m)/4R]
The difference in energy is the required energy.
Difference = TE’ – TE
= -[(G*M*m)/4R] – [-[(G*M*m)/3R]]
= -[(G*M*m)/4R] + (G*M*m)/3R]
= (G*M*m)/12R.

4. A satellite revolves around the earth in a circular orbit with a velocity “v”. What is the total energy of the satellite? Let “m” be the mass of the satellite.
a) – (m*v2)/2
b) (m*v2)/2
c) (3*m*v2)/2
d) – (m*v2)

Answer: a
Clarification: Total energy (E) = KE + PE
KE = (1/2) x m x v2
PE = – 2 x (KE) [For a satellite]
= – m x v2
E = (1/2) x m x v2 + (- m x v2)
= – (1/2) x m x v2.

5. For a satellite with an elliptical orbit and not a circular orbit, the kinetic energy varies with time.
a) True
b) False

Answer: a
Clarification: The kinetic energy of a satellite is;
KE = (1/2) x (G x m x M)/(R+h)
h: Height above the surface of the planet
For an elliptical orbit, “h” changes with time. Hence, the kinetic energy also varies with time.

6. For a satellite with an elliptical orbit and not a circular orbit, the potential energy varies with time.
a) True
b) False

Answer: a
Clarification: The potential energy of a satellite is;
PE = – (G x m x M)/(R+h)
h: Height above the surface of the planet
For an elliptical orbit, “h” changes with time. Hence, the potential energy also varies with time.

7. Assume that a satellite executes perfectly circular orbits around a perfectly circular sphere. The work done by the satellite in 1 complete revolution is _____
a) (1/2) x (G x m x M)/(R+h)
b) – (G x m x M)/(R+h)
c) – (1/2) x (G x m x M)/(R+h)
d) 0

Answer: d
Clarification: The net displacement of the satellite in one complete revolution is zero. Hence, the total work which is done by the satellite in one complete revolution also amounts to zero.
Work done = Force x Displacement.

8. If the kinetic energy of a satellite is halved, its orbital radius becomes _____
a) 1/2 times
b) 2 times
c) 4 times
d) infinity

Answer: b
Clarification: The kinetic energy of a satellite is;
KE = (1/2) x (G x m x M)/r
r: Orbital radius
Let new kinetic energy be KE’
KE’ = KE/2
= [(1/2) x (G x m x M)/r]/2
= (1/2) x (G x m x M)/(2r)
= (1/2) x (G x m x M)/r’
r’: New orbital radius
r’ = 2r.

9. For a satellite executing an elliptical orbit around a planet, its minimum potential energy is at the perigee.
a) True
b) False

Answer: a
Clarification: Perigee is the point in the orbit closest to the surface of the planet. Hence, it has a maximum velocity. Therefore, at perigee, the kinetic energy is maximum and potential energy is minimum.