Category: Class 11 Physics Objective Questions

250+ TOP MCQs on Dynamics of Rotational Motion about a Fixed Axis | Class 11 Physics

Physics Multiple Choice Questions on “Dynamics of Rotational Motion about a Fixed Axis”.

1. A rigid body is rotating about an axis. One force F1 acts on the body such that its vector passes through the axis of rotation. Another force F2 acts on it such that it is perpendicular to the axis of rotation and at a point 5cm from the axis. This force F2 is perpendicular to the radius vector at its point of application. Find the net torque on the body. Let F1 = 10N & F2= 5N.
a) 0
b) 10.25Nm
c) 0.25Nm
d) 10Nm
Answer: c
Clarification: The force F1 passes through the axis of rotation, so it will not produce a torque.
The force F2 is perpendicular to axis and radius, so it will provide a torque = r*F2 about the axis of rotation, where ‘r’ is the distance of point of application of force F2 from the axis of rotation.
Therefore, torque = r* F2= 0.05 * 5 Nm
= 0.25Nm.

2. A ring is rotating about a diameter. The radius = 5cm & mass of ring = 1kg. A force is applied on the ring such that it is perpendicular to the axis and vector AB as shown in the figure. The magnitude of force is 10N. Find the work done by the torque, when the ring rotates by 90°.

a) 0.25π J
b) 0.15π J
c) 0.2π J
d) 0
Answer: c
Clarification: Work done is given by Tθ, where T is the torque & θ is the angular displacement = π/2 rad. From the given figure, OA = 3cm & OB = 5cm,
therefore AB2 = 52 – 32= 25 – 9 = 16 OR AB = 4cm.
Torque due to force F will be AB*F = 0.04*10 = 0.4Nm.
∴ Work = Tθ = 0.4*π/2 = 0.2π J.

3. A ring of radius 7cm is rotating about the central axis perpendicular to its plane. A force acts on it, tangentially, such that it does a work of 10J in a complete rotation. Find the value of that force. `
a) 0.5/7π N
b) 35π N
c) 500/7π N
d) 0.7 N
Answer: c
Clarification: Let the force be ‘F’. Work done will be = Tθ,
where T is the torque & is the angular displacement = 2π rad.
Work = Tθ = 10
∴ T(2π) = 10
∴ T = 5/π Nm.
Also, T = r*F
∴ 5/π = 0.07*F
∴ F = 500/7π N.

4. A disc of radius 10cm is rotating about the central axis perpendicular to its plane. A force of 5N acts on it tangentially. The disc was initially at rest. Calculate the value of power supplied by the force when the disc has rotated by 30°.The mass of the disc is 2kg.
a) 1.28W
b) 1.8W
c) 3.9W
d) 2.63W
Answer: b
Clarification: Work done = Tθ, where T is torque & θ is angular displacement.
T = r*F = 0.1*5 = 0.5Nm
& θ = 30° = π/6 rad.
∴ Work = 0.5*π/6 = π/12 J.
Moment of inertia about spinning axis = MR2/2 = 2*0.01/2 = 0.01kgm2.
Using, T = Ia, where ‘I’ is moment of inertia & ‘a’ is angular acceleration, we get:
a = T/I
∴ a = 0.5/0.01 = 50 rad/s2.
Now, θ = w0t + (1/2)at2 to calculate the time for 30° rotation, where wo is initial angular velocity.
∴ π/6 = 0 + 0.5*50*t2.
∴ t = √(π/150) = 0.145 s.
∴ Power = Work/time = (π/12) / (0.145) = 1.8 W.

5. A rod is rotating about one end. If a force F1 acting on the other end produces a torque T & supplies power P, find the value of force F2 that will produce the same amount of power when it acts at the midpoint of the rod. Assume that all forces are perpendicular to the axis of rotation & axis of rod. The rod starts from rest.
a) = F1
b) > F1
c) < F1
d) No force acting at any other point will produce the same power
Answer: b
Clarification: Power is defined as the work done per unit time. For the same power, the amount of rotation should be the same in a given time. For the th angular acceleration should be the same in both the cases & therefore the torque should be the same.
Thus, F2* I/2 = F1* I.
F2 = F1*2 OR F2 > F1

250+ TOP MCQs on Solids Mechanical Properties – Stress and Strain | Class 11 Physics

Physics Multiple Choice Questions on “Solids Mechanical Properties – Stress and Strain”.

1. Stress in a solid body is defined as ___________ per unit area.
a) external force applied
b) strain
c) pressure
d) internal forces developed due to externally applied forces
Answer: d
Clarification: When a force is applied on a solid body, internal forces develop inside it. Pressure is the external force per unit area while stress is the internal force per unit area. Note that pressure and stress aren’t the same.

2. Stress, like pressure, is always perpendicular to a plane. True or False?
a) True
b) False
Answer: b
Clarification: Stress can act along any direction to a plane. If it is parallel to the plane it is called shearing stress & if it is perpendicular to the plane it’s called normal stress. Pressure is always perpendicular to a unit area.

3. A wire with a radius of 5mm is hung freely from the ceiling. A load of 5N is applied to its free end. Find the elongation in the wire if its volume is 7.85*10-5m3 & young’s modulus is 1011N/m2.
a) 6.21*10-7m
b) 7.00*10-7m
c) 6.36*10-7m
d) 8.00*10-9m
Answer: c
Clarification: The initial length of wire is Vol / πr2 = 7.85*10-5/ π*0.0052 = 1m.
Stress = Y*strain. F/A = Y*Δl / l.
Δl = F/A * l/Y = (5/πr2)*(1/1011)
= 6.37*10-7m.

4. A wire has a young’s modulus of 105N/m2, length 1m & radius 3mm. Assuming a uniform cross sectional area, find the radius of wire after it is under a force of 1N from both ends.
a) 2.58m
b) 2.30m
c) 3.54m
d) 2.24m
Answer: a
Clarification: Force = 1N. Initial area = πr2 = 2.82*10-5m2.
Stress = Y*Strain
Δl = F/A * l/Y = (1/2.82*10-5)*(1/105) = 0.35m
As volume will remain same (we can also say that product of l & r2 will be constant as other terms in expression of volume are constants).
1*32 = 1.35*R2
⇒R = 2.58m.

5. Strain can be negative. True or False?
a) True
b) False
Answer: a
Clarification: Strain is defined as change in length or volume divided by initial length or volume respectively. There can be cases when length or volume decreases like when rod is compressed or body is inserted in fluid (it volume decreases due to pressure from all sides). Hence, strain can be negative.

6. In the given system, masses are released from rest. The young’s modulus of wire is 1011N/m2, length = 1m & radius = 2mm. Find elongation in wire when masses are moving. Assume pulley to be frictionless.

a) 1.05*10-5m
b) 2*10-5m
c) 3*10-5m
d) 0.5*10-5m
Answer: a
Clarification: Let the tension in rope be ‘T’ & acceleration of masses be ‘a’.
2g-T=2a & T-1g=1a.
On solving these equations we get, T = 4g/3 = 1.33g.
For rope, Stress = Y*Strain.
∴ T/A = Y*Δl / l (where A is area of rope & l is initial length)
∴ Δl = (1.33g/πr2)*(1/1011) = 1.05*10-5 m.

250+ TOP MCQs on Thermal Properties of Matter – Change of State | Class 11 Physics

Physics Problems for Class 11 on “Thermal Properties of Matter – Change of State”.

1. Change of phase occurs at a constant temperature, which implies that even on addition of heat to a substance temperature remains constant. True or False?
a) True
b) False
Answer: a
Clarification: The temperature change with heat supplied is shown below. During change of phase heat supplied is used to make changes in intermolecular distances while the temperature remains constant.

2. Melting point depends on pressure. True or False?
a) True
b) False
Answer: a
Clarification: Melting point is less for higher pressure. For eg: we can keep a small weight of 1kg on a block of ice and we will see that the ice underneath the block starts melting at a lower temperature.

3. What is regelation?
a) Refreezing of ice on addition of impurities
b) Refreezing of ice on reduction in pressure
c) Melting of ice at lower temperature than its melting point due to increase in pressure
d) Melting of ice at higher temperature than its melting point
Answer: b
Clarification: When increase in pressure causes water to melt. It may refreeze on removal of this extra pressure. This refreezing is called relegation.

4. What happens to the boiling point of water in a container when the container is completely closed by a lid?
a) Increases
b) Decreases
c) Remains same
d) Boiling point is characteristic of a substance
Answer: a
Clarification: Boiling occurs when vapour pressure of liquid is the same as atmospheric pressure. The vapour pressure increases with increase in temperature. So when temperature has been increased sufficiently boiling starts. But with increase in pressure at higher vapour pressure has to be reached which is obtained by increasing temperature and therefore we say boiling point increases.

5. Select the correct statement
a) On heating continuously a substance changes from solid to gas via liquid phase only
b) Change of phase occurs at variable temperature
c) Water can exist in all 3 phases
d) Boiling point increases with increase in pressure
Answer: c
Clarification: Water can exist in all 3 phases that are in thermal equilibrium with each other. It is referred to as the triple point of water. When a substance is heated it may change directly from solid to gas. This is called sublimation. Boiling point increases with decrease in pressure. Change of phase occurs at constant pressure.

6. Getting burnt by steam is less dangerous than boiling water. True or False?
a) True
b) False
Answer: b
Clarification: To make steam, water has to be boiled and then more heat energy has to be added to convert it into steam. So, steam is more dangerous than boiling water.

7. If 5g of a solid requires 30cal of heat to change into liquid at the same temperature, calculate the latent heat of fusion in cal/g.
a) 150
b) 0.167
c) 6
d) 3
Answer: c
Clarification: For changing a solid into liquid, heat has to be supplied and this change occurs at constant temperature. Q = mL
⇒ 30 = 5L
⇒ L = 6 cal/g,

8. The diagram given below shows phase changes for a substance based on experimental observations. Select the correct statement.

a) Heat is being removed from body while increasing temperature
b) Heat is constant during phase change
c) Temperature is in Kelvin
d) The two horizontal lines in the graph represent phase change as the temperature is constant
Answer: d
Clarification: The diagram represents phase change from solid to liquid to gas. Heat removal leads to decrease in temperature. The temperature becomes zero at an instant, so it can’t be in Kelvin as 0K has never been achieved. The two horizontal lines represent phase change as temperature remains constant at that time.

9. Condensation temperature is the same as evaporation temperature. True or False?
a) True
b) False
Answer: a
Clarification: Condensation is the conversion of vapour to liquid, while evaporation is the conversion of liquid to vapour. Both occur at the same temperature, the only difference being that heat is removed in the former and added in the latter. The graph as shown below is a straight line at constant temperature.

10. Select the correct statement for melting of any substance.
a) It occurs at the same temperature as that of vapourization
b) It is exothermic in nature
c) It occurs at 100°C
d) It requires heat to be supplied to the substance
Answer: d
Clarification: Melting is the conversion of solid to liquid at constant temperature. It is endothermic and the melting temperature varies from substance to substance and also on atmospheric conditions like pressure.

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250+ TOP MCQs on Kinetic Theory of an Ideal Gas | Class 11 Physics

Physics Multiple Choice Questions on “Kinetic Theory of an Ideal Gas”.

1. Consider a gas enclosed in a box. A molecule of mass m, having a velocity -2i+3j+4k collides with a wall parallel to the xz plane. What will be its velocity after collision and its change in momentum? i,j & k are unit vectors along the x,y & z axis.
a) -2i-3j+4k, -6mj
b) 2i+3j-4k, -6mj
c) -2i-3j+4k, 6mj
d) 2i+3j-4k, 6mj
Answer: a
Clarification: On collision with a wall parallel to the xz plane, only the y-component will change.
Thus, new velocity will be -2i-3j+4k.
The change in momentum will be -3mj-3mj = -6mj.

2. Consider a gas enclosed in a box. Let the walls have an area of 0.01m2. Let’s say a few molecules having their x-component of velocity =2m/s, hit a wall parallel to the yz plane. What will be the value of force on the wall in time dt? Let the number of molecules per unit volume be 10 & mass of each molecule be 10-20kg.
a) 4*10-19N
b) 4*10-21N
c) 4/3*10-21N
d) 4/3*10-19N
Answer: b
Clarification: According to the assumptions of KTG, we will assume that the molecules collide elastically.
The total momentum transferred to the wall is 2mv*(1/2)nAvdt.
1/2 has been included because, on average, half the molecules near that wall will be moving away from the wall.
Force will be momentum/time
= nmAv2
= 10*10-20*0.01*4
= 4*10-21N.

3. Consider a gas inside a box. It exerts some pressure on its walls. Find this value. GIven that number of molecules per unit volume be 20, mass of each molecule 10-20kg, vx2 = 1m2/s2.
a) 20*10-20Pa
b) 20/3*10-20Pa
c) 20/3*10-19Pa
d) 20*10-19Pa
Answer: a
Clarification: Pressure is given by (1/3)nmv2.
v2 = 3vx2 = 3m2/s2.
P = (1/3)*20*10-20*3
= 20*10-20Pa.

4. Which of the following is not an assumption of KTG?
a) Time spent during collision is negligibly small
b) All gases are made up of molecules moving randomly in all directions
c) Molecules do not collide with each other
d) Molecules collide elastically with the wall
Answer: c
Clarification: According to KTG, molecules collide elastically with each other and with the walls of the container. The other given assumptions are true. These assumptions help to develop a model of molecular behavior of an ideal gas.

5. The number of molecules per unit volume in the sample is 20. The mass of each molecule is 10-20kg. The mean of speed squared is 4m2/s2. What is the value of internal energy of the gas? Assume volume of container is 0.02m3.
a) 0.8*10-20J
b) 0.53*10-20J
c) 0.01*10-20J
d) 0.45*10-20J
Answer: a
Clarification: Internal energy is given by: E = (1/2)Nmv2,
where N is the total number of molecules = nV.
E = 0.5*20*0.02*10-20*4
= 0.8*10-20J.

6. Average kinetic energy of a gas depends on which of the following factors?
a) Nature of gas
b) Temperature
c) Pressure
d) Volume
Answer: b
Clarification: Average kinetic energy is given by (3/2)kBT, which shows that it only depends on temperature.

7. A container has 3 gases whose mass ratio is 1:3:5. What is the ratio of mean square speed of the molecules of two gases? Their atomic masses are 20u, 30u & 40u corresponding to the order in which the ratios are given.
a) 2:3:4
b) 4:3:2
c) 2:√3:√2
d) √2:√3:2
Answer: b
Clarification: Their average kinetic energies will be the same. Thus, 1/2mv2will be the same.
v12:v22:v32 =m3:m2:m1
= 40:30:20
= 4:3:2.

8. What is the number of molecules in 3 cubic metre of a gas at 3atm & 27°C?
a) 2.17*10-20
b) 6.38*10-20
c) 3.86*10-20
d) 4.58*10-20
Answer: a
Clarification: Using, PV=NkT we get: N = PV/kT
= (3*3*105)/(1.38*10-23*300)
= 2.17*10-20.

9. Boyle’s law can be deduced from kinetic theory. True or False?
a) True
b) False
Answer: a
Clarification: From the kinetic theory we get: PV=(1/3)mNvrms2. Vrms2 depends only on temperature. So, if temperature is constant PV will be constant, which is Boyle’s law.

250+ TOP MCQs on Waves – Beats | Class 11 Physics

Physics Aptitude Test for Class 11 on “Waves – Beats”.

1. Consider the phenomenon of beats & select the correct statement.
a) It occurs when frequencies are close or equal
b) It correspond to the change in intensity we hear due to interference of 2 waves having close frequencies
c) It corresponds to a constant intensity sound heard during wave interference
d) Beat frequency is variable
Answer: b
Clarification: When two waves with close, but not equal, frequencies interfere, beats are produced. They have a frequency equal to the average frequency of the two interfering waves and we recognize them as the variable intensity sound we hear.

2. Two interfering waves have a frequency of 2Hz & 6Hz. What is the beat frequency?
a) 8Hz
b) 4Hz
c) 2Hz
d) 0
Answer: b
Clarification: Beat frequency is equal to the difference between frequencies of interfering waves. In this case, beat frequency
= 6-2 = 4Hz.

3. Let’s say we can hear a beat frequency of 2Hz from two tuning forks. The first tuning fork has a frequency of 100Hz. If wax is applied on the 2nd tuning fork, the beat frequency becomes 4Hz, what is the frequency of the 2nd tuning fork?
a) 100Hz
b) 98Hz
c) 106Hz
d) 104Hz
Answer: b
Clarification: As the initial beat frequency is 2Hz, the 2nd tuning fork can have a frequency of 98Hz or 102Hz. When wax is applied to the second tuning fork its frequency should decrease. On doing so the beat frequency has also increased, so the initial value of the frequency of the 2nd tuning fork must be lower than that of the first tuning fork.

4. Two longitudinal waves: s1 = 3cos(11t) & s2 = 3cos(10t) interfere. What is the amplitude of the resulting wave?
a) 6cos(0.5t)
b) 6sin(0.5t)
c) 6cos(10.5t)
d) 6sin(10.5t)
Answer: a
Clarification: s = s1+s2 = 2(3)*cos(1/2)t*cos(21/2)t
= 6cos(0.5t)cos(10.5t).
The term 6cos(0.5t) corresponds to the amplitude.

250+ TOP MCQs on Accuracy, Precision of Instruments and Errors in Measurement | Class 11 Physics

Physics Quiz on “Accuracy, Precision of Instruments and Errors in Measurement”.

1. In an experiment, it is found that the experimental value is very close to actual value, hence the experimental value can be called _____
a) Accurate
b) Precise
c) Suitable
d) Mean
Answer: a
Clarification: Accuracy is the degree to which a value is near to the actual or standard value. Hence if the experimental value is very close to the actual value, the measurement can be called accurate.

2. What is the reason for the occurrence of systematic errors in an instrument ?
a) No use for a long time
b) High use
c) Manufacturing fault
d) Delivery fault
Answer: b
Clarification: Systematic errors arise due to high use of the instrument. It can also be present because of careless handling.

3. How are systematic errors removed usually for an instrument?
a) By replacing it
b) By re-calibrating it
c) By using a repairing service
d) By not using it for some time
Answer: b
Clarification: Systematic errors arise due to careless or overuse of an instrument. It can easily be removed by re-calibrating the instrument and maintaining it properly.

4. In 5 experiments with the same objective, the values obtained are very near to each other. These values can be called ____
a) Precise
b) Accurate
c) Average
d) Invalid
Answer: a
Clarification: Precision refers to the how close the data points or numbers in a set are. In this case the values obtained are very close to each other, hence they are precise.

5. Range of an instrument is _____
a) The minimum value that can be measured
b) The maximum value that can be measured
c) All values starting from the minimum to the maximum that can be measured
d) The average of all values that can be measured
Answer: c
Clarification: Range refers to all values starting from the minimum to the maximum that can be measured. Range is sometimes also referred to as span.

6. Usually how many types of errors are present in scientific measurements?
a) 2
b) 3
c) 4
d) 5
Answer: b
Clarification: There are three types of scientific errors – Random errors, Systematic errors and Blunders. Out of the three, systematic errors are the easiest to remove.

7. What is least count of an instrument?
a) The lowest value that can be measured accurately
b) The greatest value that can be measured accurately
c) Half of the lowest value that can be measured accurately
d) Mean of the lowest and highest values
Answer: a
Clarification: The least count of an instrument is the lowest value that can be measured accurately. For example, if the lowest value that can be measured by a ruler is 1 mm, then the least count is 1 mm.