Category: Class 11 Physics Objective Questions

250+ TOP MCQs on Simple Harmonic Motion | Class 11 Physics

Physics Multiple Choice Questions on “Simple Harmonic Motion”.

1. Which of the following variables has zero value at the extreme position in SHM?
a) Acceleration
b) Speed
c) Displacement
d) Angular frequency
Answer: b
Clarification: At the extreme position in SHM, the body comes to instantaneous rest. The force, and therefore acceleration, is maximum at this point according to the force eqn: F = -kx.

2. A particle is initially at the centre and going towards the left. Let T be the time period of the SHM it is undergoing. What will be its position and velocity at time 3T/4, if it starts from the centre at t=0?

a) At right extreme, zero velocity
b) at centre, maximum speed towards left
c) at centre, maximum speed towards right
d) Mid-way between centre and -A
Answer: a
Clarification: The time period of the given motion is T. It will go from centre to -A in T/4 secs. Then back to centre at 2T/4 secs. And then towards the right extreme, A, at 3T/4 secs. At this extreme position its velocity will be zero.

3. A particle is undergoing SHM with amplitude 10cm. The maximum speed it achieves is 1m/s. Find the time it takes to reach from the mean position to half the amplitude.
a) π/60 s
b) π/30 s
c) π/15 s
d) π/40 s
Answer: a
Clarification: Let the equation of motion be: x = 0.1sin(wt) where w is the angular frequency.
On derivating this equation w.r.t time we get: v = 0.1wcos(wt).
Given that maximum speed is 1m/s, we get 0.1w = 1.
∴ w = 10s-1.
Now our equation of motion has become: x = 0.1sin(10t).
Assume we start from x=0 at t=0. Then by putting x=0.05 in the equation we can find the time. 0.05 = 0.1 sin(10t) ∴ sin(10t) = 0.5 ∴ 10t = π/6 ∴ t = π/60 s.

4. The displacement vs time graphs of 2 SHMs are given below. Which parameter is the same for both of them?

a) Angular frequency
b) Amplitude
c) Maximum speed
d) Phase constant
Answer: d
Clarification: The time period is different for both as they both complete one cycle at different times. So, their angular frequencies will be different. The maximum displacement, as seen from the graph, is different for both so their amplitudes are different. SHMs have maximum speed at their mean positions. Both the curves have different slopes at their mean positions so their maximum velocities are different. At t=0, both are at their mean positions and going towards their positive extremes, therefore their phase constants will be the same.

5. 2 particles undergoing SHM start from the mean position and go in opposite directions. Particle 1 starts with a speed of 10m/s and particle 2 starts with a speed 0f 5m/s. If the amplitude(=10cm) is the same. At what position will they first meet?
a) 0.0866m
b) -0.0633m
c) 0
d) 0.0633m
Answer: a
Clarification: Let the first particle go towards negative displacement. Its equation will be: x1 = 0.1sin(w1t + π) & maximum speed = 0.1w1=10
W1 = 100s-1.
Let the second particle go towards positive displacement. Its equation will be: x2 = 0.1sin(w2t) & maximum speed = 0.1w2 = 5
W2 = 50s-1.
Now, in their phasor diagram: w1t + w2t =2π.
∴ t = 2π/150.
At t = π/150
x1 = 0.1sin(50*2π/150) = 0.0866m.

6. A particle has an equation of motion given by: x = cos2wt – sin2wt. Select the correct statement regarding the same.
a) It is not a SHM
b) It is a SHM with T = π/w
c) It is an SHM with T = 2π/w
d) Amplitude of motion is 1/√2 m
Answer: b
Clarification: x = cos2wt – sin2wt = cos 2wt.
Thus, time period T = 2π/2w = π/w.
Also the amplitude of motion is 1m.

7. What is the relation between time periods of the two given SHMs.

a) 4T1 = T2
b) T1 = 2T2
c) 2T1 = T2
d) T1 = 4T2
Answer: c
Clarification: Particle 1 completes two cycles when 2 completes one cycle.
So, T1 = (1/2)T2
OR 2T1 = T2.

8. A particle starts from the extreme position, at t = 0, in a SHM. If the time period of motion is 2s & maximum speed is 5m/s, find the equation of motion.
a) x = 1.59cos(πt)
b) x = 1.59sin(πt)
c) x = 2.5sin(2t)
d) x = 2.5cos(2t)
Answer: a
Clarification: T = 2s. ∴ w = 2π/T = s-1.
Aw = 5. ∴ A = 5/π metre.
x = 5/π sin(πt + a), where is phase constant.
At t = 0, x = A. Thus, a = π/2.
x = 5/π sin(πt + π/2)
= 5/π cos(πt) = 1.59cos(πt).

9. What is the amplitude of motion for x = 2sin(2t) + 4sin2t ?
a) 2√2 m
b) 4 m
c) 2 m
d) The given equation is not that of an SHM
Answer: a
Clarification: x = 2sin(2t) + 4sin2t
= 2sin2t + 2(1-cos2t)
= 2sin2t – 2cos2t + 2
= 2√2 sin(2t – π/4) + 2.
Thus, the amplitude of motion is 2√2 m.

10. In an SHM the time taken to go from mean position to A/2 is the same as that from A/2 to A. True or False? Here, A is the amplitude of motion.
a) True
b) False
Answer: b
Clarification: In a SHM, the speed is maximum at the mean position and goes on decreasing towards the extreme position. So the speed at every position from 0 to A/2 is greater than that from A/2 to A. Thus, the first half is covered faster.

11. In SHM, what is the phase difference between velocity and acceleration?
a) 0
b) π
c) π/2
d) π/3
Answer: c
Clarification: Let the displacement of a particle be given by: x = Asin(wt).
Velocity will be = Awsin(wt + π/2)
acceleration will be = -Aw2sin(wt) = Aw2sin(wt + π).
Thus, the phase difference is π/2.

Physics,

250+ TOP MCQs on Dimensional Analysis and its Applications | Class 11 Physics

Physics Multiple Choice Questions on “Dimensional Analysis and its Applications”.

1. Identify the primary quantity from the following.
a) Mass
b) Density
c) Speed
d) Volume
Answer: a
Clarification: Mass is one of the 7 primary quantities in the SI system. Density is dependent on mass and volume. Speed is dependent on length and time. And volume is dependent on length as its cube.

2. Which of the following is a use of dimensional analysis?
a) To check the dimensional correctness of an equation
b) To solve the equation dimensionally
c) To get the number of dimensional constants
d) To understand the dimensional equation
Answer: a
Clarification: Dimensional analysis is basically used for two purposes. First, to check the dimensional correctness of an equation. Second, to convert a physical quantity from one system of units to another.

3. The dimension whose unit does not depend on any other dimension’s unit is known as ________
a) Fundamental dimension
b) Dependent dimension
c) Independent dimension
d) Absolute dimension
Answer: a
Clarification: A fundamental dimension is one, whose units do not depend on the units of any other dimension. There are 7 fundamental dimensions in the SI system and rest all are derived from them.

4. Which one of the following is a dimensionless quantity?
a) Mass
b) Weight
c) Specific weight
d) Reynold’s number
Answer: d
Clarification: Reynold’s number is a dimensionless quantity. Its formula is Re = ρvD/μ. On solving the associated dimensional equation, it will turn out that, Reynold’s number is dimensionless.

250+ TOP MCQs on Uniform Circular Motion | Class 11 Physics

Physics Multiple Choice Questions on “Uniform Circular Motion”.

1. The force that keeps the body moving in circular motion is _______
a) Centripetal force
b) Centrifugal force
c) Force of gravity
d) Reaction forces
Answer: a
Clarification: Centripetal force is responsible for keeping bodies moving in circular motion. The centripetal force is directly proportional to the square of the tangential velocity and inversely to the radius of circle.

2. The mathematical expression for centripetal force is ______
a) mv2/r
b) mv/r
c) v2/r
d) mv3/r
Answer: a
Clarification: The centripetal force is directly proportional to the square of the tangential velocity and inversely to the radius of circle. Hence, on calculating, we get the expression as mv2/r.

3. A body of mass 10 kg, is moving with a velocity of 5 m/s in a circle of radius 5 m, what is the centripetal acceleration of the body?
a) 5m/s2
b) 25m/s2
c) 0.5 m/s2
d) 50 m/s2
Answer: a
Clarification: The formula for centripetal force is mv2/r. Hence the centripetal acceleration is given by v2/r. Here, v = 5, r= 5. Hence, on solving we will get the centripetal acceleration as 5m/s2.

4. The centrifugal force always acts _____
a) Towards the center
b) Away from the center
c) In tangential direction
d) Outside of the plane of motion
Answer: b
Clarification: The centrifugal force always acts away from the center of the circle in which the body is moving. In contrast to this, the centripetal force always acts towards the center of the circle, keeping the body moving in the circle.

5. A ball is being rotated in a circle of radius 5 m with a constant tangential velocity of 20 m/s. A stone is also being rotated in a circle of radius 4 m with a constant tangential velocity of 16 m/s. Which one of the following choices is true about both the circular motions?
a) Both have same angular velocity
b) Both have different angular velocity
c) Angular velocity of ball > angular velocity of stone
d) Angular velocity of stone > angular velocity of ball
Answer: a
Clarification: Angular velocity = Tangential velocity/Radius. When we put in the values and calculate the angular velocities for each of the circular motions, we see that both the angular velocities are equal to each other and the value is 4 rad/s.

6. The angular velocity of a stone being rotated is 11 rad/s. What is the angular displacement covered in 0.5s?
a) 5.5 rad
b) 0.55 rad
c) 55 rad
d) 0.5 rad
Answer: a
Clarification: Angular velocity is the rate of change of angular displacement. Here the angular velocity is 11 rad/s. This implies that in one second the stone rotates by an angle of 11 rad. Hence in 0.5 s it will cover an angular displacement if 5.5 rad.

7. A body is moving in a vertical circular motion. Which one of the following forces does it not experience?
a) Force of gravity
b) Centripetal force
c) Normal reaction force
d) Centrifugal force
Answer: c
Clarification: A body moving in vertical circular motion experiences these four forces – centripetal force, centrifugal force, force of gravity, and resistance offered by the medium in which it is moving. Apart from these, it does not experience any other force in normal conditions.

8. A body of weight 20 N, mass 2 kg is moving in vertical circular motion with the help of a string of radius 1 m and with a velocity of 2 m/s. What is the tension in the string at the lowest point?
a) 28 N
b) 20 N
c) 8 N
d) 15 N
Answer: a
Clarification: At the lowest point, the body experiences centrifugal force and the weight in the same direction, opposite to the direction of tension in the string. Centrifugal force can be calculated as mv2/r = 8 N. Hence the tension = weight + centrifugal force = 28 N.

9. A body of weight 20 N, mass 2 kg is moving in vertical circular motion with the help of a string of radius 1 m and with a velocity of 5 m/s. What is the tension in the string at the highest point?
a) 30 N
b) 50 N
c) 20 N
d) 25 N
Answer: a
Clarification: At the highest point, the tension and the weight are in the same direction but the centrifugal force is in the opposite direction. Centrifugal force can be calculated as mv2/r = 50 N. Hence the tension = centrifugal force-weight = 30 N.

10. A body of weight 20 N, mass 2 kg is moving in vertical circular motion with the help of a string of radius 1 m and with a velocity of 5 m/s. What is the tension in the string when is horizontal?
a) 30 N
b) 50 N
c) 20 N
d) 25 N
Answer: b
Clarification: When the string is horizontal, the tension in the string and the centrifugal force are opposite to each other. Centrifugal force can be calculated as mv2/r = 50 N. Hence the tension = centrifugal force = 50 N.

11. Which one of the following devices acts on the principle of circular motion?
a) Centrifuge
b) Screw Gauge
c) Ruler
d) Vernier calipers
Answer: a
Clarification: The centrifuge utilizes the centrifugal force produced during circular motion. Centrifuge is commonly used to separate platelets from blood samples. It is an important equipment for studying blood. Other variations of centrifuge include devices like cream separator etc.

12. At which position in vertical circular motion is the tension in the string minimum?
a) At the highest position
b) At the lowest position
c) When the string is horizontal
d) At an angle of 35° from the horizontal
Answer: a
Clarification: When the body is at the highest point of its motion. This is because at the highest point the tension = centrifugal force – weight. This is the minimum value for tension throughout the motion.

250+ TOP MCQs on The Potential Energy of a Spring | Class 11 Physics

Physics Multiple Choice Questions on “The Potential Energy of a Spring”.

1. The potential energy possessed by a spring is also known as _____
a) Elastic potential energy
b) Extensive potential energy
c) Compressive potential energy
d) Deflection potential energy

Answer: a
Clarification: The potential energy stored in a spring is the consequence of its elastic property and hence it is also termed “elastic potential energy”. It can be stored when the spring is extended or compressed.

2. How much should a spring of indefinite length be compressed to have a potential energy equivalent to a ball of mass 6 kg raised to a height of 120 m above the ground? Let the spring have a stiffness of k = 100 N/m and assume g = 10 m/s2.
a) 6m
b) 12m
c) 20m
d) 144m

Answer: b
Clarification: PE of ball = m x g x h
= 6 x 10 x 120
= 7200 J
PE of spring = 1/2 x k x d2
Given;
PE of ball = PE of spring
7200 = 1/2x k x d2
= 1/2 x 100 x d2
d2 = 144
d = 12 m.

3. Assume a spring extend by “d” due to some load. Let “F” be the spring force and “k’ the spring constant. Then, the potential energy stored is _____
a) 2d/F2
b) F2/2k
c) 2k/T2
d) F2/2d

Answer: b
Clarification: F = k*d
PE = 1/2*k*d2
= 1/2*F*d
= 1/2*F*F/k
= F2/2k.

4. A spring of length 1m has two cars connected to both of its ends. The two cars move towards eachother such that the spring is compressed to 0.5m. If the spring constant is 500 N/m, what is the elastic potential energy stored?
a) 125 J
b) -125 J
c) 62.5 J
d) -62.5 J

Answer:c
Clarification: PE = 1/2*k*d2
= 1/2 x 500 x (-0.5) 2
= 62.5 J.
The elastic potential energy is positive even for negative displacement.

5. What is the increase in potential energy storage when the compression distance is doubled in a spring obeying Hooke’s law?
a) No increase
b) 100% increase
c) Cannot be determined
d) 4 times

Answer: d
Clarification: PE = 1/2 x k x d2
The elastic potential energy stored is directly proportional to the square of the compression. Hence, a two-fold increase in compression will result in potential energy storage to increase by four times.

6. The elastic potential energy varies linearly with displacement.
a) True
b) False

Answer: b
Clarification: PE = 1/2 x k x d2; where d = Displacement
Equation of parabola: y = 4x2
By comparing the two equations, we can conclude that the elastic potential energy varies parabolically with displacement and not linearly.

250+ TOP MCQs on Universal Law of Gravitation | Class 11 Physics

Physics Multiple Choice Questions on “Universal Law of Gravitation”.

1. For which of the following does the graph denote the variation of force of gravity “F” along a distance “r”?

a) Solid sphere
b) Spherical shell
c) Plate
d) Point
Answer: b
Clarification: The force of gravity is zero withing a spherical shell and varies similar to a point source at a distance greater than or equal to its radius.

2. For which of the following does the graph denote the variation of force of gravity “F” along a distance “r”?

a) Solid sphere
b) Spherical shell
c) Plate
d) Point
Answer: a
Clarification: The gravitational force is proportional to distance “r” until r=radius. For values of “r” greater than the radius of the sphere, the force is inversely proportional to the square of the distance.

3. What is the gravitational force experienced by an object of 10kg 200m away from an object weighing 1 ton?
a) 1.6675 N
b) 2.6675 N
c) 3.6675 N
d) 4.6675 N
Answer: a
Clarification: From Newton’s law of gravitation, we have;
F = (G*M1*M2*)/R2
G = 6.67 x 10-11 N m2/kg2
M1 = 10kg
M2 = 1000kg
R = 200m
F = (6.67 x 10-11 x 10 x 1000) / 2002
= 1.6675 N.

4. Which scientist introduced the universal law of gravitation?
a) Albert Einstein
b) Isaac Newton
c) Stephen Hawking
d) Nikola Tesla
Answer: b
Clarification: The universal law of gravitation is a part of Isaac Newton’s work “Philosophiæ Naturalis Principia Mathematica (the Principia)”, first published on 5 July 1687.

5. What will be the value of acceleration due to gravity on the surface of the earth if the radius of the earth suddenly decreases to 60% of its present value, keeping the mass of the earth unchanged?
a) 9.81 m/s2
b) 5.89 m/s2
c) 16.35 m/s2
d) 27.25 m/s2
Answer: d
Clarification: From Newton’s law of gravitation, we have;
g = (G*M1)/R2
Since the radius is reduced to 60% of its original value;
The new radius R’ = 0.6 x R
Therefore;
g’ = (G*M1)/(0.6*R) 2
g’ = g/(0.6) 2
= 9.81 / (0.6 x 0.6)
g’ = 27.25 m/s2.

6. The value of acceleration due to gravity of earth at the equator is less than that of the poles due to _____
a) shape and rotation of the earth
b) mass of the sun
c) mass of the earth
d) mass of the moon
Answer: a
Clarification: The gravitational force is a central force. Acceleration due to gravity has different values for different points on the earth’s surface.

7. The weight of an object can be zero but the mass of an object can never be zero.
a) True
b) False
Answer: a
Clarification: The weight of an object is zero when the net gravitational force acting on the object is zero. However, the mass of an object can never be zero since mass is a property of matter.

8. Gravitational force is the strongest fundamental force.
a) True
b) False
Answer: b
Clarification: Gravitational force is the weakest fundamental force. The strong nuclear force is the strongest fundamental force of nature.

9. Gravitational force is _____
a) an imaginary force
b) a long-range force
c) a short-range force
d) the strongest fundamental force
Answer: b
Clarification: Gravitational force is a long-range force which is inversely proportional to the square of the distance. The strong nuclear force is the strongest fundamental force and is a short-range force.

10. The universal law of gravitation becomes more inapplicable as the size and distance between objects decreases.
a) True
b) False
Answer: a
Clarification: As the size and distance between objects decrease, nuclear forces become stronger and the law of gravitation cannot be applied. This calls for a new branch of physics known as “quantum physics”.

11. What is the force of gravity experienced by an object at the centre of the earth? (Assume that the earth is perfectly spherical)
a) 0 g-force
b) 1 g-force
c) 9.81 g-force
d) 10 g-force
Answer: a
Clarification: Since a body at the centre of the earth would experience equal gravitational force from all sides, the vector addition of all of these forces amount to zero. Therefore, the force experienced by an object at the centre of the earth is zero.

Physics,

250+ TOP MCQs on Fluids Mechanical Properties Pressure | Class 11 Physics

Physics Multiple Choice Questions on “Fluids Mechanical Properties Pressure – 1”.

1. When the area decreases _____
a) pressure increases
b) pressure decreases
c) pressure remains constant
d) the change in pressure cannot be determined

Answer: a
Clarification: Pressure (P) = Force (F)/Area (A)
From the above equation, we can see that the pressure is inversely proportional to area.
Hence, the pressure will increase when the area is decreased.

2. Pressure decreases when _____
a) only the force is increased but not the area
b) only the area is decreases but not the force
c) either area decreases or force increases
d) the force decreases and/or area increases

Answer: d
Clarification: Pressure (P) = Force (F)/Area (A)
From the above equation, we can infer that the pressure is inversely proportional to the area and directly proportional to the applied force.
Hence, the pressure will decrease when force is decreased and/or area is increased.

3. A force F is applied on a uniform rod of cross-section A and a force F’ is applied on a uniform rod of cross-section 3A. What is the relation between F and F’ if the pressure on both is the same?
a) F/F’ = 1/3
b) F/F’ = 3
c) F’/F = 1/3
d) F/F’ = 1/9

Answer: a
Clarification: Pressure on first rod (P) = F/A
Pressure on the second rod (P’) = F’/3A
Given;
P = P’
F/A = F’/3A
F/F’ = 1/3.

4. A force is applied on a cube of side 3m. Another force, double the magnitude of the previous force is applied on a cube of 1m. What is the ratio of pressure on the first cube (P) to the pressure on the second cube (P’)?
a) 1/9
b) 18
c) 1/18
d) 1/2

Answer:c
Clarification: P = F/32
P = F/9
P’ = 2F/12
P’ = 2F/1
P/P’ = (F/9)/(2F/1)
= 1/18.

5. Pascal’s Law is valid only for _____
a) water
b) metals
c) fluids
d) gases

Answer: c
Clarification: Pascal’s law is valid for water, gases and even other liquids. However, it is not applicable to any other form of matter. Hence, we can conclude that Pascal’s Law is valid only for fluids.

6. What is Pascal’s Law?
a) For every action, there is an equal and opposite reaction
b) Force is the time rate of change of momentum
c) For an ideal gas, the pressure is directly proportional to temperature and constant volume and mass
d) A pressure change at any point in the fluid is transmitted throughout the fluid such that the same change occurs everywhere

Answer: d
Clarification:For every action, there is an equal and opposite reaction – Newton’s Third Law
Force is the time rate of change of momentum – Newton’s Second Law
For an ideal gas, the pressure is directly proportional to temperature and constant volume and mass – Ideal Gas Law
A pressure change at any point in the fluid is transmitted throughout the fluid such that the same change occurs everywhere – Pascal’s Law.

7. From which of the following is the working of hydraulic machines based on?
a) Pascal’s Law
b) Newton’s Law of Cooling
c) Law of Gravitation
d) Ideal Gas Law

Answer: a
Clarification: Pascals Law states that a pressure change at any point in the fluid is transmitted throughout the fluid such that the same change occurs everywhere. This is employed in hydraulic machines by varying cross-sectional areas at required places.

8. The SI unit of pressure is _____
a) Newton
b) Joule
c) Pascal
d) Watt

Answer: c
Clarification: The SI unit of pressure is “Pascal”. It is denoted by “Pa”.
1 Pa = 1 N/m2
It is named after the famous scientist Blaise Pascal.

9. Which of the following is used to measure pressure?
a) Ammeter
b) Speedometer
c) Barometer
d) Voltmeter

Answer: c
Clarification: A “barometer” is a scientific instrument used to measure air pressure. The barometer can also be used to measure the altitude of a region because pressure varies with altitude.