Category: Class 11 Physics Objective Questions

Physics Multiple Choice Questions on “Reflection of Waves”.

1. If a string wave encounters a completely fixed end, what happens to its phase?
a) Stays the same
b) Changes by π
c) Changes by π/2
d) Waves gets destroyed
Answer: b
Clarification: The reason for the phase to change by π is that the boundary displacement should be zero because it is rigid & by principle of superposition it is only possible if the reflected wave differs by a phase of π.

2. The equation of a wave is: y = 2sin(4x-3t). What will be the equation of the reflected wave from a free surface?
a) y = 2sin(4x+3t)
b) y = 2sin(3t-4x)
c) y = 2cos(4x+3t)
d) y = 2sin(4x-3t)
Answer: a
Clarification: At the free surface reflected wave suffers no phase change.
Only the direction of wave propagation gets reversed.
So, the equation will be y = 2sin(4x+3t).

3. In a standing wave the amplitude of a particle is fixed, but varies from particle to particle. True or False?
a) True
b) False
Answer: a
Clarification: A standing wave neither moves left or right, all particles between two nodes are in the same phase. The equation of a standing wave is of the form Asin(kx)cos(wt),
where Asin(kx) is the amplitude which is different for different values of x.
This shows that for each value of x we get a different value of amplitude.

4. What is the minimum distance between a node & an antinode in a standing wave?
a) λ
b) λ/2
c) 2λ
d) λ/4
Answer: d
Clarification: The amplitude in a standing wave is given by Asin(kx).
Node is a point where amplitude is zero and antinode is a point where amplitude is maximum.
For a node, sin(kx) = 0
kx = 2nπ OR (2π/λ)x = nππ
x = nλ/2.
For an antinode, sin(kx) = 1
kx = (2n+1)π/2
(2π/λ)x = (2n+1)π/2
x = (2n+1) λ/4.
∴ node is at λ/2, λ, 3λ/2 …
& antinode at λ/4, 3λ/4, 5λ/4 …
∴ minimum distance = λ/2 – λ/4 = λ/4.

5. If a standing wave is vibrating in the fourth harmonic and the wavelength is λ, what is the length of the string.
a) 2λ
b) λ
c) 4λ
d) λ/4
Answer: a
Clarification: If the length of string is L, then the wavelength of the standing wave is 2L/n.
Or we can say L = nλ/2, where n corresponds to nth harmonic.
∴ L = 4λ/2
= 2λ.

6. Consider the case of normal modes of vibration for the air column with only one end closed. What is the length of the air column for the fundamental harmonic?
a) λ/2
b) λ
c) 2λ
d) λ/4
Answer: d
Clarification: For the given condition wavelength is related to length by: λ = 4L/(2n+1).
∴ For fundamental harmonic n=0, & L = (2n+1)λ/4
= (0 + 1)λ/4
= λ/4.

7. A pipe is open at both ends. What should be its length such that it resonates a 10Hz source in the 2nd harmonic? Speed of sound in air = 340m/s.
a) 34m
b) 68m
c) 17m
d) 51m
Answer: a
Clarification: For a pipe open at both ends frequency ‘f’ = nv/2L, where n is speed of sound.
Given: f = 10 & n=2.
∴ 10 = 2*340/2L
∴ L = 680/20
= 34m.

8. Consider standing waves in an air column with one end closed. What is a pressure node?
a) Pressure variation is maximum
b) Displacement variation is minimum
c) Same as displacement node
d) Least pressure change
Answer: d
Clarification: The point where there is an antinode corresponds to maximum amplitude of displacement. This also implies that here, pressure changes are minimum and it is therefore called a pressure node.

9. A pipe open at both ends has a fundamental frequency of 2Hz. If the pipe is closed at one end and half filled with water, what will be the fundamental frequency?
a) 0.5Hz
b) 2Hz
c) 4Hz
d) 1Hz
Answer: b
Clarification: The fundamental frequency in the first case is given by: f1 = v/2L.
In the second case length becomes half & one end is closed,
so fundamental frequency f2 = v/(4L/2)
= v/2L = f1.
So the value of f2 = 2Hz.

10. Standing waves can be produced when two identical waves, having a phase difference of π, are travelling in the same direction. True or False?
a) True
b) False
Answer: b
Clarification: The equation of standing waves is given by; y = Asin(kx)cos(wt).
This is formed by adding two waves of the form asin(kx-wt) & bsin(kx+wt)
which shows that they both must be travelling in opposite directions.

Physics Multiple Choice Questions on “Measurement of Time”.

1. Which of the following are NOT units of time?
a) Second
b) Light year
c) Tropical year
d) Minute
Answer: b
Clarification: Light year is a unit of measurement of distance. Tropical year is the amount of time sun takes to return to the same position in the season cycle.

2. Which of the following unit is related to sound frequency?
a) Hertz
b) Decibel
c) Second
d) Meter
Answer: a
Clarification: Decibel is the unit of sound intensity. Hertz is the unit for measuring sound frequency. Hertz is the inverse of second.

3. How many milliseconds make one second?
a) 10
b) 60
c) 45
d) 80
Answer: a
Clarification: Ten milliseconds make one second. 60 seconds make one minute. And 600 minutes make one hour.

4. How many decades are there in 1 century?
a) 10
b) 20
c) 4
d) 5
Answer: a
Clarification: One decade is made up of hundred years. One century is made up of 100 years. Hence, one century contains 10 decades.

5. How many years are there in one millennium?
a) 1000
b) 100
c) 10
d) 70
Answer: a
Clarification: On millennium is equivalent to 1000 years. One millennium contains 100 centuries. It is used to represent huge time periods.

6. Which of the following gives the most accurate result for time measurement?
a) Wall clock
b) Digital watch
c) Quartz clock
d) Atomic clock
Answer: d
Clarification: Atomic clocks are the most accurate devices for time measurement. The most accurate atomic clock is the caesium atomic clock situated in Switzerland.

7. Which of the following is equivalent to one hour?
a) 36000 milliseconds
b) 3600 milliseconds
c) 3600000 milliseconds
d) 360000 milliseconds
Answer: a
Clarification: One hour is equal to 60 minutes. 60 minutes are equal to 3600 seconds. And 3600 seconds are equal to 36000 milliseconds as one second equals to 10 milliseconds.

Physics Multiple Choice Questions on “Motion in a Plane”.

1. How many variables are required to define the motion of a body in a plane?
a) 1
b) 2
c) 3
d) 4
Answer: b
Clarification: The motion of a body in a plane can be defined by using two variables. These are the two defining axes of the Coordinate system, x and y. Once we have the displacement as a function of these to variables, any other quantity can be found out.

2. A body is exhibiting circular motion. What kind of motion can this be termed as?
a) Motion along a line
b) Motion in a plane
c) Motion in space
d) Motion along a point
Answer: b
Clarification: Circular motion is an example of motion in a plane. As a circle is a 2-dimensional entity, the body moving in a circle is also moving in a plane. Hence the motion can be termed as motion in a plane.

3. Which of the following is not an example of motion in a plane?
a) A car moving in a rectangular path
b) A bicycle moving in a circular path
c) A rocket moving into space
d) A truck moving in an infinite spiral
Answer: c
Clarification: A rocket moving into space is an example of 3-dimensional motion. Hence, it cannot be out in the category of motion in a plane. Rest all examples of motion in a plane. The truck moving in an infinite spiral is also an example of motion in a plane as spiral is a 2-dimensional entity.

4. In how many independent directions can a vector in a plane be resolved?
a) 1
b) 2
c) 3
d) 4
Answer: b
Clarification: A vector in a plane will be defined by the two governing axes, X and Y. Hence, any vector can be resolved in two independent directions. If it is resolved in any other direction, that component of it will be dependent on the existing independent directions.

5. Which of the following is not true about projectile motion?
a) It is an example of motion in a plane
b) It is an example of motion along a curve
c) It is not an example of motion in space
d) The acceleration keeps changing in projectile motion
Answer: a
Clarification: The only acceleration in projectile motion is the acceleration due to gravity. If the maximum height of the projectile motion is not large, and can be neglected with respect to the radius of earth (which usually is the case), the acceleration due to gravity remains constant. Hence, the acceleration in projectile motion remains constant.

Physics Aptitude Test on “Work Done by a Variable Force”.

1. The length of a smooth inclined plane of 30-degree inclination is 5 m. The work done in moving a 10 kg mass from the bottom of the inclined plane to the top is _____ Joules. (Assume g = 10m/s²)
a) 250
b) 1000
c) 1250
d) 500

Answer: a
Clarification: Length (l) = 5 m
Height = Displacement = 5 x sin30
= 5 x (1/2)
= 2.5
Force on object = m x g
= 10 X 10
= 100 N
Work = Force x Displacement
= 100 x 2.5
= 250 Joules.

2. A rope has a uniform mass density of 0.4 kg/m. The rope is 10 m long and is lowered into a 10 m deep pit. The bottom part of the rope just touches the bottom of the pit. What is the work done to pull the rope out of the well completely? (Assume g = 10m/s²)
a) 200 J
b) 400 J
c) 2000 J
d) 4000 J

Answer:a
Clarification: Mass density (p) = 0.4 kg/m
Length (l) = 10 m
Force = f(x)
= m(x)*g
= 0.4x*g
= 4x
Displacement (d) = 10 m
Work = (int_{0}^{l})f(x)dx
= (int_{0}^{10})4xdx
= 200 J.

3. A bucket filled with water weighing 20 kg is raised from a well of depth 20 m. If the linear density of the rope is 0.2 kg/m, the work done is _____ (Assume g = 10m/s²)
a) 4000 J
b) 4040 J
c) 4400 J
d) 4800 J

Answer: c
Clarification: Mass density (p) = 0.2 kg/m
Length (l) = 20 m
Force = f(x)
= m(x)*g
= 0.2x*g
= 2x
Displacement (d) = 20 m
Work = (int_{0}^{l})f(x)dx+ (20*g*20)
= (int_{0}^{20})2xdx + 4000
= 4400 J.

4. A force F = (5i – 3j +2k) N moves a particle from r1 = (2i + 7j + 4k) m to r2 = (5i + 2j + 8k) m. What is the work done by the force?
a) 18 J
b) 28 J
c) 38 J
d) 48 J

Answer: c
Clarification: Displacement (d) = r2 – r1
= (3i – 5j + 4k) m
Work = Force . Displacement; [Dot product of vectors]
= (5i – 3j +2k) .(3i – 5j + 4k)
= 38 J.

5. A person of mass 50 kg carrying a load of 20 kg walks up a staircase. The width and height of each step are 0.25 m and 0.2 m respectively. What is the work done by the man in walking 20 steps?(Assume g = 10m/s²)
a) 800 J
b) 1600 J
c) 2000 J
d) 2800 J

Answer: d
Clarification: Total mass (m) = 50 + 20 = 70 kg
Vertical displacement = 0.2 x 20 = 4m
Horizontal displacement = 0.25 x 20 = 5m
Displacement = (5i + 4j) m
Force = m x acceleration
= 70 x (0i + 10j)
= (700j) N
Work = Force .Displacement
= (0i + 700j) . (5i + 4j)
= 2800 J.

6. A chain of length 4 m is kept on a table. 2 m of the chain hangs freely from the table’s edge. The mass of the whole chain is 8 kg. What is the work done to pull the whole chain onto the table? (Assume g = 10m/s²)
a) 20 J
b) 40 J
c) 60 J
d) 80 J

Answer: b
Clarification: Mass per length of chain (p) = 8/4 = 2 kg/m
Mass of ‘x’ m of length = 2x kg
Force = (int_{0}^{2})(2x * g)dx
= (int_{0}^{2})20xdx
= 40 J.

7. The area under the force (F) versus displacement (x) graph gives work.
a) True
b) False

Answer: a
Clarification: The area under this graph would be (int_{x1}^{x2})F * dx, which is work. Here x1 & x2 are two positions on the x-axis. Hence the statement is true.

8. An object of mass 1kg is subjected to a variable force in x-direction. The force function is F = x² N. What is the work done in moving the object from x = 2 to 5?
a) 40
b) 39
c) 38
d) 69

Answer: b
Clarification: The work is given by formula W = (int_{x1}^{x2})F * dx. So, W = (int_{2}^{5})x² * dx = (5³-2³)/3 = 39 J.

9. What is the SI unit of variable force?
a) Newtons
b) Joules
c) Ampere
d) Mole

Answer: a
Clarification: The SI unit of force is Newtons. While Joules, Ampere, and Mole are SI units of energy, current and amount of substance.

10. A spring is stretched by a length x, then the potential energy stored in it changes by 0.5*k*x², where k is the spring constant.
a) True
b) False

Answer: a
Clarification: The potential energy of a stretched spring is given by 0.5*k*x² where x is the elongation in the spring.

11. A force of 600 N elongates a spring from its natural length of 18 cm to a length of 20 cm. What is the work done?
a) 116J
b) 114J
c) 118J
d) 120J

Answer: b
Clarification: The force on a spring stretched by x meters is given by F = k*x where k is the spring constant. So, 600 = k*(2/100). Now k = 3*10⁴, and hence F = 30000*x. W = (int_{x1}^{x2})F * dx = (int_{0.18}^{0.2})30000 * x * dx = 114 J.

12. Which of the following is not a variable force?
a) F = x
b) F = y²
c) F = 2
d) F = 2*t

Answer: c
Clarification: F = x, F = x² varies with x and y co-ordinate respectively. F = 2*t varies with time but F = 2 is always constant.

13. Which of the following is not a conservative force?
a) Gravitational force
b) Spring force
c) Friction force
d) Magnetic force

Answer: c
Clarification: Friction force causes energy losses in form of heat so it is a non-conservative force while gravitational, spring and magnetic forces are all conservative.

14. A force F = x + 1 N acts on a particle in x direction. What is the work done from x = 2m to 5 m?
a) 10.5 J
b) 14.5 J
c) 13.5 J
d) 0.5 J

Answer: c
Clarification: Work done is given by W = (int_{x1}^{x2})F * dx. So, W = (int_{2}^{5})(x + 1) * dx = [(5²/2 + 5) – (2²/2+2)] = 13.5.

15. What is the value of spring constant if force of 200 N is acting on it and the spring is compressed by 2m?
a) 100 kgs^-2
b) 100 kgs^-1
c) 100 kg²s²
d) 100 Kgs

Answer: a
Clarification: Force on a elongated/compressed spring is given by F = k*x where k is spring constant and x is elongation/compression. So, 200 = k*2 which gives k = 100 kgs^-2.

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 w_o 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

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.