Torque MCQ Quiz in தமிழ் - Objective Question with Answer for Torque - இலவச PDF ஐப் பதிவிறக்கவும்

Last updated on Apr 21, 2025

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Latest Torque MCQ Objective Questions

Top Torque MCQ Objective Questions

Torque Question 1:

Which of the following are examples of torque in day to day life?

  1. Opening of bottle cap

  2. Rotation of steering wheel of a car
  3. Movement of doors
  4. All of the above

Answer (Detailed Solution Below)

Option 4 : All of the above

Torque Question 1 Detailed Solution

The correct answer is option 4) i.e. All of the above

CONCEPT:

  • Torque: Torque is the turning effect produced on an object due to a force acting on it at a distance from a fixed point. The object tends to turn about this fixed point.
    • It is defined mathematically as the cross product of force and the perpendicular distance to an axis of rotation.
    • Consider a fore F acting at a distance r from a fixed point O as shown.

The torque produced is given by:

τ = r × F = rFsinθ 

  • The SI unit of torque is N-m.

  • The torque due to a couple:
    • couple is defined as two equal and oppositely directed parallel forces that act on a rigid body. As a result, the net force on the body is zero.
    • Each force contributes to the turning effect of the object. Since the net force is zero, it can only cause rotation without translation.

The torque due to a couple is given by:

τ = d × F

Where d is the distance between the two forces.

EXPLANATION:

  • Opening of bottle cap: While opening the cap of a bottle, a force is applied tangentially to the circumference of the bottle cap as shown in the figure. The two forces form a couple which in turn produces a turning effect.

  • The steering wheel of a car: The steering wheel of the car also rotates due to the torque as force is applied tangentially through our hands on the rim of the wheel.

  • Movement of doors: Doors have a fixed axis of rotation for their movement. The hinges provided on the edges of the doors help in the turning motion. When a force is applied on the doorknob, the perpendicular distance between the hinges and the doorknob acts as the moment arm. Thus, it is an example of torque in action.

  • Therefore, all the given cases are examples of the turning effect of force i.e. torque.

Torque Question 2:

120 N of force is required to open a nut using a spanner of length 10 cm. If another spanner of length 6 cm is used to open the same nut, what amount of force has to applied?

  1. 130 N
  2. 200 N
  3. 50 N
  4. 450 N

Answer (Detailed Solution Below)

Option 2 : 200 N

Torque Question 2 Detailed Solution

The correct answer is option 2) i.e. 200 N

CONCEPT:

  • Torque: Torque is the turning effect produced on an object due to a force acting on it at a distance from a fixed point. The object tends to turn about this fixed point.
    • It is defined mathematically as the cross product of force and the perpendicular distance to an axis of rotation.
    • Consider a fore F acting at a distance r from a fixed point O as shown.

The torque produced is given by:

τ = r × F = rFsinθ 

  • The SI unit of torque is N-m.

CALCULATION:

  • The opening of a nut using a spanner works on the principle of torque.
  • When a force is applied through a spanner, the fixed point lies where the nut is positioned. As a result, a turning effect is produced on the nut which helps in its opening or tightening.
  • The angle between force and moment arm is 90 when maximum torque is produced.

Given that:

Length of the spanner = moment arm, r = 10 cm = 0.1 m

Force applied = 120 N

Torque = rF = 0.1 × 120 = 12 N-m

To open the nut using another spanner, the same torque has to be produced.

Lenght of the second spanner, r = 6 cm = 0.06 m

τ = rF

12 = 0.06 × F

⇒ F = 200 N

Torque Question 3:

The torque acting on a body is the rotational analogue of :

  1. Mass of the body
  2. Linear kinetic energy of the body
  3. Linear velocity of the body
  4. Force in linear motion

Answer (Detailed Solution Below)

Option 4 : Force in linear motion

Torque Question 3 Detailed Solution

Concept:

Torque:

  • The action of a force that acts on a point with some distance from the action of the force is called Torque. If a force F is applied at a distance r from the point of action, then torque τ is 

τ = Force × distance. 

  • A common example of torque is applying force on the hinge of door and door rotates. 

  • Torque is a vector quantity which is basically the cross product of force and distance from applied point to point of action.
  • Torque plays the same role for a rotational motion as force plays for linear motion.
  • Torque is required for a body to change its angular acceleration.
  • So, the correct option is Force in linear motion.

Additional Information

Other analogs relating to Linear and translation motion are shown in the table.

Linear Motion Rotational Motion Analogous
Velocity v  Angular velocity ω = v / r 
Acceleration a  Angular acceleration α = a / r
Momentum p = mv Angular momentum p = m ω r

Torque Question 4:

A couple of 105 N-m is applied to a fly wheel of mass 10 kg and radius of gyration 50 m. The angular acceleration of the wheel in rad/s2 is:

  1. 200
  2. 50
  3. 10
  4. 4

Answer (Detailed Solution Below)

Option 4 : 4

Torque Question 4 Detailed Solution

The correct answer is option 4) i.e. 4

CONCEPT:

​Relationship between Torque and Moment of Inertia:

For the linear motion: From Netwon's 2nd law:

F = ma ⇒ a =  

Where a is the acceleration.

For the circular motion: Consider an object of mass m moving in a circular path of radius r.

Angular displacement = r.dθ

Acceleration =     

Torque = Force × perpendicular distance = F × r = ma × r = m(rα) × r = mr2α      ----(1)

We know, the moment of inertia, I = mr2      ----(2)

Substituting (2) in (1) we get,

Torque, τ = Iα 

CALCULATION:

Given that:

Torque, τ = 105 N-m

Mass, m = 10 kg

The radius of gyration, r = 50 m

Moment of inertia, I = mr2 = 10 × 502 = 25000 kg m2

Torque, τ = Iα 

⇒ Angular acceleration, α =  = 4 rad/s2

Torque Question 5:

A triangular plate is shown. A force  is applied at point P. The torque at point P with respect to point 'O' and 'Q' are:

  1. -15 - 20, 15 - 20
  2. 15 + 20, 15 - 20
  3. 15 - 20, 15 + 20
  4. -15 + 20, 15 + 20

Answer (Detailed Solution Below)

Option 1 : -15 - 20, 15 - 20

Torque Question 5 Detailed Solution

CONCEPT:

The torque is defined as the force applied to the body which tends of rotating and it is written as;

Here r is the distance and F is the force.

CALCULATION:

Given: Force, 

The distance is written as;

and 

Now, the torque at P about point 'O'

As we know,  and,  , therefore,

⇒ 

Similarly, at the position Q is written as;

 and,  , therefore,

⇒ 

Hence, option 1) is the correct answer.

Torque Question 6:

A torque meter is calibrated to reference standards of mass, length and time each with 5% accuracy. After calibration, the measured torque with this torque meter will have net accuracy of

  1. 15%
  2. 25%
  3. 75%
  4. 5%

Answer (Detailed Solution Below)

Option 2 : 25%

Torque Question 6 Detailed Solution

Calculation:

The torque is τ = r × F = [M1 L2 T-2]

The formula for the propagation of error in a product of quantities is:

Δτ / τ = (ΔM / M) + (2 * ΔL / L) + (2 * ΔT / T)

Where:

Δτ is the uncertainty in torque

τ is the measured torque

ΔM is the uncertainty in mass

M is the mass

ΔL is the uncertainty in length

L is the length

ΔT is the uncertainty in time

T is the time

Since each reference standard (mass, length, and time) has an uncertainty of 5%, we substitute ΔM / M = 5%, ΔL / L = 5%, and ΔT / T = 5% into the formula:

Δτ / τ = 5% + 2 * 5% + 2 * 5%

Δτ / τ = 5% + 10% + 10%

Δτ / τ = 25%

Thus, the net uncertainty in the measured torque is 25%. Therefore, the correct answer is:

Option 2: 25%

Torque Question 7:

Rotational analogue of force is :

  1. Gyration
  2. moment of inertia
  3. angular momentum
  4. Torque

Answer (Detailed Solution Below)

Option 4 : Torque

Torque Question 7 Detailed Solution

Concept:

  • Torque is a force that tends to cause rotation.
  •  Torque is a vector quantity.
  • Rational analogue: It is a force in a linear motion is torque.

where τ = torque, = radius of the rotatory motion, and  = force applied

Explanation:

Torque is necessary for a body to do rotational motion.

Additional Information

  • Moment of Inertia is the tendency of a body in rotational motion which opposes the change in its rotational motion due to external forces.
  • Moment of Inertia behaves as angular mass and is called rotational inertia. 
  • Angular momentum is defined as the product of the moment of inertia I and the angular velocity ω.
  • L = Iω
  • The radius of gyration is the distance from an axis of a body to the point in the body whose moment of inertia is equal to the moment of inertia of the entire body.

Torque Question 8:

A couple produces:

  1. no motion
  2. linear and rotational motion
  3. purely rotational motion
  4. purely linear motion

Answer (Detailed Solution Below)

Option 3 : purely rotational motion

Torque Question 8 Detailed Solution

Concept:

Torque:

  • Torque is a physical quantity that can cause an object to rotate about an axis.
  • Force is what causes an object to accelerate in linear kinematics. Similarly, torque is what causes angular acceleration. Hence, torque can be defined as the rotational equivalent of linear force.
  • Torque is a vector quantity.
  • Its SI unit is N-m.
  • If F force is acting at a distance r from the axis of rotation at an angle θ as shown in the figure, then the torque is given as,

​⇒ τ = F.r.sinθ

Couple:

  • pair of forces of equal magnitude but acting in opposite directions with different lines of action is known as a couple or torque.
  • couple produces rotation without translation.
  • Examples:
    1. When we open the lid of a bottle by turning it, our fingers are applying a couple to the lid.

Torque Question 9:

A 200 turn closely wound circular coil, of radius 10 cm, carrying a current of 1.6 A, is placed in a vertical plane. It is free to rotate about a horizontal axis that coincides with its diameter. A uniform magnetic field of 0.72 T exits in the region and initially the coil axis is parallel to this field. The coil rotates through an angle of 90° under the influence of magnetic field. If the moment of inertia of the coil is 0.1 kg m2, the angular speed acquired by the coil is (nearly)

  1. 12 rad/s
  2. 20 rad/s
  3. 32 rad/s
  4. 42 rad/s

Answer (Detailed Solution Below)

Option 1 : 12 rad/s

Torque Question 9 Detailed Solution

CONCEPT:

Torque: For a body rotating in a circular motion with a moment of inertia I and angular acceleration α 

the torque τ = Iα 

The magnetic moment of coil: The magnetic moment of a current-carrying coil M  = niA

Where, n = number of turns of the coil, A = area of the coil and i = current passing through the coil

EXPLANATION:

Given

n = 200, i = 1.6 A, r = 10 cm = 0.1 m, B = 0.72 T, I = 0.1 kg m2

Now, M = niA = 200×1.6×π(0.1)2 = 10.05 Am2

Where I and α are the moment of inertia and angular acceleration respectively

then the torque τ = Iα 

Again we know, τ = MB sinθ (M = magnetic moment)

So, Iα = MB sinθ

 = 12.03  rad/s

Hence the correct answer is option 1.

Torque Question 10:

Two forces F̅ and -3 F̅ act at a Perpendicular distance ‘d’ from each other. This is equivalent to which one of the following ?

  1. Net linear force -2 F̅ only
  2. Net linear force zero only
  3. A net couple of moment 2Fd only
  4. A net linear force of -2 F̅ and a net couple of moment Fd

Answer (Detailed Solution Below)

Option 4 : A net linear force of -2 F̅ and a net couple of moment Fd

Torque Question 10 Detailed Solution

Concept:

A force can be transferred over a distance by replacing it with equal force and a moment, as shown in fig

Calculation:

Given:

Hence the equivalent of the system is a net linear force of -2 F̅ and a net couple of moment Fd

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