Free NEET Physics multiple-choice questions on Torque & Equilibrium. Attempt each question and reveal the answer with a full explanation.
A force F = 4 i + 5 j - 6 k acts on a point r = i + j . The torque acting on the point about the origin is: -6 i + 6 j + k -6 i + 6 j - k 6 i - 6 j + k 6 i + 6 j - k A particle of mass m is moving with constant velocity v along a line parallel to the x -axis at a distance y = h . The torque acting on the particle about the origin is: Zero mvh -mvh mvx Two bodies have their moments of inertia I and 2I respectively about their axis of rotation. If their kinetic energies of rotation are equal, their angular momentum will be in the ratio - 1 : 2 2 : 1 1 : 2 2 : 1 A uniform rod of length and mass m is free to rotate in a vertical plane about A. The rod initially in horizontal position is released. The initial angular acceleration of the rod is (Moment of inertia of rod about A is m 2 3 ): 3g 2 2 3g 3g 2 2 mg 2 A wheel has angular acceleration of 3.0 rad/sec 2 and an initial angular speed of 2.00 rad/sec. In a time of 2 sec it has rotated through an angle (in radian) of 10 12 4 6 If F is the force acting on a particle having position vector r and be the torque of this force about the origin, then : r =0 and F 0 r 0 and F =0 r >0 and F <0 r =0 and F =0 A uniform circular disc of radius 50 cm at rest is free to turn about an axis which is perpendicular to its plane and passes through its centre. It is subjected to a torque which produces a constant angular acceleration of 2.0 rad s -2 . Its net acceleration in m s -2 at the end of 2.0 s is approximately : 8.0 7.0 6.0 3.0 A solid sphere is rotating freely about its symmetry axis in free space. The radius of the sphere is increased keeping its mass same. Which of the following physical quantities would remain constant for the sphere? Rotational kinetic energy Moment of inertia Angular velocity Angular momentum Find the torque about the origin when a force of 3 N acts on a particle whose position vector is 2 k m . 6 Nm -6 Nm 6 k Nm 6 Nm The angular speed of a fly wheel moving with uniform angular acceleration changes from 1200 rpm to 3120 rpm in 16 seconds. The angular acceleration in rad/s 2 is 4 12 104 2 The angular acceleration of a body, moving along the circumference of a circle, is Along the radius, away from centre Along the radius towards the centre Along the tangent to its position Along the axis of rotation Which of the following conditions is necessary for a rigid body to be in mechanical equilibrium? F ext = 0 and ext = 0 F ext = 0 only ext = 0 only P = 0 only The angular speed of a flywheel is increased from 600 rpm to 1200 rpm in 10 s. The number of revolutions completed by the flywheel during this time is : 600 300 900 150 What is the torque of a force F = -3 i + j + 5 k acting at a point r = 7 i + 3 j + k ? 14 i - 38 j + 16 k 4 i + 4 j + 6 k -21 i + 3 j + 5 k 14 i + 38 j + 16 k A force F = i + 2 j + 3 k acts at a point whose position vector is r = i - j + k . The torque of the force about the origin is: -5 i - 2 j + 3 k -5 i + 2 j + 3 k 5 i - 2 j - 3 k 5 i + 2 j - 3 k A uniform rod of length 200 cm and mass 500 g is balanced on a fulcrum placed at 40 cm mark. A mass of 2 kg is suspended from the rod at 20 cm mark and another unknown mass m is suspended from the rod at 160 cm mark. Find the value of m such that the rod is in equilibrium. ( g = 10 m/s 2 ) 1 12 kg 1 6 kg 1 3 kg 1 2 kg A force F = 2 i + 3 j - 4 k acts at a point P(1, -1, 2) . The torque of this force about the point Q(2, 0, -1) is: -5 i - 10 j - 10 k 5 i + 10 j + 10 k -5 i + 10 j - 10 k 10 i + 5 j + 10 k A ladder is leaning against a smooth vertical wall and resting on a rough horizontal floor. The ladder is in equilibrium. Which of the following forces provides the torque to counter the torque produced by gravity about the point of contact with the floor? Normal force from the wall Friction force from the floor Normal force from the floor Tension in the ladder A torque of 100 N m is applied to a wheel of moment of inertia 10 kg m 2 . The work done by the torque in 2 seconds , starting from rest, is: 2000 J 1000 J 500 J 4000 J Find the torque of a force F = 2 i - 3 j + 4 k acting at the point r 1 = 3 i + 2 j + 3 k about the point r 2 = i + j + k . 7 i - 4 j - 13 k 2 i - j + 2 k -7 i + 4 j + 13 k 5 i + 3 j - 2 k A particle of mass m moves in the XY plane with a velocity v along the straight line AB. If the angular momentum of the particle with respect to origin O is L A when it is at A and L B when it is at B, then L A = L B the relationship between L A and L B depends upon the slope of the line AB L A < L B L A > L B A thin circular ring of mass M and radius R is rotating in a horizontal plane about an axis vertical to its plane with a constant angular velocity , If two objects each mass m be attached gently to the opposite ends of a diameter of the ring, the ring, will then rotate with an angular velocity : M M+m (M-2m) M+2m M M+2m (M+2m) M A force F = α i + 3 j + 6 k is acting at a point r = 2 i - 6 j - 12 k . The value of α for which the torque about the origin is zero is: -1 1 -2 2 A solid cylinder of mass 50 kg and radius 0.5 m is free to rotate about the horizontal axis. A massless string is wound round the cylinder with one end attached to it and other hanging freely. Tension in the string required to produce an angular acceleration of 2 revolutions s -2 is :- 25 N 50 N 78.5 N 157 N The angular momentum of a rigid body of mass m about an axis is n times the linear momentum (P) of the body. Total kinetic energy of the rigid body is : n 2P 2 2 P 2[1+n 2] 2m n 2P 2 2m Cannot be determined (moment of inertia not given) The moment of the force, F = 4 i + 5 j - 6 k at (2, 0, -3) , about the point (2, -2, -2) , is given by -7 i - 8 j - 4 k -4 i - j - 8 k -8 i - 4 j - 7 k -7 i - 4 j - 8 k A uniform rod of length 200 cm and mass 500 g is balanced on a wedge placed at 40 cm mark. A mass of 2 kg is suspended from the rod at 20 cm and another unknown mass 'm' is suspended from the rod at 160 cm mark as shown in the figure. Find the value of 'm' such that the rod is in equilibrium. ( g = 10 m/s 2 ) 1 3 kg 1 6 kg 1 12 kg 1 2 kg The Sun rotates around its centre once in 27 days. What will be the period of revolution if the Sun were to expand to twice its present radius without any external influence? Assume the Sun to be a sphere of uniform density. 108 days 100 days 105 days 115 days A rod of length L and weight W is supported by two parallel knife edges A and B at distance x from each end. The center of gravity of the rod is at a distance d from end A . The normal reaction at A is: W(L - x - d) L - 2x Wd L W(L - x) d W(d - x) L - 2x The torque of a force F = -2 i + 3 j + 4 k acting at a point r = 3 i + 2 j + 3 k about the point (1, 1, 1) is: i - 10 j + 8 k i + 10 j - 8 k 2 i - 6 j + 12 k -2 i + 6 j - 12 k A force F = (2 i + 3 j + 4 k ) N acts on a rigid body and the body rotates with an angular velocity = (3 i + 2 j + 4 k ) rad/s . The power delivered is: Cannot be determined from given data 28 W 24 W 12 W A solid cylinder of mass 2 kg and radius 4 cm is rotating about its axis at the rate of 3 rpm . The torque required to stop after 2 revolutions is 2 10 6 N m 2 10 -6 N m 2 10 -3 N m 12 10 -4 N m Find the torque about the origin when a force of 3 j N acts on a particle whose position vector is 2 k m . -6 i N m 6 i N m 6 j N m 6 k N m A couple produces: Purely rotational motion Purely translational motion Both translational and rotational motion No motion The torque required to produce an angular acceleration of 2 rad/s 2 in a body of moment of inertia 2.5 kg m 2 is: 5.0 N m 1.25 N m 0.8 N m 4.5 N m The work done by a torque in rotating a body is given by W = d . If the torque acting on a wheel is ( ) = k 2 , where k is a constant, the work done in rotating the wheel from = 0 to = is: k 3 3 k 2 k 2 2 2k 3 A force F = (2 i + 3 j ) N acts on a particle. The torque of this force about the origin is zero. If the particle is at (x, 3) , then the value of x is: 2 -2 3 0 A uniform rod of mass M and length L is pivoted at one end. A constant force F is applied at the other end perpendicular to the rod. The angular acceleration of the rod at the start is: 3F/(ML) F/(ML) 2F/(ML) 6F/(ML)