Question 1

For an inextensible string connecting two bodies, the constraint condition is that:

Accepted Answer

The components of the velocities of its two ends along the string are equal. Inextensibility means the length is constant, so the rate at which the string is fed in at one end equals that at the other — i.e. equal velocity components along the string.

Question 2

The virtual-work (constraint) method uses the fact that the total work done by the internal string tensions in an ideal system is:

Accepted Answer

Zero. For ideal (massless, inextensible) strings the net work of the tension pairs is zero: Σ T·v = 0, giving the acceleration/velocity constraint relations.

Question 3

A load hangs from a single movable pulley and the free end of the string is pulled upward. The load moves:

Accepted Answer

At half the speed of the free end. A movable pulley is supported by two string segments, so the load moves half as fast (and half as far) as the free end. Pitfall: correspondingly the effort force is half the load.

Question 4

The string-constraint equation for a system of connected bodies is obtained by requiring that:

Accepted Answer

The total length of the (inextensible) string is constant, so the sum of the rates of change of its segments is zero. Differentiating "total length = constant" gives Σ(dℓ_i/dt) = 0 (and a second derivative gives the acceleration constraint).

Question 5

A block slides on the incline of a wedge that is itself free to move on the floor. The constraint is that:

Accepted Answer

The block's acceleration relative to the wedge is directed along the incline surface. The block stays in contact with the wedge, so its acceleration relative to the wedge lies along the incline; its absolute acceleration is the vector sum of that and the wedge's acceleration.

Question 6

For two blocks connected by a single inextensible string over a fixed ideal pulley, the magnitudes of their accelerations are:

Accepted Answer

Equal. A single string over a fixed pulley gives both blocks the same speed and the same magnitude of acceleration.

Question 7

For a rigid rod, the constraint linking the velocities of its two ends A and B is:

Accepted Answer

The components of v_A and v_B along the rod are equal. A rigid rod cannot stretch, so the velocity components of its two ends along the rod must be equal.

Question 8

A block rests on a wedge that can slide on the ground. If the wedge accelerates, the block's motion is constrained by:

Accepted Answer

Both the contact with the incline and (if applicable) the string over the wedge pulley. Contact with the wedge surface (and any connecting string) provides the geometric constraints that relate the block's and wedge's accelerations.

Question 9

An ideal movable-pulley arrangement in which the load is supported by n string segments gives a mechanical advantage (ideal) of:

Accepted Answer

n. With n supporting segments, the effort = load/n, so the ideal mechanical advantage is n (velocity ratio is also n).

Question 10

In the virtual-work method, one assigns small virtual displacements to the bodies that are:

Accepted Answer

Consistent with the constraints, then sets the net work of the constraint forces to zero. Virtual displacements must respect the constraints; since constraint forces do no net work, ΣF·δr for the active forces gives the required relation.

Question 11

Two blocks are connected by a string over a fixed ideal pulley. If one block descends at 3 m/s, the other block:

Accepted Answer

Rises at 3 m/s. Single string over a fixed pulley ⇒ equal speeds. The other block rises at 3 m/s.

Question 12

A load hangs from a movable pulley whose string's free end is pulled upward at 6 m/s. The load rises at:

Accepted Answer

3 m/s. Movable pulley ⇒ load speed = half the free-end speed = 6/2 = 3 m/s.

Question 13

A load on a movable pulley rises at 2 m/s. The free end of the string is being pulled at:

Accepted Answer

4 m/s. Free-end speed = 2 × load speed = 2 × 2 = 4 m/s.

Question 14

In an ideal single-movable-pulley system, the effort needed to just support a 200 N load is:

Accepted Answer

100 N. Two supporting segments ⇒ effort = load/2 = 200/2 = 100 N.

Question 15

A ladder leans against a wall. Its foot (end A) slides along the floor at 5 m/s. When the ladder makes 37° with the floor (sin37° = 0.6, cos37° = 0.8), the top end B slides down at:

Accepted Answer

≈ 6.67 m/s. Equal velocity components along the rod: v_A cosθ = v_B sinθ ⇒ v_B = v_A cotθ = 5 × (0.8/0.6) ≈ 6.67 m/s.

Question 16

A ladder's foot slides at 4 m/s along the floor. When the ladder makes 53° with the floor (cot53° = 0.75), the speed of its top end (on the wall) is:

Accepted Answer

3 m/s. v_B = v_A cotθ = 4 × 0.75 = 3 m/s.

Question 17

A block on the floor is pulled by a rope passing over an overhead pulley; the rope is drawn in at 5 m/s and makes 60° with the horizontal at the block. The speed of the block is:

Accepted Answer

10 m/s. Rate of shortening of the rope = component of block velocity along the rope: 5 = v cos60° ⇒ v = 5/0.5 = 10 m/s. Pitfall: the block moves faster than the rope is pulled.

Question 18

A block is dragged by a rope over a pulley, the rope being pulled in at 6 m/s while it makes 30° with the block's (horizontal) motion. The block's speed is (cos30° ≈ 0.866):

Accepted Answer

≈ 6.93 m/s. 6 = v cos30° ⇒ v = 6/0.866 ≈ 6.93 m/s.

Question 19

Two blocks connected by a string over a fixed pulley: one has a downward acceleration of 4 m/s². The other block's acceleration is:

Accepted Answer

4 m/s² upward. Single string, fixed pulley ⇒ equal-magnitude accelerations in opposite senses: 4 m/s² upward.

Question 20

The free end of the string of a movable-pulley system is pulled with an acceleration of 8 m/s². The load's acceleration is:

Accepted Answer

4 m/s². Load acceleration = half the free-end acceleration = 8/2 = 4 m/s².

Constrained Motion (Kinematic Linkages) — JEE Main Physics MCQs with Solutions

Questions with solutions