JEE Advanced · 2021 · 03 Oct 2021 · Paper 2 · Rotational Mechanics

One end of a horizontal uniform beam of weight

W

and length

L

is hinged on a vertical wall at point

O

and its other end is supported by a light inextensible rope. The other end of the rope is fixed at point

Q

, at a height

L

above the hinge at point

O

. A block of weight

α W

is attached at the point

P

of the beam, as shown in the figure (not to scale). The rope can sustain a maximum tension of

(22) W

. Which of the following statement(s) is(are) correct ?

(A)The vertical component of reaction force at $O$ does not depend on $α$ ✓ Correct
(B)The horizontal component of reaction force at $O$ is equal to $W$ for $α = 0.5$ ✓ Correct
(C)The tension in the rope is $2 W$ for $α = 0.5$
(D)The rope breaks if $α > 1.5$ ✓ Correct

Answer: (A), (B), (D)

Solution

Taking torque about point O, the beam is in equilibrium. From geometry, the rope makes 45° with the horizontal. Let

R_{x}

and

R_{y}

be the horizontal and vertical components of the reaction at O, and

T

be the tension in the rope.

Torque balance about O:

W \frac{ℓ}{2} + α W ℓ = \frac{T}{2} ℓ

T = 2 (\frac{W}{2} + α W)

Taking torque about P:

R_{y} ℓ = W \frac{ℓ}{2}

R_{y} = \frac{W}{2}

This is independent of

α

, so (A) is correct.

Force balance:

R_{x} = \frac{T}{2} = \frac{W}{2} + α W

For

α = 0.5

R_{x} = \frac{W}{2} + 0.5 W = W

, so (B) is correct.

For

α = 0.5

T = 2 (\frac{W}{2} + 0.5 W) = 2 W \neq = 2 W

, so (C) is incorrect.

Maximum tension:

22 W = 2 (\frac{W}{2} + α W)

2 W = \frac{W}{2} + α W

α = \frac{3}{2} = 1.5

So the rope breaks if

α > 1.5

, making (D) correct.

JEE Advanced · 2021 · 03 Oct 2021 · Paper 2 · One or more correct options