JEE Main · 2019 · 09 Jan 2019 · Shift 2 · Laws of Motion

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A mass of 10 kg is suspended vertically by a rope from the roof. When a horizontal force is applied on the rope at some point, the rope deviated at an angle of

4 5^{\circ}

at the roof point. If the suspended mass is at equilibrium, the magnitude of the force applied is (

g = 10 ms^{- 2}

)

(A)200 N
(B)140 N
(C)70 N
(D)100 N✓ Correct

Answer: (D)

Solution

The hanging mass is held by three forces: its weight

m g

downward, the rope tension

T

4 5^{\circ}

, and the applied horizontal force

F

. Resolving the tension, the vertical balance is

T cos 4 5^{\circ} = m g

and the horizontal balance is

T sin 4 5^{\circ} = F

. Dividing one by the other,

tan 4 5^{\circ} = \frac{F}{m g}

, and since

tan 4 5^{\circ} = 1

we get

F = m g = 10 \times 10 = 100 N

JEE Main · 2019 · 09 Jan 2019 · Shift 2 · Single correct option

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