A mass of 10 kg is suspended vertically by a rope of length 5 m from… | JEE Main 2022 (27 Jun, Shift 2)

Q15JEE Main 2022 · 27 JUN · SHIFT 2Medium

Equilibrium, Third Law & Frames

A mass of 10 kg is suspended vertically by a rope of length 5 m from the roof. A force of 30 N is applied at the middle point of rope in horizontal direction. The angle made by upper half of the rope with vertical is $\theta$ = tan ${}^{-1}$ (x $\times$ 10 $-$ 1). The value of x is ____________.

(Given, g = 10 m/s2)

Numerical answer type. Work it out, then open the answer.

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Answer: 3

At the midpoint: $T\sin\theta = F = 30\text{ N}$ and $T\cos\theta = mg = 100\text{ N}$ .
$\tan\theta = \frac{30}{100} = 0.3 = 3 \times 10^{-1}$
So $x = 3$ .

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