A mass of 10kg is suspended vertically by a rope from the roof. When a horizontal force is applied on the rope at some point, the rope is deviated by an angle of 30^o at the roof point. If the suspended mass is at equilibrium, the magnitude of the force applied is (g=10ms^-2)

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140 N

70 N

$\frac{100}{\sqrt{3}} N$

200 N

answer is C.

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Detailed Solution

At equilibrium
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$\begin{array}{l} Tcos θ = mg; T = \frac{mg}{\cos θ} \\ \begin{aligned} F = Tsin θ = \frac{mg}{\cos θ} \times \sin θ \\ = \frac{10 \times 10}{\cos 45^{\circ}} \times \sin 45^{\circ} = 100 \end{aligned} \end{array}$

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