Q. 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 $45^{\circ}$ at the roof point. If the suspended mass is at equilibrium, the magnitude of the force applied is $(g = 10\; ms^{ -2})$


at equation
$\tan 45^{\circ} = \frac{100}{F}$
F = 100 N

Solition Image

You must select option to get answer and solution

Questions from JEE Main 2019

Physics Most Viewed Questions