A mass of 10   kg is suspended vertically by a rope of length 5   m from the roof. A force of 30…

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 α=tan-1x×10-1. The value of x is _____ .

(Given, g=10 m s-2)

Solution

Tsinθ=100 and Tcosθ=30

Therefore, tanθ=103

Then, tanα=3 10α=tan-13×10-1

Hence, value of x=3.

Asked in: JEE Main 2022 (27 Jun Shift 2)

Practice more Laws of Motion questions on Aicharya