Question

A mass of M kg is suspended by a weight-less string, the horizontal force that is required to displace it until the string makes an angle of 60° with the initial vertical direction is 

(A) Mg/√3 

(B) Mg.√2 

(C) Mg/√2 

(D) Mg.√3

Answer 1

Correct option: (A) Mg/√3

Explanation:

PE = U = mgh

W = F × d

From figure

ℓ – ℓ cos θ = h

h = ℓ (1 – cos θ)

hence PE = mgh

= mgℓ (1 – cos θ)

Workdone = F × ℓ sin θ

as WD = PE

Fℓ sin θ = mgℓ (1 – cos θ)

= [{mg(1 – cos θ)} / {sin θ}]

= [{mg(1 – [1/2])} / {(√3) / 2} = {(mg) / (√3)}