Question

An inclined plane making an angle of 30° with the horizontal is placed in an uniform electric field E = 100Vm^–1. A particle of mass 1 kg and charge 0.01 c is allowed to slide down from rest from a height of 1m. If the coefficient of friction is 0.2 the time taken by the particle to reach the bottom is …….. sec 

(A) 2.337 

(B) 4.337 

(C) 5 

(D) 1.337

Answer 1

The correct option (D) 1.337   

Explanation:

Different forces acting on the charged particle are shown.

The net forces is down the plane.

F = mg sin 30 – μR – qE cos 30

F = mg sin 30 – μ(mg cos 30 + qE sin30) – qEcos 30

Dividing by m,

∴ (F/m) = g sin 30 – μ[g cos 30 + (qE/m)sin 30] – (qE/m)cos 30

But F = ma hence (F/m) = a = acceleration

∴ a = [9.8 × (1/2)] – 0.2[9.8 × (√3/2) + {(0.01 × 100)/1} × (1/2)] – [{(0.01)/1} × 100 × (√3/2)]

= 4.9 – 0.2(8.48 + 0.5) – (0.86)

a = 2.24 m/s   (1)

here q = 0.01 c

m = 1kg

 E = 100V/m

Now distance travelled along the plane = s

 sin 30 = (h/s)

∴ s = [h/(sin30)] = [1/(1/2)] = 2m

as   s = ut + (1/2)at²

∴    2 = 0 + (1/2)at²

∴   (4/a) = t²

t² = [4 / (0.24)] = 1.785

t = 1.33 sec