UPSC Prelims 1997 · General Studies

Question 134: Science and Technology

A smooth inclined plane is inclined at an angle θ with the horizontal as shown in the figure. A body starts from rest and slides down the inclined surface. The time taken by the body to reach the bottom is

a(2h / g)

b(2l / g)

c(2l / (g sin θ))

d(1/ sin θ) (h / 2g)

AnswerThe answer is (c).

Along a smooth incline of length l at angle θ, the pull along the slope gives an acceleration of g sin θ. Starting from rest, l = ½ (g sin θ) t², so t = √(2l / (g sin θ)). Since l = h / sin θ, this can also be written as (1 / sin θ) × √(2h / g). Remember that the acceleration down a smooth slope is g sin θ.

Why each option is right or wrong

(a) √(2h/g) is the time of free fall through the height h and ignores the slope, which slows the body.
(b) √(2l/g) treats the body as falling freely through the length l and ignores sin θ.
(c) Right: √(2l / (g sin θ)).
(d) Has h/2g under the root where the correct working gives 2h/g.

Explanation: written by UPSC Grade, not by UPSC.

Practise this paper timedPrevious question Next question

Source: UPSC, Civil Services (Preliminary) Examination 1997, General Studies. UPSC's own paper isn't online; the words are from a published compilation. The paper on upsc.gov.in.