Question

A person standing on the floor of a lift drops a coin. The coin reaches the floor of the lift in time if the lift is stationary and the time t₂ if it is accelerated in upward direction. Than 

(A) t₁ = t₂ 

(B) t₁ > t₂ 

(C) t₁ < t₂ 

(D) Cannot say anything

Answer 1

Correct option: (B) t₁ > t₂ 

Explanation:

t₁ ➙ time to reach floor when lift stationery.

d = (1 / 2) g t₁² 

t₂ ➙ time to reach floor when lift is moving up

d = (1 / 2) (g + a)t₂² 

comparing, (1 / 2) gt₁² = (1 / 2) (g + a) t₂² 

gt₁² = (g + a)t₂² 

(t₁² / t₂²) = {(g + a) / g} = 1 + (a / g)

∵ a > 0, (a / g) > 0

 (t₁² / t₂²) > 1 ⇒ (t₁ / t₂) > 1 ⇒ t₁ > t₂