A mass falls from a height 'h' and its time of fall 't' is recorded in terms of time period T of a simple pendulum. On the surface of earth it is found that t = 2T. The entire set up is taken on the surface of another planet whose mass is half of that of earth and radius the same. Same experiment is repeated and corresponding times noted as t' and T'. Then we can say

Explanation:

Acceleration due to gravity is given as, g = GM_e/R²     -----(1) 

where, G = Gravitational constant, Me = Mass of the Earth, R = Radius of the Earth

The time period of a pendulum can be defined as, T = 2π√(L/g)   -----(2)

Given:

Height = h, time of flight, t = 2T, M' = Me/2

∴ From equation (1) we get, g' = g/2

Using the 2nd equation of motion we can find the time recorded when an object drops from a height of h.

s = ut + 1/2× gt² 

As the object is released from rest hence u = 0, also s = h

∴ h = 1/2× gt² ⇒ t = √(2h/g)   -----(3)

From equations (2) and (3) we can conclude that:

T∝ 1/√g   & t ∝ 1/√g .

⇒T/t is independent of g. 

⇒t'/T' = 2 ⇒ t' = 2T'  (as t = 2T)

Hence option 2) is the correct choice.