A stone drop from height 'h' reaches to earth surface in 1 sec. If the same stone taken to moon and drop freely then it will reaches from the surface of the moon in the time:

\sqrt{6} s

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

9 s

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\sqrt{3} s

No worries! We‘ve got your back. Try BYJU‘S free classes today!

6 s

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $\sqrt{6} s$
Using, $s = u t + \frac{1}{2} a t^{2}$
On earth: $h = 0 + \frac{1}{2} g t^{2}$ where $t = 1$
$⟹ g = 2 h$
As acceleration due to gravity on moon, $g^{'} = \frac{g}{6}$
Thus on moon $h = \frac{1}{2} \frac{g}{6} t^{2} ⟹ t = \sqrt{6} s e c$

Suggest Corrections

Similar questions

Q. A stone drop from height 'h' reaches at earth surface in 1 sec. If the same stone is taken to moon and dropped freely from height h, then it will reach the surface of moon in ..... sec

Q. A stone drop from height 'h' reaches to Earth surface in

1

sec. If the same stone taken to Moon and drop freely then it will reaches from the surface of the Moon in the time(The 'g' of Moon is

1 / 6

times of Earth).

a stone dropped from a height h reaches at the earth surface in 1 second. If the same stone is taken to the moon and drop freely from a height h then in what time will it reach the ground?

Q. When a body is dropped from same height on surface of the earth and the moon the ratio of time to reach the ground will be:

Q. A stone dropped from a certain height reaches ground in

5 sec

. If it is dropped and stopped for infinitesimal time interval at

t = 3 s

, then time taken by the stone to reach the ground is

Join BYJU'S Learning Program

A stone drop from height 'h' reaches to earth surface in 1 sec. If the same stone taken to moon and drop freely then it will reaches from the surface of the moon in the time:

The correct option is A √6 sUsing, s=ut+12at2On earth: h=0+12gt2 where t=1⟹g=2hAs acceleration due to gravity on moon, g′=g6Thus on moon h=12g6t2⟹t=√6sec

The correct option is A $\sqrt{6} s$
Using, $s = u t + \frac{1}{2} a t^{2}$
On earth: $h = 0 + \frac{1}{2} g t^{2}$ where $t = 1$
$⟹ g = 2 h$
As acceleration due to gravity on moon, $g^{'} = \frac{g}{6}$
Thus on moon $h = \frac{1}{2} \frac{g}{6} t^{2} ⟹ t = \sqrt{6} s e c$