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).

\sqrt{6}

second

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9

second

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\sqrt{3}

second

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6

second

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Solution

The correct option is A $\sqrt{6}$ second
By second equation of Newton we have
$S = u t + \frac{1}{2} {a t}^{2}$
Here $u = 0 a = g S = h = 1 c m$
For the stome in moon, $g^{'} = \frac{g}{6}$
$∴ h = \frac{g}{2 \times 6} t^{2} ∴ t = \sqrt{12 \frac{h}{g}} = \sqrt{6} s e c$

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A stone drop from height 'h' reaches to Earth surface in 1sec. 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).

The correct option is A √6 secondBy second equation of Newton we haveS=ut+12at2Here u=0a=gS=h=1cmFor the stome in moon, g′=g6 ∴h=g2×6t2∴t=√12hg=√6sec

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).

The correct option is A $\sqrt{6}$ second
By second equation of Newton we have
$S = u t + \frac{1}{2} {a t}^{2}$
Here $u = 0 a = g S = h = 1 c m$
For the stome in moon, $g^{'} = \frac{g}{6}$
$∴ h = \frac{g}{2 \times 6} t^{2} ∴ t = \sqrt{12 \frac{h}{g}} = \sqrt{6} s e c$