On a planet, a body is left to fall freely from a height 2 metres and reaches the ground in 2 seconds. If the length of a simple pendulum is 2 metres on that planet, its time period is

h=12gt2 given height h=2m; time=2secsubstitute 2=12gplanet22 ⇒gp=42=2 m/sec2Time period T=2πlg ⇒T=2π22 ⇒T=2π=2(3.14)T=6.28

On a planet, a body is left to fall freely from a height 2 metres and reaches the ground in 2 seconds. If the length of a simple pendulum is 2 metres on that planet, its time period is

h=12gt2 given height h=2m; time=2secsubstitute 2=12gplanet22 ⇒gp=42=2 m/sec2Time period T=2πlg ⇒T=2π22 ⇒T=2π=2(3.14)T=6.28

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On a planet, a body is left to fall freely from a height 2 metres and reaches the ground in $\sqrt{2}$ seconds. If the length of a simple pendulum is 2 metres on that planet, its time period is

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6.28 seconds

3.14 seconds

1.57 seconds

4.17 seconds

answer is B.

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Detailed Solution

$h= \frac{1}{2} {gt}^{2} given height h=2m; time= \sqrt{2} sec$

$s u b s t i t u t e 2= \frac{1}{2} g_{planet} {(\sqrt{2})}^{2} \Rightarrow g_{p} = \frac{4}{2} {=2 m/sec}^{2}$

$Time period T=2π \sqrt{\frac{l}{g}} \Rightarrow T =2π \sqrt{\frac{2}{2}} \Rightarrow T =2π=2(3.14)$

$T=6.28$

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On a planet, a body is left to fall freely from a height 2 metres and reaches the ground in 2 seconds. If the length of a simple pendulum is 2 metres on that planet, its time period is

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On a planet, a body is left to fall freely from a height 2 metres and reaches the ground in $\sqrt{2}$ seconds. If the length of a simple pendulum is 2 metres on that planet, its time period is