A pendulum having a period of oscillation of $2$ seconds is taken to a planet where $g$ is four times that on the earth. The period of the pendulum would be :

2 s

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1 s

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4 s

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2 \sqrt{2} s

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Solution

The correct option is B $1 s$
$\frac{T_{1}}{T_{2}} = \sqrt{\frac{g_{2}}{g_{1}}}$ here $T_{1} = 2 s e c g_{2} = 4 g_{1}$
So, $\frac{T_{1}}{T_{2}} = \sqrt{4} \Rightarrow T_{2} = \frac{2}{2} = 1 s e c$

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A pendulum having a period of oscillation of 2 seconds is taken to a planet where g is four times that on the earth. The period of the pendulum would be :

The correct option is B 1sT1T2=√g2g1 here T1=2secg2=4g1So, T1T2=√4⇒T2=22=1sec

A pendulum having a period of oscillation of $2$ seconds is taken to a planet where $g$ is four times that on the earth. The period of the pendulum would be :

The correct option is B $1 s$
$\frac{T_{1}}{T_{2}} = \sqrt{\frac{g_{2}}{g_{1}}}$ here $T_{1} = 2 s e c g_{2} = 4 g_{1}$
So, $\frac{T_{1}}{T_{2}} = \sqrt{4} \Rightarrow T_{2} = \frac{2}{2} = 1 s e c$