The length of a seconds pendulum on the surface of earth is $1$ m. Its length on the surface of the moon is

\frac{1}{\sqrt{6}}

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1

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\frac{1}{36}

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36

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Solution

The correct option is B $\frac{1}{\sqrt{6}}$ m
Seconds pendulum $⟹ T = 2 s e c .$
On earth- $2 π \sqrt{\frac{l_{1}}{g}} = 2 ⟶ (1)$
On moon- $2 π \sqrt{\frac{l_{2}}{\frac{8}{6}}} = 2 ⟶ (2)$
$b y$ $d i v i d i n g$ ,
$\frac{\sqrt{l_{2}}}{{6 l}_{2}} = 1$
$⟹ \sqrt{l_{1}} = \sqrt{6 l_{2}}$
$⟹ 1 = \sqrt{6 l_{2}}$
$⟹ l_{2} = \frac{1}{\sqrt{6}}$

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