A pendulum clock that keeps correct time on the earth is taken to the moon. It runs $\sqrt{6}$ times ___.

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Solution

The time period of a simple pendulum is
$T = 2 π \sqrt{\frac{l}{g}}$ g is lesser on moon
$\Rightarrow T w i l l b e g r e a t e r \Rightarrow \frac{T_{e}}{T_{m}} = \sqrt{\frac{g_{m}}{g_{e}}} = \sqrt{\frac{\frac{g_{e}}{6}}{g_{e}}} = \frac{1}{\sqrt{6}} \Rightarrow T_{m} = \sqrt{6} T_{e}$

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A pendulum clock that keeps correct time on the earth is taken to the moon. It runs √6 times ___.

The time period of a simple pendulum is T=2π√lg g is lesser on moon ⇒T will be greater ⇒TeTm=√gmge=√ge6ge=1√6⇒Tm=√6Te

A pendulum clock that keeps correct time on the earth is taken to the moon. It runs $\sqrt{6}$ times ___.

The time period of a simple pendulum is
$T = 2 π \sqrt{\frac{l}{g}}$ g is lesser on moon
$\Rightarrow T w i l l b e g r e a t e r \Rightarrow \frac{T_{e}}{T_{m}} = \sqrt{\frac{g_{m}}{g_{e}}} = \sqrt{\frac{\frac{g_{e}}{6}}{g_{e}}} = \frac{1}{\sqrt{6}} \Rightarrow T_{m} = \sqrt{6} T_{e}$