Question

The length of a second pendulum at the surface of earth is $1m$. The length of second pendulum at the surface of moon, where $g$ is $\frac{1}{6}th$ that of earth's surface.

• A. $\frac{1}{6}m$
• B. $\frac{1}{36}m$
• C. $6m$
• D. $36m$

Answer 1

Answer: (A)

We know that the time period of a second pendulum is $2s$ .

At earth ,

$T=2\pi \sqrt{l_e/g_e}$

$2=2\pi \sqrt{l_e/g_e}$

or $g_e={\pi }^2{l}_e$ ...............................eq1

At moon ,

$T=2\pi \sqrt{l_m/g_m}$

or $2=2\pi \sqrt{l_m/\left(g_e/6\right)}$ , given $g_m=g_e/6$

or $2=2\pi \sqrt{6l_m/{\pi }^2{l}_e}$ , putting the value of $g_m$ from eq1

or $l_m=l_e/6$

Now , given $l_e=1m$

Hence , $l_m=1/6m$