Question

The length of a second's pendulum on the surface of Earth is $1m$. What will be the length of a second's pendulum on the moon, where g is ${\displaystyle \frac{1}{6}}$th that of earth’s surface.
A. ${\displaystyle \frac{1}{6}}m$
B. $6m$
C. ${\displaystyle \frac{1}{36}}m$
D. $36m$

Answer 1

HintTo find the solution we need the comparison of the lengths. For that we have to relate it with the gravitational acceleration on both the surfaces.

 Complete step-by-step solution:Length and gravitational acceleration at earth be, $l_e\mathrm{\&}{g}_e$
Length and gravitational acceleration at moon be, $l_m\mathrm{\&}{g}_m$
Now, as the time period at the earth was 2s so,
$$\begin{gathered} T=2\pi \sqrt{l_e/g_e} \\ \Rightarrow 2=2\pi \sqrt{l_e/g_e} \\ \Rightarrow g_e={\pi }^2{l}_e \end{gathered}$$
Similarly at the moon,
$$\begin{gathered} T=2\pi \sqrt{l_m/g_m} \\ \Rightarrow 2=2\pi \sqrt{l_m/(g_e/6)},as(g_m=g_e/6) \\ \Rightarrow 2=2\pi \sqrt{6l_m/{\pi }^2{l}_e} \\ \Rightarrow l_m=l_e/6 \end{gathered}$$
by putting value of $g_e$
So the answer =$l_m=1/6m,as(l_e=1m)$

Note: Here the correct interpretation of the formulas along with the right solving technique is utterly necessary. Apart from that the value should not be equated in between before simplifying as it will in most cases lead to further complexities.