Gravitation Ques 41
- The mass and the diameter of a planet are three times the respective values for the earth. The period of oscillation of a simple pendulum on the earth is $2 s$. The period of oscillation of the same pendulum on the planet would be
(a) $\frac{2}{\sqrt{3}} s$
(b) $\frac{3}{2} s$
(c) $2 \sqrt{3} s$
(d) $\frac{\sqrt{3}}{2} s$
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Answer:
Correct Answer: 41.(c)
Solution:
Formula:
Variation Of Acceleration Due To Gravity :
- Period of motion of a pendulum is given by
$$ T=2 \pi \sqrt{\frac{l}{g}} $$
On the surface of earth, let period of motion is $T _e$ and acceleration due to gravity is $g _e$
$\therefore \quad T _e=2 \pi \sqrt{\frac{l}{g _e}}$
On the another planet, let period of motion is $T _p$ and gravitational acceleration is $g _p$
$$ \therefore \quad T _p=2 \pi \sqrt{\frac{l}{g _p}} $$
( $\because$ Pendulum is same, so $l$ will be same)
From Eqs. (ii) and (iii),
$$ \frac{T _e}{T _p}=\frac{2 \pi \sqrt{\frac{l}{g _e}}}{2 \pi \sqrt{\frac{l}{g _p}}}=\sqrt{\frac{g _p}{g _e}} $$
Now,
$$ g _e=\frac{G M _e}{R _e^{2}} \text { and } g _p=\frac{G M _p}{R _p^{2}} $$
Given, $\quad M _p=3 M _e$ and $R _p=3 R _e$
$$ \begin{array}{ll} \therefore & g _p=\frac{G \times 3 M _e}{9 R _e^{2}}=\frac{1}{3} \cdot \frac{G M _e}{R _e^{2}}=\frac{1}{3} g _e \\ \Rightarrow & \frac{g _p}{g _e}=\frac{1}{3} \text { or } \sqrt{\frac{g _p}{g _e}}=\frac{1}{\sqrt{3}} \end{array} $$
From Eqs. (iv) and (v), $T _p=\sqrt{3} T _e$
or
$$ T _p=2 \sqrt{3} s $$
$$ \left(\because T _e=2 s\right) $$
BETA
