First slide

Variation of g

Question

Densities of two planets are in the ratio 1 : 2 and their radii are in the ratio 3 : 1. Then the ratio of time periods of a simple pendulum of their surfaces are in the ratio.

Easy

A

$\sqrt{6} : 1$
B

$1 : \sqrt{6}$
C

$\sqrt{2} : \sqrt{3}$
D

$\sqrt{2} : 3$

Solution

(3)
$\begin{array}{l} g = \frac{G M}{R^{2}} = \frac{G . \frac{4}{3} π R^{3} ρ}{R^{2}} = \frac{4}{3} π G ρ R \Rightarrow g \propto ρ R \\ T = 2 π \sqrt{\frac{l}{g}} \Rightarrow T \propto \frac{1}{\sqrt{g}} \Rightarrow T \propto \frac{1}{\sqrt{ρ R}} \\ ∴ \frac{T_{1}}{T_{2}} = \sqrt{\frac{ρ_{2}}{ρ_{1}} \frac{R_{2}}{R_{1}}} = \sqrt{2. \frac{1}{3}} = \sqrt{\frac{2}{3}} \end{array}$

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