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The periods of a pendulum on two planets are in the ratio 3 : 4. The acceleration due to gravity on them are in the ratio

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a

9 : 16

b

3 : 4

c

4 : 3

d

16 : 9

answer is D.

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Detailed Solution

$\begin{array}{l} T = 2 π \sqrt{\frac{ℓ}{g}} = T \propto \frac{1}{\sqrt{g}} \\ \sqrt{\frac{g_{1}}{g_{2}}} = \frac{T_{2}}{T_{1}} \Rightarrow \frac{g_{1}}{g_{2}} = \frac{T_{2}^{2}}{T_{1}^{2}} = \frac{16}{9} \\ \sqrt{\frac{g_{1}}{g_{2}}} = \frac{T_{2}}{T_{1}} \Rightarrow \frac{g_{1}}{g_{2}} = \frac{T_{2}^{2}}{T_{1}^{2}} = \frac{16}{9} \end{array}$

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