Two bodies of equal weights are kept at heights $h$ and $1.5 h$ respectively. The ratio of their potential energies is

$3 : 2$

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$1 : 1$

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$2 : 3$

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$3 : 4$

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Solution

The correct option is C
$2 : 3$

Option (C) $2 : 3$
Step 1:Given
$h_{1} = h (h e i g h t o f f i r s t b o d y) h_{2} = 1.5 h (h e i g h t o f s e c o n d b o d y) m_{1} = m_{2} = m$
Step 2: Finding P.E
We apply the formula of Potential energy $P . E . = m g h$
Let potential energy of the first and second body is $P_{1} . E, P_{2} . E$ .
$\frac{P_{1} . E}{P_{2} . E} = \frac{m g h_{1}}{m g h_{2}} = \frac{h_{1}}{h_{2}} (m a s s o f b o t h b o d i e s a r e e q u a l)$
$\frac{h_{1}}{h_{2}} = \frac{h}{1.5 h} = \frac{10}{15} = \frac{2}{3} = 2 : 3$
Hence, option (C) is the correct answer.

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