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(D) The potential energy $U$ stored in a spring stretched by a distance $l$ is given by the formula $U = \frac{1}{2} k l^2$,where $k$ is the spring co

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$25V$

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(D) The potential energy $U$ stored in a spring stretched by a distance $l$ is given by the formula $U = \frac{1}{2} k l^2$,where $k$ is the spring constant.This implies that $U \propto l^2$.Given that the initial potential energy $U_1 = V$ for an extension $l_1 = 2 \, cm$.We need to find the potential energy $U_2$ for an extension $l_2 = 10 \, cm$.Using the proportionality $U \propto l^2$,we have:$\frac{U_2}{U_1} = \left( \frac{l_2}{l_1} \right)^2$$\frac{U_2}{V} = \left( \frac{10}{2} \right)^2 = (5)^2 = 25$Therefore,$U_2 = 25V$.

If the potential energy of a spring is $V$ on stretching it by $2 \, cm$,then its potential energy when it is stretched by $10 \, cm$ will be

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