NEET Physics Question - Solid and Fluids

When a block of mass $M$ is suspended by a long wire of length $L$, the length of the wire becomes $(L + l)$. The elastic potential energy stored in the extended wire is :

(2019)

$Mgl$

$MgL$

$\frac{1}{2} Mgl$

$\frac{1}{2} MgL$

Solution:

The elastic potential energy stored in a stretched wire is given by $U = \frac{1}{2} \times \text{Force} \times \text{Extension}$. Here, the applied force is the weight of the block $Mg$ and the extension is $l$. Therefore, $U = \frac{1}{2} Mgl$.

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Solution:

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