A uniform chain of mass m and length l is suspended vertically. The lower end of the chain is lifted up to the point of suspension. The work done by the gravitational force is?

\frac{m g l}{3}

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\frac{m g l}{2}

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\frac{m g l}{4}

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mgl

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Solution

The correct option is A $\frac{m g l}{4}$
The centre of mass uniform chain is at its mid point i.e., $\frac{m}{2}$
When, the lower end is lifted up to the point of suspension, the l changes to $\frac{l}{2}$ . The work done will be stored as a potential energy of chain.
So, $W = P E = m^{'} g h = \frac{m}{2} g \frac{l}{2}$
$\Rightarrow W = \frac{m g l}{4}$ .

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A uniform chain of mass m and length l is suspended vertically. The lower end of the chain is lifted up to the point of suspension. The work done by the gravitational force is?

The correct option is A mgl4The centre of mass uniform chain is at its mid point i.e., m2When, the lower end is lifted up to the point of suspension, the l changes to l2. The work done will be stored as a potential energy of chain.So, W=PE=m′gh=m2gl2⇒W=mgl4.