A uniform chain of mass $m$ and length $L$ hangs on a table with two thirds of its length on the table. The work done by a person to put the hanging part back on the table will be

The correct option is A $\frac{mgL}{18}$
On applying,
$W_{\text{ext}}=\Delta U=U_f-U_i...\left(1\right)$

Let us take a small mass $dm$ of length $dx$ at a distance $x$ below reference level (table) , then its potential energy is $dU=-dmgx$
$\frac{L}{3}$ length of the chain is hanging from the table.
On integrating both sides by taking limits of $x=0\text{to}x=\frac{L}{3}$
$\underset{0}{\overset{U}{\int }}dU=\underset{0}{\overset{L/3}{\int }}-dmgx$
$\Rightarrow U=-\underset{0}{\overset{L/3}{\int }}\left(\frac{m}{L}dx\right)gx$
[ $\because dm=\lambda dx=\frac{m}{L}dx$]
$\Rightarrow U=-\frac{mg}{L}{\left[\frac{x^2}{2}\right]}_0^{L/3}$
$\therefore U=\frac{-mg}{2L}[\frac{L^2}{9}-0]=\frac{-mgL}{18}$

When $\frac{L}{3}$ length of the chain is hanging from the table, then potential energy
$U_i=\frac{-mgL}{18}$
When the hanging part is pulled back on the table, potential energy of the system is
$U_f=0$ as $h=0$ from reference level.
Substituting in Eq.$\left(1\right)$,
$W_{\text{ext}}=U_f-U_i$
$W_{\text{ext}}=0-\left(\frac{-mgL}{18}\right)$
$\therefore W_{\text{ext}}=\frac{mgL}{18}$
Hence work done by the person to put the hanging part back on the table is $\frac{mgL}{18}$