Question

A uniform chain of length $L$ and mass $M$ is lying on a smooth table and $\dfrac{2}{3}$ if its length is hanging down over the edge of the table. If $g$ is the acceleration due to gravity, the work done to pull the hanging part on the table is:
(A) $MgL$
(B) $\dfrac{{MgL}}{3}$
(C) $\dfrac{{MgL}}{9}$
(D) $\dfrac{{2MgL}}{9}$

Answer 1

Hint
The mass of the hanging part of the chain is not constant throughout the pull. For a uniform chain, the linear mass density is constant through the entire length of the chain. So the work done will be calculated by the formula $W = \int {Fdl}$ where the force will be the weight of the chain and the length will be the part hanging.

Formula used: In this solution we will be using the following formula,
$\Rightarrow W = \int {Fdl}$, where $W$ is the work done,$F$ is force and $l$ is distance

Complete step by step answer
Work is done when a force is applied on a body through a distance. When a force is applied on a body but for some reason it does not move i.e. distance moved by the body is zero, then the work done by the force, irrespective of how large it may be, is zero.
In our question above, we see that two-third of the length of a chain is hanging down, we are to calculate the work done for pulling this two-third length on the table.
Generally for work done, we have
$\Rightarrow W = \int {Fdl}$
For a body under gravity, the force required to pull such body upward is given as
$\Rightarrow F = mg$ where $m$ is the mass of the body and $g$ is acceleration due to gravity.
For a uniform body of length $l$, we can say that the linear mass density is
$\Rightarrow \lambda = \dfrac{m}{l}$
So we can calculate the mass as
$\Rightarrow m = \lambda l$
Hence, replacing $m$ with $\lambda l$ we have the force as,
$\Rightarrow F = g\lambda l$
We do this because the mass is a function of the length. In this case, when a part of the mass gets pulled to the table, that part is no longer hanging, thus the total mass hanging below reduces as the length hanging below reduces.
Then, inserting into work done formula we have
$\Rightarrow W = \int {g\lambda ldl}$.
$g\lambda$ is a constant and thus can be factorized out of the integration sign. Hence,
$\Rightarrow W = g\lambda \int {ldl}$
Now, only two-third of its length was hanging. Thus we must integrate the function from
$\Rightarrow l = \dfrac{2}{3}L$ to $l = 0$.
Hence on substituting the limits we get,
$\Rightarrow W = g\lambda \int_0^{\dfrac{2}{3}L} {ldl}$
Using the mathematical principles we get on integrating,
$\Rightarrow \int_0^{\dfrac{2}{3}L} {ldl} = \left[ {\dfrac{{{l^2}}}{2}} \right]_0^{\dfrac{2}{3}L} = \dfrac{1}{2}\left[ {{{\left( {\dfrac{2}{3}L} \right)}^2} - 0} \right]$
So on calculating we get,
$\Rightarrow \int_0^{\dfrac{2}{3}L} {ldl} = \dfrac{1}{2}\left[ {\dfrac{4}{9}{L^2}} \right] = \dfrac{2}{9}{L^2}$
Inserting into work done formula, we have that
$\Rightarrow W = g\lambda \left( {\dfrac{2}{9}{L^2}} \right) = \dfrac{{2g\lambda {L^2}}}{9}$
Now, for this chain, $\lambda = \dfrac{M}{L}$, then on substituting,
$\Rightarrow W = \dfrac{{2gM{L^2}}}{{9L}}$
Cancelling $L$, we have
$\Rightarrow W = \dfrac{{2gML}}{9}$
$\therefore W = \dfrac{{2MgL}}{9}$
Hence, the correct option is (D).

Note
However, since the chain is uniform, using the above general method, it can be proven that the work done $W = mg \times {l_c}$ where $m$ in this case, is the mass of the hanging chain, and ${l_c}$ is the distance of the center of gravity from the top of the table.
Hence, alternatively we can find the work down by substituting,
$W = \dfrac{2}{3}Mg \times \dfrac{2}{6}L$ (the center of gravity for a uniform length is halfway down the hanging length)
Thus on calculating,
$W = \dfrac{{2MgL}}{9}$.