Question

A uniform chain has mass $M$ and length $L$ . It is lying on a smooth horizontal table with half of its length hanging vertically downward. The work done in pulling the chain up the table:
A. $\dfrac{{MgL}}{2}$
B. $\dfrac{{MgL}}{4}$
C. $\dfrac{{MgL}}{8}$
D. $\dfrac{{MgL}}{{16}}$

Answer 1

To conclude the work done in pulling the chain up the table, we will first mention the length and mass of the half of the chain that is hanging vertically downward from the table. And then we will find the length of the centre mass of the chain that is half the length of the part of the chain that is hanging. And then we will apply the formula of work done in the terms of change in potential energy.

Formula used: The formula of work done in the terms of mass of the half of chain, force exerted vertically downward and the length of the centre of mass of the hanging part:
$W = mg\Delta H$

Complete step by step answer:
As per the question, half part of a uniform chain is lying on the table and the other half part is hanging vertically downward from the edge of the table.As the mass of the whole chain is $'M'$ and the length of the chain is $'L'$. So, the length of the half of the chain is $\dfrac{L}{2}$ , as well as its mass is $\dfrac{M}{2}$. Similarly, the length and mass of the other half part of the chain are $\dfrac{L}{2}\,and\,\dfrac{M}{2}$ respectively.

Now, we have to conclude the work done in pulling the part of the chain up the table that is hanging downward from the table.As we know that, we can pull the part of the chain that is the centre of the mass of the hanging part: We will pull the half of the hanging part from the table, so the length of the half of hanging part is $\dfrac{{\dfrac{L}{2}}}{2}\,or\,\dfrac{L}{4}$. Therefore, the work done is equal to the change in potential energy:
$W = \Delta E$ ………(i)
where, $W$ is the work done to pull the chain up the table and, $\Delta E$ is the change in the potential energy.

Now, the change is potential energy is equal to-
$\Delta E = mg\Delta H$ ……….(ii)
where, $m$ is the mass of the part of the chain that is hanging and to be pulled up the table,
$g$ is the force vertically downward, $\Delta H$ is the length of the centre of mass that is to be pulled i.e.. $\dfrac{L}{4}$ .
Now, from eq(i) and eq(ii), we get the formula of work done:-
$W = mg\Delta H \\\ \Rightarrow W = \dfrac{M}{2}.g.\dfrac{L}{4} \\\ \therefore W = \dfrac{{MgL}}{8}$
Therefore, the work done in pulling the chain up the table is $\dfrac{{MgL}}{8}$.

Hence, the correct option is C.

Note: Depending on the object's shape and structure, its centre of mass can be inside or outside it. For a given gravitational field strength, the greater the mass of the object, the greater will be its weight. Here we have considered the mass is concentrated at one point i.e. center of mass in order to solve this question. By finding the center of mass, we can easily calculate the work done by a rigid object.