QJEEM19Phf7e8: JEE Main 2019 Physics - Class 11 JEEM19Phf7e8 Solution: A uniform cable of mass 'M' and length 'L' is placed on a…

To solve this problem, we need to determine the work done to lift the hanging part of the cable up to the surface. The work done lifting a small element of the cable will be the weight of the element times the distance it has to be lifted.

Let's take a small section of the cable at a depth $x$ below the surface. This section has a length of $dx$ , so its mass is $(M/L)dx$ where $(M/L)$ is the linear mass density of the cable.

The work $dW$ done to lift this small section up to the surface is the weight of the section times the distance it has to be lifted :

$dW = (M/L)gdx \times x$ .

Integrating this expression from 0 to L/n (the length of the hanging part of the cable) gives the total work done :

$$W=\underset{0}{\overset{L/n}{\int }}(M/L)gxdx$$

$$=(Mg/L)\underset{0}{\overset{L/n}{\int }}xdx$$

$$=(Mg/L)\times [x^2/2{]}_0^{L/n}$$

$$=(Mg/L)\times [L^2/(2n^2)]$$

$$=MgL/(2n^2)$$

So the work done to lift the hanging part of the cable up to the surface is $MgL/(2n^2)$ .

Therefore, the correct answer is Option C :

$$\frac{MgL}{2n^2}$$