Answer

Verified

51.9k+ views

**Hint:**On the hanging part of the chain, gravitational force is acting. To pull the string back to the table we need to apply a force which is opposite to the gravitational force. This force will have the same magnitude as the gravitational force, but the direction will be opposite. The work which we need to calculate in this question is the work done against the gravitational force.

**Formula used:**

Potential energy,

$U = mgh$ (where $U$ stands for the potential energy stored in an object placed at a height, $m$ stands for the mass of the object, $g$ is the acceleration due to gravity ($g = 9.8m{s^{ - 2}}$) and $h$ stands for the height at which the object is stored)

**Complete step by step solution:**

It is given that one by fourth of the length $L$ is hanging over the edge of the table.

Consider small elements of length $dy$ on the hanging part of the string

Let $m$ be the mass of the string.

Mass per unit length of the string $ = \dfrac{m}{L}$

Mass of each small element of length $L = \left( {\dfrac{m}{L}} \right)dy$

Let the string segment be hanging at a distance $y$below the table-top

Since the origin is at the table top and the string is hanging downwards $y$ will be negative.

The potential energy is $U = mgh$

Substituting the values of $m$ and $h$ in the above equation

The change in potential energy for a small element,

$dU = - \left( {\dfrac{m}{L}} \right)gydy$

The change in potential energy for the string can be obtained by integrating the above equation

$U = \int {dU} $

Since the length of the segment hanging is $\dfrac{L}{4}$ and it is hanging downwards, to find the potential we have to integrate from $ - \dfrac{L}{4}$ to $0$.

$U = \int\limits_{\dfrac{{ - L}}{4}}^0 {\dfrac{{ - mg}}{L}} ydy$

Taking the constants out,

$U = \dfrac{{ - mg}}{L}\int\limits_{\dfrac{{ - L}}{4}}^0 {ydy} $

Integrating the above equation,

$U = \dfrac{{ - mg}}{L}\left[ {\dfrac{{{y^2}}}{2}} \right]_{\dfrac{{ - L}}{4}}^0$

Applying the limits,

$U = \dfrac{{ - mg}}{{2L}}\left[ {0 - {{\left( {\dfrac{{ - L}}{4}} \right)}^2}} \right]$

Opening the brackets, we get

$U = \dfrac{{ - mg}}{{2L}}\left[ {0 - \dfrac{{{L^2}}}{{16}}} \right]$

$U = \dfrac{{mg}}{{2L}}\left( {\dfrac{{{L^2}}}{{16}}} \right)$

Solving the above equation, we get

$U = \dfrac{{mgL}}{{32}}$……………………(A)

The values of $m,g,\& L$ is given in the question

$m = 0.012kg$, $L = 28cm$, $g = 9.8m{s^{ - 2}}$

Substituting these values into the equation (A)

$U = \dfrac{{0.012 \times 9.8 \times 0.28}}{{32}} = 1.029 \times {10^{ - 3}}$

The work done will be equal to this change in potential energy

**The answer is: $1.029 \times {10^{ - 3}}J$.**

**Note:**The work done in pulling the string will be equal to the change in potential energy as the potential energy will change with the change in height. Since the length of the hanging segment of the string will change from $\dfrac{L}{4}$ to $0$ the work done will also change. The gravitational potential depends on the weight of the object and the height of the object. All the values should be converted into MKS values before calculation.

Recently Updated Pages

Which is not the correct advantage of parallel combination class 10 physics JEE_Main

State two factors upon which the heat absorbed by a class 10 physics JEE_Main

What will be the halflife of a first order reaction class 12 chemistry JEE_Main

Which of the following amino acids is an essential class 12 chemistry JEE_Main

Which of the following is least basic A B C D class 12 chemistry JEE_Main

Out of the following hybrid orbitals the one which class 12 chemistry JEE_Main

Other Pages

Which of the following sets of displacements might class 11 physics JEE_Main

Electric field due to uniformly charged sphere class 12 physics JEE_Main

Differentiate between homogeneous and heterogeneous class 12 chemistry JEE_Main

The resultant of vec A and vec B is perpendicular to class 11 physics JEE_Main

when an object Is placed at a distance of 60 cm from class 12 physics JEE_Main

If a wire of resistance R is stretched to double of class 12 physics JEE_Main