A uniform chain of length L and mass M is lying on a smooth table and one third of its length is hanging vertically down over the edge of the table. If g is acceleration due to gravity, work required to pull the hanging part on to the table is

Question

A uniform chain of length L and mass M is lying on a smooth table and one third of its length is hanging vertically down over the edge of the table. If g is acceleration due to gravity, work required to pull the hanging part on to the table is (A) MgL (B)  (MgL/3) (C)  (MgL/9) (D)  (MgL/18)

Tardigrade Admin · Accepted Answer

Mass of hanging portion is M / 3 (one-third) and centre of mass c, is at a distance h=(L/6) below the table top. Therefore, the required work done is, <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/060944463caa4971427cb3dfea9e5a76-.png" /> W=m g h=((M/3))(g)((L/6))=(M g L/18)

Q. A uniform chain of length $L$ and mass $M$ is lying on a smooth table and one third of its length is hanging vertically down over the edge of the table. If $g$ is acceleration due to gravity, work required to pull the hanging part on to the table is

$M gL$

$\frac{M gL}{3}$

$\frac{M gL}{9}$

$\frac{M gL}{18}$

Mass of hanging portion is $M /3$ (one-third) and centre of mass $c$ , is at a distance $h = \frac{L}{6}$ below the table top.
Therefore, the required work done is,

$W = m g h = (\frac{M}{3}) (g) (\frac{L}{6}) = \frac{M gL}{18}$

Q. A uniform chain of length L and mass M is lying on a smooth table and one third of its length is hanging vertically down over the edge of the table. If g is acceleration due to gravity, work required to pull the hanging part on to the table is

MgL

3MgL​

9MgL​

18MgL​