Question

01. A uniform rope of linear density d and length 1
is hanging from the edge of table. The work done
in pulling the rope on to the table is
1) dg l/2
2) dg l² / 2
3) dg l²
4) d²gl

Answer 1

Question:-

A uniform rope of linear density $\displaystyle\sf {d}$ and length $\displaystyle\sf {l}$

is hanging from the edge of table. The work done

in pulling the rope on to the table is,

$\displaystyle\sf {(1)\quad\dfrac {dgl}{2}}$

$\displaystyle\sf {(2)\quad\dfrac {dgl^2}{2}}$

$\displaystyle\sf {(3)\quad dgl^2}$

$\displaystyle\sf {(4)\quad d^2gl}$

Answer:-

$\displaystyle\large\boxed {\sf {(2)\quad\dfrac {dgl^2}{2}}}$

Solution:-

Given the linear density,

$\displaystyle\longrightarrow\sf {d=\dfrac{m}{l}\quad\implies\quad m=ld}$

where $\displaystyle\sf {m}$ is the whole mass of the rope.

The amount of force given to the rope to pull it above is equal to the weight of the rope, i.e.,

$\displaystyle\longrightarrow\sf {F=mg}$

$\displaystyle\longrightarrow\sf {F=dgl}$

The work done to raise the rope by a small distance $\displaystyle\sf {dl}$ is,

$\displaystyle\longrightarrow\sf {dW=dgl\ dl}$

Then the total work done to raise the whole rope is,

$\displaystyle\longrightarrow\sf {W=\int\limits_0^ldgl\ dl}$

$\displaystyle\longrightarrow\sf {W=dg\int\limits_0^ll\ dl}$

$\displaystyle\longrightarrow\sf {W=dg\left [\dfrac {l^2}{2}\right]_0^l}$

$\displaystyle\longrightarrow\sf {\underline {\underline {W=\dfrac {dgl^2}{2}}}}$