Question

A wire elongates by 3 mm when a load W is hung from it. If the wire goes over a massless and frictionless pulley and two weight W and 2W are hung at two ends, the elongation in the wire will be
A 2 mm B 3 mm C 1 mm D 4 mm

Answer 1

$\Large\boxed{\sf{\quad(D)\quad\!4\ mm\quad}}$

We have, extension,

$\longrightarrow\sf{\Delta L=\dfrac{FL}{AY}}$

Here the extension produced is 3 mm if the wire is extended by the load of weight W. So we have,

$\longrightarrow\sf{\dfrac{WL}{AY}=3\ mm\quad\quad\dots(1)}$

Now the wire goes over a massless and frictionless pulley and two loads of weights W and 2W are hung from its two ends. Let T be the tension in the wire and the net acceleration of the pulley system be 'a' in $\sf{m\ s^{-2}}$ which acts towards the 2W weight load.

Consider the load of weight W hung from the wire.

$\setlength{\unitlength}{1mm}\begin{picture}(5,5)\put(0,0){\circle*{1}}\put(0,0){\vector(0,-1){10}}\multiput(0,0)(10,-5){2}{\vector(0,1){10}}\put(-5,8){ \sf{T} }\put(-6,-8.5){ \sf{W} }\put(6,0){ \sf{a} }\end{picture}$

• The net acceleration of the system acts upwards on this weight since it is towards the 2W weight load.

Hence the net force acting on this load is given by,

$\longrightarrow\sf{T-W=\dfrac{W}{g}\ a}$

$\longrightarrow\sf{2T-2W=2\cdot\dfrac{W}{g}\ a\quad\quad\dots(2)}$

Consider the load of weight 2W hung from the wire.

$\setlength{\unitlength}{1mm}\begin{picture}(5,5)\put(0,0){\circle*{1}}\put(0,0){\vector(0,1){10}}\multiput(0,0)(10,5){2}{\vector(0,-1){10}}\put(-5,8){ \sf{T} }\put(-8,-8.5){ \sf{2W} }\put(6,0){ \sf{a} }\end{picture}$

• The net acceleration of the system acts downwards on this weight.

Hence the net force acting on this load is given by,

$\longrightarrow\sf{2W-T=2\cdot\dfrac{W}{g}\ a\quad\quad\dots(3)}$

From (1) and (2),

$\longrightarrow\sf{2T-2W=2W-T}$

$\longrightarrow\sf{T=\dfrac{4}{3}\ W}$

Hence the extension in the wire is,

$\longrightarrow\sf{\Delta L=\dfrac{TL}{AY}}$

$\longrightarrow\sf{\Delta L=\dfrac{\dfrac{4}{3}\ WL}{AY}}$

$\longrightarrow\sf{\Delta L=\dfrac{4}{3}\times3\ mm}$

$\longrightarrow\sf{\underline{\underline{\Delta L=4\ mm}}}$