Question

A mass of 0.5 kg is suspended from wire, then length of wire increases by 3 mm then work done -
[RPMT 2000]

• A. $4.5\times {10}^{-3}Joule$
• B. $7.5\times {10}^{-3}Joule$
• C. $9.3\times {10}^{-3}Joule$
• D. $2.5\times Joule$

Answer 1

The correct option is B $7.5\times {10}^{-3}Joule$

Consider a wire of length L, area of cross-section A. A load is hung which creates a force f and elongates the wire by l.

Then,

$y=\frac{stree}{strain}=\frac{f/A}{x/L}=\frac{fL}{Ax}$

$f=\frac{yAx}{L}$ ………$\left(1\right)$

work done by the external force in elongating the wire

$dW=fdx$ ……….$\left(2\right)$

$dW=\frac{yAx}{L}dx$

$\int dW=\frac{yA}{L}{\int }_o^{l}xdx$

$W=\frac{yA}{L}{\left[\frac{x^2}{2}\right]}_o^l$

$W=\frac{yA}{L}[\frac{l^2}{2}-0]$

$W=\frac{yAl}{L}\frac{l}{2}$ ……….$\left(3\right)$

Now from equation $\left(1\right)$ we have $f=\frac{yAx}{L}$

when, wire is fully stretched $x=1$

$\Rightarrow f=Mg$

$Mg=\frac{yAl}{L}$ ……..$\left(4\right)$

From equation $\left(3\right)$ and $\left(4\right)$

$W=\left(mg\right)\frac{l}{2}$

$W=\frac{1}{2}Fl$.

Using this formula from proof

$W=\frac{1}{2}Fl$

$W=\frac{1}{2}\times (0.5\times 10)\times 3\times {10}^{-3}=\frac{15}{2}\times {10}^{-3}$

$W=7.5\times {10}^{-3}$J.