Question

A metal rod of length $2$m has its cross-section $1c{m}^2$. By raising its temperature from $0^o$C to ${100}^o$C and holding it so that it is not permitted to expand or bend, the force developed is? (Given, $Y={10}^{12}$ dyne$/c{m}^2,\alpha ={10}^{-5}/{}^{o}C$)

• A. ${10}^3$N
• B. $2\times {10}^2$N
• C. ${10}^4$N
• D. ${10}^6$N

Answer 1

The correct option is C ${10}^4$N

$\because Y=\frac{Fl}{A\Delta l}$
So, $\Delta l=\frac{Fl}{AY}$ .........(i)
When, the rod is permitted neither to expand nor to bend then linear expansion of the rod is given by
$\Delta l=\alpha l\Delta T$ ..........(ii)
From Eqs. (i) and (ii), we get
$\frac{Fl}{AY}=\alpha l\Delta T$
$\therefore F=AY\alpha \Delta T$
$=1\times {10}^{12}\times {10}^{-5}(100-0)$
$={10}^9$ dyne
$={10}^4$N.