Question

A uniform steel rod of $2.5m{m}^2$ cross sectional area is heated through ${40}^o$C. The force exerted on it is _________. $({\alpha }_{steel}=1.2\times {10}^{-5}{/}^{o}C,Y_{steel}=2\times {10}^{11}N/m^2)$

• A. $480$N
• B. $340$N
• C. $240$N
• D. $140$N

Answer 1

The correct option is C $240$N

$\Delta l=\alpha l\Delta T$

$Y=\frac{F/A}{\Delta l/l}$

$\Rightarrow F=AY\frac{\Delta l}{l}=AY\frac{\alpha l\Delta T}{l}=AY\alpha \Delta T\Rightarrow F=2.5\times {10}^{-6}\times 2\times {10}^{11}\times 1.2\times {10}^{-5}\times 40=240N$