Question

The ratio of lengths of two rods is $1:2$ and the ratio of coefficient of expansions is $2:3$ The first rod is heated through ${60}^{o}C.$ The temperature through which the second rod is to be heated so that its expansion is twice that of first is:

• A. ${60}^{o}C$
• B. ${40}^{o}C$
• C. ${30}^{o}C$
• D. ${10}^{o}C$

Answer 1

The correct option is B ${40}^{o}C$

We know that,

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

For first rod,

$\Delta l_1=l_1{\alpha }_1\Delta T_1$ …(i)

Similarly for second rod,

$\Delta l_1=l_2{\alpha }_2\Delta T_2$ …(ii)

Taking ratio of (i) by (ii)

$\Rightarrow \left(\frac{\Delta l_1}{\Delta l_2}\right)=\left(\frac{l_1}{l_2}\right)\left(\frac{{\alpha }_1}{{\alpha }_2}\right)\left(\frac{\Delta T_1}{\Delta T_2}\right)$

It is given that,

$\frac{\Delta l_1}{\Delta l_2}=\frac{1}{2},\frac{l_1}{l_2}=\frac{1}{2},\frac{{\alpha }_1}{{\alpha }_2}=\frac{2}{3}$ and $\Delta T={60}^o$C

So, $\frac{1}{2}=\left(\frac{1}{2}\right)\left(\frac{2}{3}\right)\frac{\left({60}^{o}C\right)}{\Delta T_2}$

$\Delta T_2={40}^{o}C$

Option B is correct.