Question

There is some change in length when a $33,000\text{N}$ tensile force is applied on a steel rod of area of cross-section ${10}^{-3}{\text{m}}^2$. The change of temperature required to produce the same elongation, if the steel rod is heated, is -
[Take modulus of elasticity $Y=3\times {10}^{11}{\text{N/m}}^2$ and the coefficient of linear expansion of steel is $\alpha =1.1\times {10}^{-5}{/}^{\circ }\text{C}$]

• A. ${20}^{\circ }\text{C}$
• B. ${15}^{\circ }\text{C}$
• C. ${10}^{\circ }\text{C}$
• D. $0^{\circ }\text{C}$

Answer 1

The correct option is C ${10}^{\circ }\text{C}$

From the definition,
Modulus of elasticity $=\frac{\text{Force}}{\text{Area}}\times \frac{l}{\Delta l}....\left(1\right)$
Given, Tensile force $\left(F\right)=33,000\text{N}$
Area of cross-secton $\left(A\right)={10}^{-3}{\text{m}}^2$
Modulus of elasticity $Y=3\times {10}^{11}{\text{N/m}}^2$
and
Coefficient of linear expansion of steel is $\alpha =1.1\times {10}^{-5}{/}^{\circ }\text{C}$
Using the given data in $\left(1\right)$, we get
$3\times {10}^{11}=\frac{33,000}{{10}^{-3}}\times \frac{l}{\Delta l}$
$\Rightarrow \frac{\Delta l}{l}=\frac{33,000}{{10}^{-3}}\times \frac{1}{3\times {10}^{11}}=11\times {10}^{-5}.....\left(2\right)$
We know that, change in length of the wire due to heating is,
$\Delta l=l\left(\alpha \Delta T\right)....\left(3\right)$
From $\left(2\right)$ and $\left(3\right)$,
$\alpha \Delta T=11\times {10}^{-5}$
$\Rightarrow 11\times {10}^{-5}=1.1\times {10}^{-5}\times \Delta T$
$\Rightarrow \Delta T=10\text{K or}{10}^{\circ }\text{C}$
Thus, option (c) is the correct answer.