Properties Of Matter Ques 22
- The pressure that has to be applied to the ends of a steel wire of length $10 \mathrm{~cm}$ to keep its length constant when its temperature is raised by $100^{\circ} \mathrm{C}$ is (For steel, Young’s modulus is $2 \times 10^{11} \mathrm{Nm}^{-2}$ and coefficient of thermal expansion is $1.1 \times 10^{-5} \mathrm{~K}^{-1}$ )
(2014 Main)
(a) $2.2 \times 10^8 \mathrm{~Pa}$
(b) $2.2 \times 10^9 \mathrm{~Pa}$
(c) $2.2 \times 10^7 \mathrm{~Pa}$
(d) $2.2 \times 10^6 \mathrm{~Pa}$
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Answer:
Correct Answer: 22.( a )
Solution:
- If the deformation is small, then the stress in a body is directly proportional to the corresponding strain. According to Hooke’s law i.e.
Young’s modulus $(Y)=\frac{\text { Tensile stress }}{\text { Tensile strain }}$
So, $ Y=\frac{F / A}{\Delta L / L}=\frac{F L}{A \Delta L} $
If the rod is compressed, then compressive stress and strain appear. Their ratio $Y$ is same as that for tensile case.
Given, length of a steel wire $(L)=10 \mathrm{~cm}$ Temperature $(\theta)=100^{\circ} \mathrm{C}$
As length is constant.
$\therefore$ Strain $=\frac{\Delta L}{L}=\alpha \Delta \theta$
Now, pressure $=$ stress $=Y \times$ strain
[Given, $Y=2 \times 10^{11} \mathrm{~N} / \mathrm{m}^2$ and $\alpha=1.1 \times 10^{-5} \mathrm{~K}^{-1}$ ]
$ =2 \times 10^{11} \times 1.1 \times 10^{-5} \times 100=2.2 \times 10^8 \mathrm{~Pa} $
BETA
