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What is the percentage increase in length of a wire of diameter $2.5 \,mm$, stretched by a force of $100 \,kg$ wt is .................. $\%$ ( Young's modulus of elasticity of wire $=12.5 \times 10^{11} \,dyne / cm ^2$ )
$0.16$
$0.32$
$0.08$
$0.12$
Solution
(a)
$Y=\frac{F L}{A \Delta L} \Rightarrow \text { Percentage increase } \frac{\Delta L}{L} \times 100=\frac{F}{A Y} \times 100$
$\text { Diameter }=2.5 \,mm$
$d=\frac{2.5}{1000} \,m$
Area $=\frac{\pi d^2}{4} \quad Y=12.5 \times 10^{11} \,dyne / cm ^2$ $\left\{\frac{1 \text { dyne }}{ cm ^2}=\frac{0.1 \,N }{ m ^2}\right\}$
$\Rightarrow A=\frac{\pi \times(2.5)^2}{4} \quad F=100 \times 10=1000 \,N$
$\Rightarrow \frac{1000 \times L}{\frac{3.14 \times(2.5)^2}{4 \times(1000)^2} \times 12.5 \times 10^{11} \times 0.1}=\frac{\Delta L}{L} \times 100$
$=0.16 \%$
