One end of the steel rod is clamped to the roof and the other end is attached to mass of $1000 \mathrm{~kg}$…
One end of the steel rod is clamped to the roof and the other end is attached to mass of $1000 \mathrm{~kg}$ as shown in the figure. The length of the rod is $50 \mathrm{~cm}$ and its cross-sectional area is $1000 \mathrm{~mm}^2$. The change in the length of the rod due to the weight of the mass is (Young's modulus of steel $=2 \times 10^{11} \mathrm{Nm}^{-2}$ and acceleration due to gravity $=10 \mathrm{~ms}^{-2}$ )
$0.025 \mathrm{~mm}$
$0.10 \mathrm{~mm}$
$0.050 \mathrm{~mm}$
$0.075 \mathrm{~mm}$
Solution
As, young's modulus of a rod under tension,
$
Y=\frac{F / A}{\Delta l / l}=\frac{F l}{A \cdot \Delta l}
$
We have, change in length $\Delta l=\frac{F . l}{A . Y}$
Here, force on rod, $F=1000 \times 10=10,000 \mathrm{~N}$
Length of rod, $l=50 \mathrm{~cm}=50 \times 10^{-2} \mathrm{~m}$
Area of rod, $A=1000 \mathrm{~mm}^2=1000 \times 10^{-6} \mathrm{~m}^2$
Substituting these values in eq. (i), we get hange of length of rod,
$
\begin{aligned}
\Delta l & =\frac{10000 \times 50 \times 10^{-2}}{1000 \times 10^{-6} \times 2 \times 10^{11}} \\
& =25 \times 10^{-6} \mathrm{~m} \\
& =0.025 \mathrm{~mm}
\end{aligned}
$
