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A crane with a steel cable of length $11 \mathrm{~m}$ and radius $2.0 \mathrm{~cm}$ is employed to lift a block of concrete of mass 40 tons in a building site. If the Young's modulus of steel is $2.0 \times 10^{11} \mathrm{~Pa}$, what will be roughly the increase in the length of the cable while lifting the block?
(Take, $g=10 \mathrm{~ms}^{-2}$ )
Options:
(Take, $g=10 \mathrm{~ms}^{-2}$ )
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Verified Answer
The correct answer is:
$1.75 \mathrm{~cm}$
Given, length, $l=11 \mathrm{~m}$ Radius, $\quad r=2 \mathrm{~cm}$ Radius, $\quad=2 \times 10^{-2} \mathrm{~m}$
Mass, $m=40$ tons
$$
=4 \times 10^{4} \mathrm{~kg}
$$
Increase in length, $\Delta l=$ ?
We know that, Young's modulus,
$$
\begin{aligned}
Y &=\frac{m g l}{\pi r^{2}(\Delta l)} \\
\Rightarrow \quad \Delta l &=\frac{m g l}{\pi r^{2} Y} \\
&=\frac{4 \times 10^{4} \times 10 \times 11}{3.14 \times\left(2 \times 10^{-2}\right)^{2} \times 2 \times 10^{11}} \\
&=1.75 \times 10^{-2} \mathrm{~m} \\
&=1.75 \mathrm{~cm}
\end{aligned}
$$
Mass, $m=40$ tons
$$
=4 \times 10^{4} \mathrm{~kg}
$$
Increase in length, $\Delta l=$ ?
We know that, Young's modulus,
$$
\begin{aligned}
Y &=\frac{m g l}{\pi r^{2}(\Delta l)} \\
\Rightarrow \quad \Delta l &=\frac{m g l}{\pi r^{2} Y} \\
&=\frac{4 \times 10^{4} \times 10 \times 11}{3.14 \times\left(2 \times 10^{-2}\right)^{2} \times 2 \times 10^{11}} \\
&=1.75 \times 10^{-2} \mathrm{~m} \\
&=1.75 \mathrm{~cm}
\end{aligned}
$$
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