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A one metre steel wire of negligible mass and area of cross-section $0.01 \mathrm{~cm}^2$ is kept on a smooth horizontal table with one end fixed. A ball of mass $1 \mathrm{~kg}$ is attached to the other end. The ball and the wire are rotating with an angular velocity of $\omega$. If the elongation of the wire is $2 \mathrm{~mm}$, then $\omega$ is (Young's modulus of steel $=2 \times 10^{11} \mathrm{Nm}^{-2}$ )
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Verified Answer
The correct answer is:
$20 \mathrm{rad} \mathrm{s}^{-1}$
Given, elongation of the wire, $\Delta l=2 \mathrm{~mm}$
$$
=2 \times 10^{-3} \mathrm{~m}
$$
Mass of the ball, $m=1 \mathrm{~kg}$
Length of wire, $l=1 \mathrm{~m}$
Area of cross-sectional of wire,
$$
A=0.01 \mathrm{~cm}^2=0.01 \times 10^{-4} \mathrm{~m}
$$
Young's modulus of steel, $Y=2 \times 10^{11} \mathrm{Nm}^{-2}$
$\because$ Tension force in wire, $T=m \omega^2 l$
$\because$ Stress $=\frac{\text { Tension }}{\text { Area }}=\frac{m \omega^2 l}{A}$
Strain $=\frac{\Delta l}{l}=\frac{\text { stress }}{\text { Young's modulus }}$
or $\quad \Delta l=\frac{m \omega^2 l^2}{Y A}$
or $\quad \omega=\sqrt{\frac{Y A \Delta l}{m l^2}}$
Putting the given values, we get
$$
\begin{aligned}
& =\sqrt{\frac{2 \times 10^{11} \times 0.01 \times 10^{-4} \times 2 \times 10^{-3}}{1 \times(1)^2}} \\
\omega & =20 \mathrm{rad} / \mathrm{sec}^{-1}
\end{aligned}
$$
$$
=2 \times 10^{-3} \mathrm{~m}
$$
Mass of the ball, $m=1 \mathrm{~kg}$
Length of wire, $l=1 \mathrm{~m}$
Area of cross-sectional of wire,
$$
A=0.01 \mathrm{~cm}^2=0.01 \times 10^{-4} \mathrm{~m}
$$
Young's modulus of steel, $Y=2 \times 10^{11} \mathrm{Nm}^{-2}$
$\because$ Tension force in wire, $T=m \omega^2 l$
$\because$ Stress $=\frac{\text { Tension }}{\text { Area }}=\frac{m \omega^2 l}{A}$
Strain $=\frac{\Delta l}{l}=\frac{\text { stress }}{\text { Young's modulus }}$
or $\quad \Delta l=\frac{m \omega^2 l^2}{Y A}$
or $\quad \omega=\sqrt{\frac{Y A \Delta l}{m l^2}}$
Putting the given values, we get
$$
\begin{aligned}
& =\sqrt{\frac{2 \times 10^{11} \times 0.01 \times 10^{-4} \times 2 \times 10^{-3}}{1 \times(1)^2}} \\
\omega & =20 \mathrm{rad} / \mathrm{sec}^{-1}
\end{aligned}
$$
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