Question

A $15kg$ mass fastened to the end of a steel wire of unstretched length $1.0m$ is whirled in a vertical circle with an angular velocity of $2rev$ $s^{-1}$ at the bottom of the circle. The cross-section of the wire is $0.05{cm}^2$. The elongation of the wire when the mass is at the lowest point of its path is
(Take $g=10m{s}^{-2},Y_{steel}=2\times {10}^{11}N{m}^{-2}$)

• A. $0.52mm$
• B. $1.52mm$
• C. $2.52mm$
• D. $3.52mm$

Answer 1

The correct option is

C

$2.52mm$

Mass $\left(m\right)=15kg$

Length of wire $\left(L\right)=1m$

Area of cross section of wire $\left(A\right)=0.05cm²=5\times {10}^{-6}{m}^2$

Angular frequency $\left(\nu \right)=2rev/s$

So, Angular velocity, $\left(\omega \right)=2\pi \nu =2\pi \left(2\right)=4\pi rad/s$

Young's modulus for steel $\left(Y\right)=2\times {10}^{11}N/m^2$

At the lowest point of the vertical circle,

$T-mg=m{\omega }^{2}L$

$T=mg+m{\omega }^{2}L$

$=15\left[9.8+\left(4\pi \right)²\times 1\right]$

$=15\left(9.8+16\pi ²\right)$

$=15\times 167.72N$

$=2515.8N$

Now,

Youngs modulus $=\frac{\text{stress}}{\text{strain}}$

$Y=\frac{TL}{A\Delta L}$

$\Delta L=\frac{FL}{AY}$

$=\frac{2515.8\times 1}{(5\times {10}^{-6})\times (2\times 10¹¹)}$

$=2.52\times {10}^{-3}m$

$\Delta L=2.52mm$

Hence, option $\left(C\right)$ is correct.