Question

A mass of 6.5kg in hanging from the end of a 60 cm long steel wire $(Y=2\times {10}^{11}Pa)$ with area of corss-section 0.05 cm2. When it is revolving in a vertical circle it has an angular velocity of 2 revolutions per second, at the bottom of the circle. Approximate elongation of the wire (in metres) when the mass is at its lowest point of the trajectory is:

• A. $8\times {10}^{-4}$
• B. $4\times {10}^{-4}$
• C. $8\times {10}^{-5}$
• D. $4\times {10}^{-5}$

Answer 1

Answer: (D)

Forces acting on the mass are the tension T and the weight W. At the lowest point, $T-W=\frac{m{v}^2}{r}$ or $T=W+mr{\omega }^2$ $(\because v=r\omega )$ $=6.5\times 9.8+6.5\times 0.60\times {(2)}^2$ $=63.7+15.6$ $=79.3N$ We have, $Y=\frac{TL}{Al}$ $\Rightarrow$ $l=\frac{TL}{Ay}$ $=\frac{79.3\times 0.60}{0.05\times {10}^{-4}\times 2\times {10}^{11}}$ $=475.8\times {10}^{-7}m$ $4.75\times {10}^{-5}m$ $\approx 4\times {10}^{-5}m$