Q. A steel wire has length 2 m, radius 1 mm and $Y = 2 \times 10^{11}\ Nm^{- 2}.$ A 1 kg sphere is attached to one end of the wire and whirled in a vertical circle with an angular velocity of 2 revolutions per second. When the sphere is at the lowest point of the vertical circle, the elongation of the wire is nearly (Take g = $10ms^{- 2})$
A) 1 mm
B) 2 mm
C) 0.1 mm
D) 0.01 mm
Explanation
(a) $Y = \ \frac{(mg + ml\omega)l}{\pi r^{2}\mathrm{\Delta}l}$
Or $\mathrm{\Delta}l\ = \ \frac{m\left( g + ml\omega^{2} \right)l}{\pi r^{2}Y}$
Or $\mathrm{\Delta}l = \ \frac{1\left( 10 + 2 \times 4\pi^{2} \times 4 \right)^{2}}{\pi\left( 1 \times 10^{- 3} \right)^{2} \times 2 \times 10^{11}}$
Or $\mathrm{\Delta}l = \ \frac{(20 + 64 \times 9.88)7}{2 \times 22 \times 10^{5}}$
= $\frac{4566.24}{44 \times 10^{5}} \times 10^{3}\ mm = 1\ mm$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.