Q. A load of 4.0 kg is suspended from a ceiling through a steel wire of length 2.0 m and radius 2.0 mm. It is found that the length of the wire increase by 0.031 mm as equilibrium is achieved. Taking g =$3.1\ \pi\ ms^{- 2},$ the Young’s modulus of steel is
A) $2.0\ \times 10^{8}\ Nm^{- 2}$
B) $2.0\ \times 10^{9}\ Nm^{- 2}$
C) $2.0\ \times 10^{11}\ Nm^{- 2}$
D) $2.0\ \times 10^{13}\ Nm^{- 2}$
Explanation
(c) $Y = \ \frac{Mgl}{\pi r^{2} \times l} = \ \frac{4 \times (3.1\pi) \times 2.0}{\pi \times \left( 2 \times 10^{- 3} \right)^{2} \times 0.031 \times 10^{- 3}}$
= 2 $\times 10^{11}\ Nm^{- 2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.