Q. When the length of a wire having cross-section area $10^{- 6}m^{2}$ is stretched by 0.1%, then tension in it is $100\ N$. Young’s modulus of material of the wire is
A) $10^{12}N/m^{2}$
B) $10^{2}N/m^{2}$
C) $10^{10}N/m^{2}$
D) $10^{11}N/m^{2}$
Explanation
(d) $A = 10^{- 6}m^{2}$
$Y = \frac{\left( \frac{T}{A} \right)}{\frac{\mathrm{\Delta}l}{l}} = \frac{\left( \frac{100}{10^{- 6}} \right)}{\left( \frac{0.1}{100} \right)} = \frac{100}{10^{- 6}} \times \frac{100}{0.1} = \frac{10^{4}}{10^{- 7}} = 10^{11}N/m^{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.