Q. An aluminium rod, Young’s modulus $7.0\ \times 10^{9}N\ m^{- 2}$, has a breaking strain of 0.2%. The minimum cross-sectional area of the rod in $m^{2}\ $in order to support a load of $10^{4}\ N\ is$
A) $1\ \times 10^{- 2}$
B) $1.4\ \times 10^{- 3}$
C) $1.0\ \times 10^{- 3}$
D) $7.1\ \times 10^{- 4}$
Explanation
(d) $Y = \ \frac{F/A}{Breaking\ strain}$
Or $a = \ \frac{F}{Y \times Breaking\ strain} = \ \frac{10^{4} \times 100}{7 \times 10 \times 0.2}$
= 0.71$\times 10^{- 3} = 7.1\ \times 10^{- 4}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.