Q. A wire can be broken by applying a load of 200 N. The force required to break another wire of the same length and same material, but double in diameter, is
A) 200 N
B) 400 N
C) 600 N
D) 800 N
Explanation
(d) Young’s modulus $Y = \frac{FL}{Al}$
or $F = \frac{YAl}{L}$
or $F \propto A\ or\ F \propto r^{2}\ or\ F \propto d^{2}$
$\therefore$ $\frac{F_{1}}{F_{2}} = \frac{d_{1}^{2}}{d_{2}^{2}}$
Given, $d_{1} = d,\ \ d_{2} = 2d,\ F_{1} = 200N$
$\therefore$ $\frac{200}{F_{2}} = \frac{(d)^{2}}{(2d)^{2}} = \frac{1}{4}$
or $F_{2} = 4 \times 200 = 800N$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.