Q. A force F is required to break a wire of length $l$ and radius r. What force is required to break a wire, of the same material, having twice the length and six times the radius ?
A) $F$
B) $3\ F$
C) 9 $F$
D) 36 $F$
Explanation
(d) Breaking force does not depend upon length.
Breaking force = breaking stress $\times$ area of cross-section for a given material, breaking stress in constant.
$\therefore\ \ \ \frac{F_{2}}{F_{1}} = \frac{A_{2}}{A_{1}} = \frac{\pi(6r)^{2}}{\pi r^{2}} = 36\ $
Or $F_{2} = 36F_{1} = 36F$s
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.