Q. A load $W$ produces an extension of $1mm$ in a thread of radius $r$. Now if the load is made $4W$ and radius is made $2r$ all other things remaining same, the extension will become
A) $4\ mm$
B) $16\ mm$
C) $1\ mm$
D) $0.25\ mm$
Explanation
(c) $l = \frac{FL}{AY}\ \therefore l \propto \frac{F}{r^{2}}$
$\frac{l_{1}}{l_{2}} = \frac{F_{2}}{F_{1}}\left( \frac{r_{1}}{r_{2}} \right)^{2} = (4) \times \left( \frac{1}{2} \right)^{2} = 1\therefore l_{2} = l_{1} = 1mm$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.