Q. The mass and length of a wire are $M$ and $L$ respectively. The density of the material of the wire is $d$. On applying the force $F$ on the wire, the increase in length is $l$, then the Young’s modulus of the material of the wire will be
A) $\frac{Fdl}{Ml}$
B) $\frac{FL}{Mdl}$
C) $\frac{FMl}{dl}$
D) $\frac{FdL^{2}}{Ml}$
Explanation
(d) $Y = \frac{F}{A}\frac{L}{l} = \frac{FdL^{2}}{Ml}$
As $M =$ Volume $\times$ density $= A \times L \times d\therefore A = \frac{M}{Ld}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.