Question

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. $\frac{Fd{L}^2}{Ml}$

Answer 1

The correct option is D $\frac{Fd{L}^2}{Ml}$

$Y=\frac{F}{A}\frac{L}{l}=\frac{Fd{L}^2}{Ml}AsM=volume\times density=A\times L\times \therefore A=\frac{M}{Ld}$