Question

A brass wire of length 1m and area of cross section 1mm^2 is stretched by applying a force of 20N, then work done per unit volume of the wire is
[Ybrass = 0.91 x 1011 N/m²]
(A) 2198J/m3
(B) 1049 J/m3
(C) 4422J/m3
(D) 4521 J/m3​

Answer 1

Answer:

2198J/m3

Explanation:

work done per unit volume(strain energy)=stress²/2y

stress²=(F/A)²

(20/10^-6)²=4 x 10^15

strain energy=4 x 10^15 /2 x 0.91 x 10^11

=2197.8021≈2198

Answer 2

Answer:

Work done per unit volume of the wire is equal to 2198J/m³.

Therefore, the option (A) is correct.

Explanation:

Given: The length of the brass wire $l=1m$

Area of cross section of wire, $A=1mm^{2}$ ∴$A=10^{-3}m^{2}$

Force applied to stretch the brass wire, $F=20N$

Stress (σ) =Force/Area

∴  $\sigma=\frac{F}{A}$

⇒   Stress  , $\sigma=\frac{20N}{10^{-6}m^{2}}=2*10^{7}Nm^{-2}$

Given, the Young's module of brass, $Y_{brass}=0.91*10^{11}Nm^{-2}$

Work done per unit volume, $W=\frac{(Stress)^{2}}{2Y}$

⇒    $W=\frac{(2*10^{7})^{2}}{2*0.91*10^{11}}$

⇒    $W=2.1978*10^{3}J$

∴     $W=2197.8Jm^{-3}\\W=2198Jm^{-3}$

Therefore, the work done per unit volume is 2198J/m³.