A metal rod of area of cross-section 3 ,cm^2 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The energy stored per unit volume $(u)$ in a stretched rod is given by the formula: $u = \frac{1}{2} 	imes 	ext{stress} 	imes 	ext{strain}$.We

Vedclass · Accepted Answer

$2.25 	imes 10^5 \,Jm^{-3}$

Vedclass · Answer

(C) The energy stored per unit volume $(u)$ in a stretched rod is given by the formula: $u = \frac{1}{2} 	imes 	ext{stress} 	imes 	ext{strain}$.We know that $	ext{stress} = \frac{F}{A}$ and $	ext{strain} = \frac{	ext{stress}}{Y} = \frac{F}{AY}$.Substituting these into the formula, we get $u = \frac{1}{2} 	imes \frac{F}{A} 	imes \frac{F}{AY} = \frac{F^2}{2A^2Y}$.Given: $F = 9 	imes 10^4 \,N$, $A = 3 \,cm^2 = 3 	imes 10^{-4} \,m^2$, and $Y = 2 	imes 10^{11} \,Nm^{-2}$.Calculating stress: $\sigma = \frac{9 	imes 10^4}{3 	imes 10^{-4}} = 3 	imes 10^8 \,Nm^{-2}$.Now, $u = \frac{1}{2} 	imes \sigma 	imes \frac{\sigma}{Y} = \frac{\sigma^2}{2Y}$.$u = \frac{(3 	imes 10^8)^2}{2 	imes 2 	imes 10^{11}} = \frac{9 	imes 10^{16}}{4 	imes 10^{11}} = 2.25 	imes 10^5 \,Jm^{-3}$.

$A$ metal rod of area of cross-section $3 \,cm^2$ is stretched along its length by applying a force of $9 \times 10^4 \,N$. If the Young's modulus of the material of the rod is $2 \times 10^{11} \,Nm^{-2}$, the energy stored per unit volume in the stretched rod is:

Similar Questions

Which of the following wires has the maximum elastic potential energy?

Two wires having the same length and material are stretched by the same force. Their diameters are in the ratio $1:3$. The ratio of strain energy per unit volume for these two wires (smaller to larger diameter),when stretched,is (in $:1$)

What is called elastic potential energy? Write its different formulas.

If one end of a wire is fixed with a rigid support and the other end is stretched by a force of $10 \, N$,then the increase in length is $0.5 \, mm$. The ratio of the energy stored in the wire to the work done in displacing it through $1.5 \, mm$ by the weight is

If the force constant of a wire is $K$,the work done in increasing the length of the wire by $l$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module