A metallic rod of length L and cross-sectiona | vedclass

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

Question

(C) The work done in stretching a rod is equal to the elastic potential energy stored in it.The formula for the work done $(W)$ is given by:$W = \frac

Vedclass · Accepted Answer

$y^2$

Vedclass · Answer

(C) The work done in stretching a rod is equal to the elastic potential energy stored in it.The formula for the work done $(W)$ is given by:$W = \frac{1}{2} 	imes 	ext{Force} 	imes 	ext{Elongation}$From Hooke's Law,the force $(F)$ required to produce an elongation $(y)$ in a rod of length $L$ and area $A$ is:$F = \frac{Y A y}{L}$Substituting this into the work formula:$W = \frac{1}{2} 	imes \left( \frac{Y A y}{L} ight) 	imes y$$W = \frac{1}{2} \frac{Y A}{L} y^2$Since $Y$,$A$,and $L$ are constants for a given rod,we have:$W \propto y^2$Therefore,the work done is proportional to $y^2$.

$A$ metallic rod of length $L$ and cross-sectional area $A$ is made of a material of Young's modulus $Y$. If the rod is elongated by an amount $y$,then the work done is proportional to ......

Similar Questions

An elastic material of Young's modulus $Y$ is subjected to a stress $S$. The elastic energy stored per unit volume of the material is

An $8\,m$ long copper wire and $4\,m$ long steel wire,each of cross-section $0.5\,cm^2$,are fastened end to end and stretched by a $500\,N$ force. The elastic potential energy of the system is (Young's modulus: $Y_{cu} = 1 \times 10^{11}\,N/m^2$,$Y_{steel} = 2 \times 10^{11}\,N/m^2$):

The elastic energy stored in a wire of Young's modulus $Y$ is

Calculate the work done, if a wire is loaded by $Mg$ weight and the increase in length is $l$.

The work done per unit volume in stretching a wire is :-

Vedclass Test Series

Exam Paper Generator

Online Exam Module