Two wires having the same length and material | vedclass

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

Question

(D) The strain energy per unit volume $u$ is given by the formula $u = \frac{1}{2} 	imes 	ext{stress} 	imes 	ext{strain}$.Since $	ext{stress} = \

Vedclass · Accepted Answer

$81$

Vedclass · Answer

(D) The strain energy per unit volume $u$ is given by the formula $u = \frac{1}{2} 	imes 	ext{stress} 	imes 	ext{strain}$.Since $	ext{stress} = \frac{F}{A}$ and $	ext{strain} = \frac{	ext{stress}}{Y}$,we have $u = \frac{1}{2} 	imes \frac{	ext{stress}^2}{Y} = \frac{1}{2} 	imes \frac{F^2}{A^2 Y}$.Given that the wires have the same length,material ($Y$ is constant),and are stretched by the same force $F$,the energy density $u$ is inversely proportional to the square of the area of cross-section $A^2$.Since $A = \pi r^2$,we have $A \propto d^2$ (where $d$ is the diameter),so $u \propto \frac{1}{(d^2)^2} = \frac{1}{d^4}$.Therefore,the ratio of strain energy per unit volume for the smaller diameter wire $(d_S)$ to the larger diameter wire $(d_L)$ is $\frac{u_S}{u_L} = \left( \frac{d_L}{d_S} ight)^4$.Given the ratio of diameters $d_S : d_L = 1 : 3$,we have $\frac{u_S}{u_L} = \left( \frac{3}{1} ight)^4 = 81 : 1$.

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 $:1$)

Similar Questions

$A$ wire of length $L$ and area of cross-section $A$ is made of a material with Young's modulus $Y$. It is stretched by an amount $x$. The work done in stretching the wire is:

$A$ wire suspended vertically from one of its ends is stretched by attaching a weight of $200\, N$ to the lower end. The weight stretches the wire by $1\, mm$. Then the elastic energy stored in the wire is ........ $J$.

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

$A$ wire of Young's modulus $1.6 \times 10^{12} \,N/m^{2}$ is stretched by a force so as to produce a strain of $2 \times 10^{-4}$. The energy density of the wire is

$A$ wire of length $1 \, m$ and Young's modulus $Y = 2.0 \times 10^{11} \, N/m^2$ is fixed between two rigid supports. If its temperature is increased by $100 ^\circ C$,what is the energy stored in the wire? (Given: $\alpha = 18 \times 10^{-6} \, ^\circ C^{-1}$,$A = 1 \, cm^2$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module