A steel rod of length ell,cross-sectional are | vedclass

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

Question

(B) The thermal expansion of the rod is $\Delta \ell = \ell \alpha t$.The force $F$ required to prevent this expansion (or the force exerted by the ro

Vedclass · Accepted Answer

$\frac{1}{2}(YA\alpha t) 	imes (\ell\alpha t)$

Vedclass · Answer

(B) The thermal expansion of the rod is $\Delta \ell = \ell \alpha t$.The force $F$ required to prevent this expansion (or the force exerted by the rod) is given by Hooke's Law: $F = Y A \frac{\Delta \ell}{\ell} = Y A \frac{\ell \alpha t}{\ell} = Y A \alpha t$.The work done by the rod during this expansion is equivalent to the elastic potential energy stored in the rod,which is given by $W = \frac{1}{2} 	imes 	ext{Force} 	imes 	ext{Extension}$.Substituting the values: $W = \frac{1}{2} 	imes (Y A \alpha t) 	imes (\ell \alpha t)$.

$A$ steel rod of length $\ell$,cross-sectional area $A$,Young's modulus of elasticity $Y$,and thermal coefficient of linear expansion $\alpha$ is heated so that its temperature increases by $t\,^\circ C$. The work that can be done by the rod on heating is:

Similar Questions

When a force is applied on a wire of uniform cross-sectional area $3 \times 10^{-6} \, m^2$ and length $4 \, m$,the increase in length is $1 \, mm$. The energy stored in it will be $(Y = 2 \times 10^{11} \, N/m^2)$. (in $, J$)

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

Identical springs of steel and copper are equally stretched. On which more work will have to be done?

$A$ 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