A metallic bar of Young's modulus 0.5 times 1 | vedclass

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

Question

(A) The thermal strain produced in the bar is given by $\epsilon = \alpha \Delta T$.Given $\alpha = 10^{-5} \,^{\circ}C^{-1}$ and $\Delta T = 100^{\ci

Vedclass · Accepted Answer

$50 	imes 10^3 \,N$

Vedclass · Answer

(A) The thermal strain produced in the bar is given by $\epsilon = \alpha \Delta T$.Given $\alpha = 10^{-5} \,^{\circ}C^{-1}$ and $\Delta T = 100^{\circ}C - 0^{\circ}C = 100^{\circ}C$.So, $\epsilon = 10^{-5} 	imes 100 = 10^{-3}$.Since the bar is prevented from expanding, the compressive stress $\sigma$ is given by $\sigma = Y 	imes \epsilon$, where $Y$ is Young's modulus.$\sigma = (0.5 	imes 10^{11} \,N \,m^{-2}) 	imes 10^{-3} = 0.5 	imes 10^8 \,N \,m^{-2}$.The compressive force $F$ is given by $F = \sigma 	imes A$, where $A$ is the area of cross-section.$F = (0.5 	imes 10^8 \,N \,m^{-2}) 	imes (10^{-3} \,m^2) = 0.5 	imes 10^5 \,N = 50 	imes 10^3 \,N$.

Similar Questions

$A$ wire of cross-sectional area $10^{-6} \, m^2$ is elongated by $0.1 \%$ when the tension in it is $1000 \, N$. The Young's modulus of the material of the wire is (Assume radius of the wire is constant).

$A$ uniform plank of Young's modulus $Y$ is moved over a smooth horizontal surface by a constant horizontal force $F$. The area of cross-section of the plank is $A$. The compressive strain on the plank in the direction of the force is

Two wires $A$ and $B$ are made of the same material having a ratio of lengths $\frac{L_A}{L_B} = \frac{1}{3}$ and their diameters ratio $\frac{d_A}{d_B} = 2$. If both the wires are stretched using the same force,what would be the ratio of their respective elongations?

Vedclass Test Series

Exam Paper Generator

Online Exam Module