A steel wire of length L at 40^{circ} C is su | vedclass

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

Question

(A) The extension produced by the mass $m$ is $\Delta L = \frac{FL}{AY} = \frac{mgL}{AY}$.The contraction due to cooling is $\Delta L = L \alpha \Delt

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(A) The extension produced by the mass $m$ is $\Delta L = \frac{FL}{AY} = \frac{mgL}{AY}$.The contraction due to cooling is $\Delta L = L \alpha \Delta T$.Since the wire regains its original length,the extension due to the mass must equal the contraction due to cooling:$\frac{mgL}{AY} = L \alpha \Delta T \Rightarrow mg = AY \alpha \Delta T$.Given $r = 1 \ mm = 10^{-3} \ m$,so $A = \pi r^2 = \pi 	imes (10^{-3})^2 = \pi 	imes 10^{-6} \ m^2$.Given $\Delta T = 40^{\circ} C - 30^{\circ} C = 10^{\circ} C$,$\alpha = 10^{-5} /^{\circ} C$,$Y = 10^{11} \ N/m^2$,and $g \approx 10 \ m/s^2$.Substituting the values:$m = \frac{A Y \alpha \Delta T}{g} = \frac{(\pi 	imes 10^{-6}) 	imes 10^{11} 	imes 10^{-5} 	imes 10}{10} = \pi \approx 3.14 \ kg$.Rounding to the nearest integer,$m \approx 3 \ kg$.

Similar Questions

When a load $W$ is hung from a wire of length $2L$,it just breaks. Now this wire is completely melted and a new wire of length $L$ is formed. If the load $W$ is hung from this new wire,what happens?

What force is required to stretch a wire of cross-sectional area $1 \, cm^2$ to $1.1$ times its original length? (Given: Young's modulus $Y = 2 \times 10^{11} \, N/m^2$)

For four wires made of the same material,if the same force is applied,in which wire will the increase in length be maximum?

Which of the following curves represents the correct distribution of elongation $(y)$ along a heavy rod under its own weight? ($L$ = length of rod, $x$ = distance of a point from the lower end).

Vedclass Test Series

Exam Paper Generator

Online Exam Module