Question

A brass wire of $1mm$ radius is hung from the ceiling. A small mass, $M$ is hung from the free end of the wire. The temperature of the wire is is $40^{\circ }C$ . When the wire is cooled down from $40^{\circ }C$ to $20^{\circ }C$ it regains its original length of $0.2m$ . The value of $M$ is nearly: (Coefficient of linear expansion and Young's modulus of brass are $10^{-5}\_{}^{{}^{\circ }}{C}_{}^{-1}$ and $10^{11}N{m}^{-2}$ respectively ; $g=10m{s}^{-2}$

• A. $6.5kg$
• B. $1.5kg$
• C. $0.9kg$
• D. $0.5kg$

Answer 1

Answer: (A)

Given, $T_1=40^{\circ }C$ and $T_2=20^{\circ }C$
$\Rightarrow \Delta T=T_1-T_2=40-20=20^{\circ }C$
Also, Young's modulus, $Y=10^{11}N{m}^{-2}$
Coefficient of linear expansion, $\alpha =10^{-5}{}^{\circ }{C}^{-1}$
Radius of wire $r=1mm$ .
Area of the brass wire, $A=\pi \times {\left(10^{-3}\right)}^2{m}^2$
Now, expansion in the wire due to a rise in temperature is $\Delta l=l\alpha \Delta T\Rightarrow \frac{\Delta l}{l}=\alpha \Delta T\dots \left(1\right)$
We know that Young's modulus is defined as $Y=\frac{\text{Stress}}{\text{Strain}}=\frac{F/A}{∆l/l}=\frac{Mgl}{A∆l}$ , here $F$ is force and $g$ is gravitational acceleration.
From the above equation, we get, $M=\frac{YA\Delta l}{gl}\dots \left(2\right)$
Using equation $\left(1\right)$ , we get,
$M=\frac{YA}{g}\times \alpha \Delta T$
$=\frac{10^{11}\times 22\times 10^{-6}\times 10^{-5}\times 20}{7\times 10}$
$\Rightarrow M=\frac{22\times 20}{7\times 10}=\frac{44}{7}=6.28kg$
Which is closest to $6.5kg$ , so option $(a)$ is nearly correct.