Question

A brass wire of $1\,mm$ 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 $0.2\,m$ . The value of $M$ is nearly: (Coefficient of linear expansion and Young's modulus of brass are $10^{- 5}\_{}^{^\circ }C_{}^{- 1}$ and $10^{11}\,Nm^{- 2}$ respectively ; $g = 10\, m s^{- 2}$

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

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, $Y=10^{11}Nm^{- 2}$
Coefficient of linear expansion, $\alpha =10^{- 5}{}^\circ C^{- 1}$
Radius of wire $r=1\,mm$ .
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 $\Delta l=l\alpha \Delta T\Rightarrow \frac{\Delta l}{l}=\alpha \Delta T\ldots \left(1\right)$
We know that Young's modulus is defined as $Y=\frac{\text{Stress}}{\text{Strain}}=\frac{F/A}{∆ l/l}=\frac{M g l}{A ∆ l}$ , here $F$ is force and $g$ is gravitational acceleration.
From the above equation, we get, $M=\frac{Y A \Delta l}{g l}\ldots \left(2\right)$
Using equation $\left(1\right)$ , we get,
$M=\frac{Y A}{g}\times \alpha \Delta T$
$=\frac{10^{11} \times 22 \times 10^{- 6} \times 10^{- 5} \times 20}{7 \times 10}$
$\Rightarrow M=\frac{22 \times 20}{7 \times 10}=\frac{44}{7}=6.28\,kg$
Which is closest to $6.5\,kg$ , so option $\left(\right.a\left.\right)$ is nearly correct.