Heat and Thermodynamics 2 Question 6
9. Steel wire of length $L$ at $40^{\circ} C$ is suspended from the ceiling and then a mass $m$ is hung from its free end. The wire is cooled down from $40^{\circ} C$ to $30^{\circ} C$ to regain its original length $L$. The coefficient of linear thermal expansion of the steel is $10^{-5}{ }^{\circ} C$, Young’s modulus of steel is $10^{11} N / m^{2}$ and radius of the wire is $1 mm$. Assume that $L>$ diameter of the wire. Then the value of $m$ in $kg$ is nearly.
(2011)
Analytical & Descriptive Questions
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Answer:
Correct Answer: 9. $1.105 \times 10^{11} N / m^{2}, 2 \times 10^{-5} \operatorname{per}^{\circ} C$
Solution:
- $\Delta l _1=\frac{F L}{A Y}=\frac{m g L}{\pi r^{2} Y}=$ Increase in length
$$ \Delta l _2=L \alpha \Delta \theta=\text { Decrease in length. } $$
To regain its original length, $\Delta l _1=\Delta l _2$
$\therefore \quad \frac{m g L}{\pi r^{2} Y}=L \alpha \Delta \theta \quad \Rightarrow \quad \therefore \quad m=\left(\frac{r^{2} Y \alpha \Delta \theta}{g}\right)$
Substituting the values we get, $m \simeq 3 kg$
$\therefore$ Answer is 3 .
