Properties of Matter 1 Question 7

7. A rod of length $L$ at room temperature and uniform area of cross-section $A$, is made of a metal having coefficient of linear expansion $\alpha /{ }^{\circ} C$. It is observed that an external compressive force $F$, is applied on each of its ends, prevents any change in the length of the rod, when its temperature rises by $\Delta T K$. Young’s modulus, $Y$ for this metal is

(a) $\frac{F}{2 A \alpha \Delta T}$

(b) $\frac{F}{A \alpha(\Delta T-273)}$

(c) $\frac{2 F}{A \alpha \Delta T}$

(d) $\frac{F}{A \alpha \Delta T}$

(2019 Main, 9 Jan I)

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Answer:

Correct Answer: 7. (a)

Solution:

  1. If a rod of length $L$ and coefficient of linear expansion $\alpha /{ }^{\circ} C$, then with the rise in temperature by $\Delta T K$, its change in length is given as,

$$ \Delta L=L \alpha \Delta T \Rightarrow \frac{\Delta L}{L}=\alpha \Delta T $$

Also, when a rod is subjected to some compressive force $(F)$, then its’ Young’s modulus is given as

$$ \begin{aligned} Y & =\frac{\text { Stress }}{\text { Strain }}=\frac{\frac{F}{A}}{\frac{\Delta L}{L}} \\ \frac{\Delta L}{L} & =\frac{F}{Y A} \end{aligned} $$

Since, it is given that the length of the rod does not change. So, from Eqs. (i) and (ii), we get

$$ \alpha \Delta T=\frac{F}{Y A} \Rightarrow \quad Y=\frac{F}{A \alpha \Delta T} $$



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