Properties Of Matter Ques 45
- 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: 45.(a)
Solution:
- 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
$$ \alpha \Delta T=\frac{F}{Y A} \Rightarrow \quad Y=\frac{F}{A \alpha \Delta T} $$
BETA
