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A metal rod of cross-sectional area $3 \times 10^{-6} \mathrm{~m}^{2}$ is suspended vertically from one
end has a length $0 \cdot 4 \mathrm{~m}$ at $100^{\circ} \mathrm{C}$. Now the rod is cooled upto $0^{\circ} \mathrm{C}$, but prevented
from contracting by attaching a mass ' $\mathrm{m}$ ' at the lower end. The value of ' $\mathrm{m}$ ' is
$\left(\mathrm{Y}=10^{11} \mathrm{~N} / \mathrm{m}^{2}\right.$, coefficient of linear expansion $\left.=10^{-5} / \mathrm{K}, \mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\right)$
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end has a length $0 \cdot 4 \mathrm{~m}$ at $100^{\circ} \mathrm{C}$. Now the rod is cooled upto $0^{\circ} \mathrm{C}$, but prevented
from contracting by attaching a mass ' $\mathrm{m}$ ' at the lower end. The value of ' $\mathrm{m}$ ' is
$\left(\mathrm{Y}=10^{11} \mathrm{~N} / \mathrm{m}^{2}\right.$, coefficient of linear expansion $\left.=10^{-5} / \mathrm{K}, \mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\right)$
Solution:
2153 Upvotes
Verified Answer
The correct answer is:
$30 \mathrm{~kg}$
$\Delta \mathrm{L}=\mathrm{L} \propto \Delta \mathrm{T}=\frac{\mathrm{FL}}{\mathrm{AY}}$
$\therefore \mathrm{F}=\mathrm{AY} \propto \Delta \mathrm{T}$
$\therefore \mathrm{Mg}=\mathrm{AY} \propto \Delta \mathrm{T}$
$\mathrm{M}=\frac{\mathrm{AY} \propto \Delta \mathrm{T}}{\mathrm{g}}=\frac{3 \times 10^{-6} \times 10^{11} \times 10^{-5} \times 100}{10}=30 \mathrm{~kg}$
$\therefore \mathrm{F}=\mathrm{AY} \propto \Delta \mathrm{T}$
$\therefore \mathrm{Mg}=\mathrm{AY} \propto \Delta \mathrm{T}$
$\mathrm{M}=\frac{\mathrm{AY} \propto \Delta \mathrm{T}}{\mathrm{g}}=\frac{3 \times 10^{-6} \times 10^{11} \times 10^{-5} \times 100}{10}=30 \mathrm{~kg}$
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