JEE MAIN - Physics (2023 - 31st January Morning Shift - No. 24)
A thin rod having a length of $$1 \mathrm{~m}$$ and area of cross-section $$3 \times 10^{-6} \mathrm{~m}^{2}$$ is suspended vertically from one end. The rod is cooled from $$210^{\circ} \mathrm{C}$$ to $$160^{\circ} \mathrm{C}$$. After cooling, a mass $$\mathrm{M}$$ is attached at the lower end of the rod such that the length of rod again becomes $$1 \mathrm{~m}$$. Young's modulus and coefficient of linear expansion of the rod are $$2 \times 10^{11} \mathrm{~N} \mathrm{~m}^{-2}$$ and $$2 \times 10^{-5} \mathrm{~K}^{-1}$$, respectively. The value of $$\mathrm{M}$$ is __________ $$\mathrm{kg}$$.
(Take $$\mathrm{g=10~m~s^{-2}}$$)
Answer
60
Explanation
When the rod is cooled from 210°C to 160°C, it will contract in length due to thermal contraction. The change in length of the rod is given by:
ΔL = L$$\alpha $$ΔT
where L is the original length of the rod, α is the coefficient of linear expansion, and ΔT is the change in temperature.
When a mass M is attached to the lower end of the rod, it will stretch due to the weight of the mass. The elongation of the rod is given by:
$$\Delta L = {{MgL} \over {AY}}$$
where M is the mass, g is the acceleration due to gravity, A is the cross-sectional area of the rod, Y is the Young's modulus of the rod, and L is the original length of the rod.
$$ \therefore $$ $$L\alpha \Delta T = {{MgL} \over {AY}}$$
$$ \Rightarrow $$ $$\alpha \Delta T = {{Mg} \over {AY}}$$
$$ \Rightarrow $$ $$Mg = AY\alpha \Delta T$$
$$ \Rightarrow $$ $\mathrm{M} \times 10=2 \times 10^{11} \times 3 \times 10^{-6} \times 2 \times 10^{-5} \times 50 $
$$ \Rightarrow $$ $\mathrm{M}=60 \mathrm{~kg}$
ΔL = L$$\alpha $$ΔT
where L is the original length of the rod, α is the coefficient of linear expansion, and ΔT is the change in temperature.
When a mass M is attached to the lower end of the rod, it will stretch due to the weight of the mass. The elongation of the rod is given by:
$$\Delta L = {{MgL} \over {AY}}$$
where M is the mass, g is the acceleration due to gravity, A is the cross-sectional area of the rod, Y is the Young's modulus of the rod, and L is the original length of the rod.
$$ \therefore $$ $$L\alpha \Delta T = {{MgL} \over {AY}}$$
$$ \Rightarrow $$ $$\alpha \Delta T = {{Mg} \over {AY}}$$
$$ \Rightarrow $$ $$Mg = AY\alpha \Delta T$$
$$ \Rightarrow $$ $\mathrm{M} \times 10=2 \times 10^{11} \times 3 \times 10^{-6} \times 2 \times 10^{-5} \times 50 $
$$ \Rightarrow $$ $\mathrm{M}=60 \mathrm{~kg}$
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