Question 258: Properties of Matter

Question

A thin rod having a length of $$1 \mathrm{~m}$$ and area of cross-section $$3 	imes 10^{-6} \mathrm{~m}^{2}$$ 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 $$2 	imes 10^{11} \mathrm{~N} \mathrm{~m}^{-2}$$ and $$2 	imes 10^{-5} \mathrm{~K}^{-1}$$, respectively. The value of $$\mathrm{M}$$ is __________ $$\mathrm{kg}$$.
(Take $$\mathrm{g=10~m~s^{-2}}$$)

Accepted Answer

When the rod is cooled from 210ÃÂ°C to 160ÃÂ°C, it will contract in length due to thermal contraction. The change in length of the rod is given by:
ÃÂL = L$$\alpha $$ÃÂT
where L is the original length of the rod, ÃÂ± is the coefficient of linear expansion, and ÃÂT 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.
$$ 	herefore $$ $$L\alpha \Delta T = {{MgL} \over {AY}}$$
$$ \Rightarrow $$ $$\alpha \Delta T = {{Mg} \over {AY}}$$
$$ \Rightarrow $$ $$Mg = AY\alpha \Delta T$$
$$ \Rightarrow $$ $\mathrm{M} 	imes 10=2 	imes 10^{11} 	imes 3 	imes 10^{-6} 	imes 2 	imes 10^{-5} 	imes 50 $
$$ \Rightarrow $$ $\mathrm{M}=60 \mathrm{~kg}$

Properties of Matter

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