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A rod of length $20 \,\,cm$ is made of metal. It expands by $0.075\,\, cm$ when its temperature is raised from $0^o C$ to $100^o C$. Another rod of a different metal $B$ having the same length expands by $0.045 cm$ for the same change in temperature, a third rod of the same length is composed of two parts one of metal $A$ and the other of metal $B$. Thus rod expand by $0.06 \,\,cm$.for the same change in temperature. The portion made of metal $A$ has the length .............. $cm$
$20$
$10$
$15$
$18$
Solution
Let $l_{1}$ be the length of potion of metal $A$ and $l_{2}$ be the length of portion of metal $B.$
Thus $l_{1}+l_{2}=20 \mathrm{cm}$
For the given change in temperature, change in length $\propto$ intial length of rod. Hence $\Delta l_{1}=0.075 \frac{l_{1}}{20}$
And $\Delta l_{2}=0.045 \frac{l_{2}}{20}$
Thus $\Delta l_{1}+\Delta l_{2}=0.060$
Solving for $l_{1}$ gives
$l_{1}=10 \mathrm{cm}$
