A rod of length 20 , cm is made of metal A. I | vedclass

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(B) Let $l_1$ be the length of the portion of metal $A$ and $l_2$ be the length of the portion of metal $B$.Since the total length is $20 \, cm$,we ha

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$10$

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(B) Let $l_1$ be the length of the portion of metal $A$ and $l_2$ be the length of the portion of metal $B$.Since the total length is $20 \, cm$,we have $l_1 + l_2 = 20 \, cm$.For a given change in temperature,the change in length $\Delta l$ is proportional to the initial length $l$. Thus,$\Delta l = \alpha \cdot l \cdot \Delta T$.For metal $A$,$\Delta l_A = 0.075 \, cm$ for $l = 20 \, cm$,so the expansion coefficient factor is $k_A = \frac{0.075}{20} = 0.00375$.For metal $B$,$\Delta l_B = 0.045 \, cm$ for $l = 20 \, cm$,so the expansion coefficient factor is $k_B = \frac{0.045}{20} = 0.00225$.The total expansion of the composite rod is $\Delta l_{total} = k_A l_1 + k_B l_2 = 0.06 \, cm$.Substituting $l_2 = 20 - l_1$:$0.00375 l_1 + 0.00225(20 - l_1) = 0.06$$0.00375 l_1 + 0.045 - 0.00225 l_1 = 0.06$$0.0015 l_1 = 0.015$$l_1 = \frac{0.015}{0.0015} = 10 \, cm$.

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