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(D) Let the initial length be $L$ and the initial radius be $r$. The initial area of the cross-section is $A = \pi r^2$.When the cylinder is heated, t

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$6$

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(D) Let the initial length be $L$ and the initial radius be $r$. The initial area of the cross-section is $A = \pi r^2$.When the cylinder is heated, the length increases by $3 \%$, so the new length is $L' = L(1 + 0.03)$.Since the material is uniform, the linear expansion coefficient $\alpha$ is constant. The change in length is $\Delta L = L \alpha \Delta T$, where $\Delta L / L = 0.03$.The area of the cross-section depends on the radius $r$. The change in radius is $\Delta r = r \alpha \Delta T$.The new area is $A' = \pi (r + \Delta r)^2 = \pi (r^2 + 2r \Delta r + (\Delta r)^2)$.Neglecting the higher-order term $(\Delta r)^2$, we get $A' \approx \pi r^2 + 2 \pi r \Delta r = A + 2A (\Delta r / r)$.The fractional change in area is $\Delta A / A = 2 (\Delta r / r) = 2 \alpha \Delta T$.Since $\Delta L / L = \alpha \Delta T = 0.03$, we have $\Delta A / A = 2 	imes 0.03 = 0.06$.Thus, the area of the cross-section increases by $6 \%$.

On heating a uniform metallic cylinder, its length increases by $3 \%$. The area of cross-section of its base will increase by ........... $\%$

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