An iron sphere having diameter D and mass M i | vedclass

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(D) Given,diameter of sphere $= D$. Initial surface area,$A = 4 \pi R^2 = 4 \pi (D/2)^2 = \pi D^2$. Surface area after heating by temperature $\delta

Vedclass · Accepted Answer

$\pi D^2 \cdot \alpha \delta T(\alpha \delta T+2)$

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(D) Given,diameter of sphere $= D$. Initial surface area,$A = 4 \pi R^2 = 4 \pi (D/2)^2 = \pi D^2$. Surface area after heating by temperature $\delta T$ is $A' = \pi (D')^2$,where $D'$ is the new diameter. From the equation of linear expansion,$D' = D(1 + \alpha \delta T)$. Substituting $D'$ into the expression for $A'$,we get $A' = \pi [D(1 + \alpha \delta T)]^2 = \pi D^2 (1 + 2\alpha \delta T + \alpha^2 \delta T^2)$. The change in surface area $\Delta A = A' - A = \pi D^2 (1 + 2\alpha \delta T + \alpha^2 \delta T^2) - \pi D^2$. $\Delta A = \pi D^2 (2\alpha \delta T + \alpha^2 \delta T^2) = \pi D^2 \alpha \delta T (2 + \alpha \delta T)$.

An iron sphere having diameter $D$ and mass $M$ is immersed in hot water so that the temperature of the sphere increases by $\delta T$. If $\alpha$ is the coefficient of linear expansion of the iron,then the change in the surface area of the sphere is:

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