An iron sphere having diameter $D$ and mass $M$ is immersed in hot water so that the temperature of the…

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
  1. $\pi D^2 \cdot \alpha \delta T(\alpha . \delta T-4)$
  2. $\pi D^2 \cdot \alpha \cdot \delta T(\alpha \cdot \delta T+4)$
  3. $\pi D^2 \cdot \alpha \cdot \delta T(\alpha . \delta T-2)$
  4. $\pi D^2 \cdot \alpha . \delta T(\alpha . \delta T+2)$

Solution

Given, diameter of sphere $=D$ Initial surface area, $A=4 \pi R^2$ $=4 \pi\left(\frac{D}{2}\right)^2$ $=\pi D^2$ ...(i) Surface area after heating by temperature $\delta T$, $A^{\prime}=4 \pi\left(\frac{D^{\prime}}{2}\right)^2=\pi\left(D^{\prime}\right)^2$ ...(ii) where $D^{\prime}$ is the new diameter. From the equation of linear expansion, we have $D^{\prime}=D(1+\alpha \delta T)$ ...(iii) putting the value of $D^{\prime}$ from eq. (iii) in eq. (ii) we get $A^{\prime}=\pi D^2(1+\alpha \delta T)^2$ $=\pi D^2\left(1+\alpha^2 \delta T^2+2 \alpha \delta T\right)$ ...(iv) $\begin{aligned} & =\pi D^2\left(1+\alpha^2 \delta T^2+2 \alpha \delta T\right)-\pi D^2 \\ & =\pi D^2\left[1+\alpha^2 \delta T^2+2 \alpha \delta T-1\right] \\ & =\pi D^2\left[\alpha^2 \delta T^2+2 \alpha \delta T\right]\end{aligned}$ Change in surface area $=A^{\prime}-A \doteq \pi D^{\prime 2}-\pi D^2$ $=\pi D^2 \alpha \delta T(\alpha . \delta T+2)$

Asked in: AP EAMCET 2022 (05 Jul Shift 1)

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