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\[

{\text{Let }}{r_i}{\text{ and }}{r_o}{\text{ are the inner and outer radii of the hollow sphere respectively}} \\

{\text{Given, }}\dfrac{{d{r_i}}}{{dt}} = 1{\text{ cm/sec }} \\

{\text{Volume of the hollow sphere is given by V}} = \dfrac{4}{3}\pi \left( {{r_o}^3 - {r_i}^3} \right) \\

{\text{Also, given that the volume of the hollow sphere is constant}} \\

{\text{i}}{\text{.e}}{\text{. V}} = \dfrac{4}{3}\pi \left( {{r_o}^3 - {r_i}^3} \right) = {\text{constant}} \\

{\text{Differentiating the above equation with respect to time, we get}} \\

\dfrac{{d{\text{V}}}}{{dt}} = \dfrac{4}{3}\pi \left( {3{r_o}^2\dfrac{{d{r_o}}}{{dt}} - 3{r_i}^2\dfrac{{d{r_i}}}{{dt}}} \right) \\

{\text{Since, differentiation of a constant is zero}}{\text{. Therefore, }}\dfrac{{d{\text{V}}}}{{dt}} = 0. \\

\Rightarrow 0 = \dfrac{4}{3}\pi \left( {3{r_o}^2\dfrac{{d{r_o}}}{{dt}} - 3{r_i}^2\dfrac{{d{r_i}}}{{dt}}} \right) \Rightarrow 0 = 3{r_o}^2\dfrac{{d{r_o}}}{{dt}} - 3{r_i}^2\dfrac{{d{r_i}}}{{dt}} \\

{\text{Putting the value of }}\dfrac{{d{r_i}}}{{dt}} = 1{\text{ cm/sec, we get}} \\

\Rightarrow 0 = 3{r_o}^2\dfrac{{d{r_o}}}{{dt}} - 3{r_i}^2 \Rightarrow \dfrac{{d{r_o}}}{{dt}} = \dfrac{{{r_i}^2}}{{{r_o}^2}}{\text{ }}...........{\text{(1)}} \\

{\text{Since, we have to find rate of increase of outer radii of the hollow }} \\

{\text{sphere i}}{\text{.e}}{\text{.}}\dfrac{{d{r_o}}}{{dt}}{\text{ when }}{r_i} = 4{\text{ cm and }}{r_o} = 8{\text{ cm}}{\text{.}} \\

{\text{So put }}{r_i} = 4{\text{ cm and }}{r_o} = 8{\text{ cm in equation (1), we have }} \\

\dfrac{{d{r_o}}}{{dt}} = \dfrac{{{4^2}}}{{{8^2}}} = \dfrac{1}{4}{\text{ cm/sec}}{\text{.}} \\

{\text{Since, the above rate of change of inner radii is positive that means its increasing with time}}{\text{.}} \\

\\

{\text{Note - These type of problems are always computed by somehow obtaining the relation between}} \\

{\text{the rate of change of two quantities where the value of one is given and the value of the other }} \\

{\text{is to be computed}}{\text{.}} \\

\\

\]

{\text{Let }}{r_i}{\text{ and }}{r_o}{\text{ are the inner and outer radii of the hollow sphere respectively}} \\

{\text{Given, }}\dfrac{{d{r_i}}}{{dt}} = 1{\text{ cm/sec }} \\

{\text{Volume of the hollow sphere is given by V}} = \dfrac{4}{3}\pi \left( {{r_o}^3 - {r_i}^3} \right) \\

{\text{Also, given that the volume of the hollow sphere is constant}} \\

{\text{i}}{\text{.e}}{\text{. V}} = \dfrac{4}{3}\pi \left( {{r_o}^3 - {r_i}^3} \right) = {\text{constant}} \\

{\text{Differentiating the above equation with respect to time, we get}} \\

\dfrac{{d{\text{V}}}}{{dt}} = \dfrac{4}{3}\pi \left( {3{r_o}^2\dfrac{{d{r_o}}}{{dt}} - 3{r_i}^2\dfrac{{d{r_i}}}{{dt}}} \right) \\

{\text{Since, differentiation of a constant is zero}}{\text{. Therefore, }}\dfrac{{d{\text{V}}}}{{dt}} = 0. \\

\Rightarrow 0 = \dfrac{4}{3}\pi \left( {3{r_o}^2\dfrac{{d{r_o}}}{{dt}} - 3{r_i}^2\dfrac{{d{r_i}}}{{dt}}} \right) \Rightarrow 0 = 3{r_o}^2\dfrac{{d{r_o}}}{{dt}} - 3{r_i}^2\dfrac{{d{r_i}}}{{dt}} \\

{\text{Putting the value of }}\dfrac{{d{r_i}}}{{dt}} = 1{\text{ cm/sec, we get}} \\

\Rightarrow 0 = 3{r_o}^2\dfrac{{d{r_o}}}{{dt}} - 3{r_i}^2 \Rightarrow \dfrac{{d{r_o}}}{{dt}} = \dfrac{{{r_i}^2}}{{{r_o}^2}}{\text{ }}...........{\text{(1)}} \\

{\text{Since, we have to find rate of increase of outer radii of the hollow }} \\

{\text{sphere i}}{\text{.e}}{\text{.}}\dfrac{{d{r_o}}}{{dt}}{\text{ when }}{r_i} = 4{\text{ cm and }}{r_o} = 8{\text{ cm}}{\text{.}} \\

{\text{So put }}{r_i} = 4{\text{ cm and }}{r_o} = 8{\text{ cm in equation (1), we have }} \\

\dfrac{{d{r_o}}}{{dt}} = \dfrac{{{4^2}}}{{{8^2}}} = \dfrac{1}{4}{\text{ cm/sec}}{\text{.}} \\

{\text{Since, the above rate of change of inner radii is positive that means its increasing with time}}{\text{.}} \\

\\

{\text{Note - These type of problems are always computed by somehow obtaining the relation between}} \\

{\text{the rate of change of two quantities where the value of one is given and the value of the other }} \\

{\text{is to be computed}}{\text{.}} \\

\\

\]

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