$
{\text{The volume of metal in a hollow sphere is constant}}{\text{. If the inner radius is }} \\
{\text{increasing at the rate of 1 cm/sec}},{\text{ find the rate of increase of outer radius, }} \\
{\text{when the radii are 4 cm and 8 cm respectively}}{\text{.}} \\
{\text{ }} \\
$
Answer
333.9k+ views
\[
{\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{.}} \\
\\
\]
Last updated date: 29th May 2023
•
Total views: 333.9k
•
Views today: 5.91k
Recently Updated Pages
If ab and c are unit vectors then left ab2 right+bc2+ca2 class 12 maths JEE_Main

A rod AB of length 4 units moves horizontally when class 11 maths JEE_Main

Evaluate the value of intlimits0pi cos 3xdx A 0 B 1 class 12 maths JEE_Main

Which of the following is correct 1 nleft S cup T right class 10 maths JEE_Main

What is the area of the triangle with vertices Aleft class 11 maths JEE_Main

The coordinates of the points A and B are a0 and a0 class 11 maths JEE_Main

Trending doubts
Write an application to the principal requesting five class 10 english CBSE

Tropic of Cancer passes through how many states? Name them.

Write a letter to the principal requesting him to grant class 10 english CBSE

List out three methods of soil conservation

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

Write a letter to the Principal of your school to plead class 10 english CBSE

What is per capita income

Change the following sentences into negative and interrogative class 10 english CBSE

A Short Paragraph on our Country India
