Answer
Verified
430.2k+ 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{.}} \\
\\
\]
Recently Updated Pages
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred
The branch of science which deals with nature and natural class 10 physics CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Define absolute refractive index of a medium
Find out what do the algal bloom and redtides sign class 10 biology CBSE
Prove that the function fleft x right xn is continuous class 12 maths CBSE
Trending doubts
Select the word that is correctly spelled a Twelveth class 10 english CBSE
What is the z value for a 90 95 and 99 percent confidence class 11 maths CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE
What is the past participle of wear Is it worn or class 10 english CBSE
Fill the blanks with proper collective nouns 1 A of class 10 english CBSE
Write a letter to the principal requesting him to grant class 10 english CBSE