Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

$
  {\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{ }} \\
$

seo-qna
Last updated date: 20th Apr 2024
Total views: 430.2k
Views today: 8.30k
Answer
VerifiedVerified
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{.}} \\
    \\
 \]

Recently Updated Pages