#### Question

The volume of metal in a hollow sphere is constant. If the inner radius is increasing at the rate of 1 cm/sec, find the rate of increase of the outer radius when the radii are 4 cm and 8 cm respectively.

#### Solution

\[\text { Let }r_1 \text { be the inner radius and } r_2 \text { be the outer radius and V be the volumeof the hollow sphere at any timet. Then }, \]

\[V=\frac{4}{3}\pi\left( {r_1}^3 - {r_2}^3 \right)\]

\[ \Rightarrow \frac{dV}{dt}=4\pi\left( {r_1}^2 \frac{d r_1}{dt} - {r_2}^2 \frac{d r_2}{dt} \right)\]

\[\Rightarrow {r_1}^2 \frac{d r_1}{dt} = {r_2}^2 \frac{d r_2}{dt} \left[ \because \frac{dV}{dt} = 0 \right]\]

\[\Rightarrow \left( 4 \right)^2 \times1= \left( 8 \right)^2 \frac{d r_2}{dt}\]

\[\Rightarrow\frac{d r_2}{dt}=\frac{1}{4}cm/sec\]

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Solution for question: The Volume of Metal in a Hollow Sphere is Constant. If the Inner Radius is Increasing at the Rate of 1 Cm/Sec, Find the Rate of Increase of the Outer Radius When the Radii concept: Rate of Change of Bodies Or Quantities. For the courses CBSE (Science), CBSE (Commerce), CBSE (Arts), PUC Karnataka Science