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The volume of metal in a hollo...

# 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.

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Solution :
Volume of a hollow sphere is given by,<br> V = 4/3pi(R^3 - r^3), where R is outer radius and r is inner radius.<br> Here, it is given that r = 4cm and R = 8cm.<br> Now, <br> (dV)/dt = 4/3pi[3R^2(dR)/dt - 3r^2(dr)/dt]->(1)<br> It is given that volume remains constant.<br> :. (dV)/dt = 0<br> Also, (dr)/dt = 1 cm/sec<br> So, putting these values in (1),<br> =>4/3pi[3(8)^2(dR)/dt -3(4)^2(1)] = 0<br> =>192(dR)/dt - 48 = 0<br> =>(dR)/dt = 48/192 = 1/4<br> So, rate of change of outer radius is 1/4 (cm^3)/sec.<br>

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