If the volume of a spherical ball is increasing at the rate of 4π cc/sec then the rate of increase of its radius in cm/sec, when the volume is 288π cc…

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If the volume of a spherical ball is increasing at the rate of 4π cc/sec then the rate of increase of its radius in cm/sec, when the volume is 288π cc is

MARKS Content Team · Accepted Answer

Correct answer: (D) 1/(3 6). Volume of sphere of radius r is V=4/3πr^3 Given V=288π cc ⇒288π=4/3πr^3 ⇒r^3=288×3/4=216 ⇒r=6 cm. Again, V=4/3πr^3 Differentiate w.r.t. 't' ⇒dV/dt=4πr^2dr/dt Put the given values, to get ⇒4π=4π×6^2×dr/dt ⇒dr/dt=1/36 cm/sec.

If the volume of a spherical ball is increasing at the rate of $4 π cc / \sec$ then the rate of increase of its radius $(in cm / \sec),$ when the volume is $288 π cc$ is

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