JEE Main 2014 · 19 Apr · Shift 1 · Mathematics Q73

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MathematicsSingle correct+4 / -1Question 73 of 90

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 :

\dfrac16

\dfrac19

\dfrac1{36}

\dfrac1{24}

Answer and worked solution

Official answer: C

$\frac43\pi r^3=288\pi$ gives $r=6$ . $\dfrac{dV}{dt}=4\pi r^2\dfrac{dr}{dt}$ , so
$4\pi=144\pi\frac{dr}{dt}\implies\frac{dr}{dt}=\frac1{36}\text{ cm/s}.$

Topics:Rate of change

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JEE Main 2014 · 19 Apr · Shift 1, Mathematics question 73