CUET Preparation Today
If $V=4\pi r^3/3,$ at what rate in cubic units/ sec is V increasing when r= 10, and dr/dt=0.01 ? |
$\pi $ $4\pi $ $40\pi $ $4\pi /3 $ |
$4\pi $ |
The correct answer is Option (2) → $4\pi$ $\frac{dv}{dt}=4πr^2\frac{dr}{dt}$ $⇒4π(10)^2×0.01$ $=4π$ |
