$\begin{array}{1 1} (A)\;\frac{-1}{\pi}cm/s \\ (B)\;\frac{-1}{\pi}m^2/s \\ (C)\;\frac{-1}{\pi}m/s \\(D)\;\frac{1}{\pi}cm/s \end{array} $

Want to ask us a question? Click here

Browse Questions

Ad |

0 votes

0 votes

- If $y=f(x)$,then $\large\frac{dy}{dx}$ measures the rate of change of $y$ w.r.t $x$.
- $\big(\large\frac{dy}{dx}\big)_{x=x_0}$ represents the rate of change of $y$ w.r.t $x$ at $x=x_0$

Step 1:

Given :$\large\frac{dv}{dt}=$$900cm^3/sec$

Radius $r=15cm$

Volume of the sphere is $v=\large\frac{4}{3}$$\pi r^3$

Differentiating w.r.t $t$ on both sides we get,

$\large\frac{dv}{dt}$$=\large\frac{4}{3}$$\pi3r^2\large\frac{dr}{dt}$

Step 2:

Substituting the values of $r$ and $\large\frac{dr}{dt}$ we get,

$900=\large\frac{4}{3}$$\times \pi\times 3\times(15)^2\times \large\frac{dr}{dt}$

On simplifying we get,

$\large\frac{dr}{dt}$$=\large\frac{900\times 3}{4\times \pi\times 3\times 15\times 15}$

$\quad\;=\large\frac{1}{\pi}$$cm/s$

The rate at which the radius of the balloon increases is $\large\frac{1}{\pi}$$cm/s$

Ask Question

Take Test

x

JEE MAIN, CBSE, NEET Mobile and Tablet App

The ultimate mobile app to help you crack your examinations

...