A spherical balloon is filled with 4500 \pi cubic meters of helium gas. If a leak in the balloon causes the gas to escape at the rate of 72 \pi cubic meters per minute, then the rate (in meters per minute) at which the radius of the balloon decreases 49 minutes after the leakage began is

 

  • Option 1)

    \frac{9}{7}

  • Option 2)

    \frac{7}{9}

  • Option 3)

    \frac{2}{9}

  • Option 4)

    \frac{9}{2}

 

Answers (1)
V Vakul

As we learnt in

Rate Measurement -

Rate of any of variable with respect to time is rate of measurement. Means according to small change in time how much other factors change is rate measurement:

\Rightarrow \frac{dx}{dt},\:\frac{dy}{dt},\:\frac{dR}{dt},(linear),\:\frac{da}{dt}


\Rightarrow \frac{dS}{dt},\:\frac{dA}{dt}(Area)


\Rightarrow \frac{dV}{dt}(Volume)


\Rightarrow \frac{dV}{V}\times 100(percentage\:change\:in\:volume)

- wherein

Where dR / dt  means Rate of change of radius.

 

 

 

\frac{dv}{dt}= -72\pi\ \, \, \, \, \, \, \, v_{0}= 4500\pi

v= \frac{4}{3}\pi r^{3}

\frac{dv}{dt} = \frac{4}{3}\pi r^{2} \times 3. \frac{dr}{dt}

After\: 49\: min\: ,v= v_{0}+49\cdot \frac{dv}{dt}

= 4500\pi -49\times 72\pi

= 4500\pi -3528\pi = 972\pi

\Rightarrow 972\pi = \frac{4}{3}\pi r^{3}

  r = 9 

\therefore -72\pi = 4\pi \times 81\times \frac{dr}{dt}

\therefore \frac{dr}{dt}=\frac{-2}{9}

Thus, radius decreases at a rate of \frac{2}{9}\: m/min


Option 1)

\frac{9}{7}

Incorrect

Option 2)

\frac{7}{9}

Incorrect

Option 3)

\frac{2}{9}

Correct

Option 4)

\frac{9}{2}

Incorrect

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