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AIEEE 2012 Solved Paper

Section: Mathematics

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Question : 72 of 89

Marks: +1, -0

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 :

$\frac{9}{7}$
$\frac{7}{9}$
$\frac{2}{9}$
$\frac{9}{2}$

Solution:

Volume of spherical balloon

=V=\;\frac{4}{3} \pi r^3

\Rightarrow 4500 \pi=\;\frac{4 \pi r^3}{3}

( as Given, volume

=4500 \pi \text{m}^3

)

Differentiating both the sides,

w \cdot r \cdot t' t'

we get,

\;\frac{dV}{dt}=4 \pi r^2 \left(\;\frac{dr}{dt}\right)

Now, it is given that

\;\frac{dV}{dt}=72 \pi

\therefore

After 49 min, Volume -

=(4500-49 \times 72) \pi

=(4500-3528) \pi

=972 \pi \text{m}^3

\;\Rightarrow V=972 \pi \text{m}^3

\;\therefore 972 \pi=\;\frac{4}{3} \pi r^3

\;\Rightarrow r^3=3 \times 243=3 \times 3^5=3^6=(3^2)^3 \Rightarrow r=9

Also, we have

\;\frac{dV}{dt}=72 \pi

\therefore 72 \pi=4 \pi \times 9 \times 9 \left(\;\frac{dr}{dt}\right) \Rightarrow \;\frac{dr}{dt}= \left(\;\frac{2}{9}\right)

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