JEE MAIN - Mathematics (2012 - No. 14)
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 :
$${{9 \over 7}}$$
$${{7 \over 9}}$$
$${{2 \over 9}}$$
$${{9 \over 2}}$$
Explanation
Volume of spherical balloon $$ = V = {4 \over 3}\pi {r^3}$$
$$ \Rightarrow 4500\pi = {{4\pi {r^3}} \over 3}$$
( as Given, volume $$ = 4500\pi {m^3}$$ )
Differentiating both the sides, $$w.r.t't'$$ we get,
$${{dV} \over {dt}} = 4\pi {r^2}\left( {{{dr} \over {dt}}} \right)$$
Now, it is given that $${{dV} \over {dt}} = 72\pi $$
$$\therefore$$ After $$49$$ min, Volume -
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {4500 - 49 \times 72} \right)\pi $$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {4500 - 3528} \right)\pi $$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 972\pi {m^3}$$
$$ \Rightarrow V = 972\,\,\pi {m^3}$$
$$\therefore$$ $$972\pi = {4 \over 3}\pi r{}^3$$
$$ \Rightarrow {r^3} = 3 \times 243 = 3 \times {3^5} = {3^6} = {\left( {{3^2}} \right)^3} \Rightarrow r = 9$$
Also, we have $${{dV} \over {dt}} = 72\pi $$
$$\therefore$$ $$72\pi = 4\pi \times 9 \times 9\left( {{{dr} \over {dt}}} \right) \Rightarrow {{dr} \over {dt}} = \left( {{2 \over 9}} \right)$$
$$ \Rightarrow 4500\pi = {{4\pi {r^3}} \over 3}$$
( as Given, volume $$ = 4500\pi {m^3}$$ )
Differentiating both the sides, $$w.r.t't'$$ we get,
$${{dV} \over {dt}} = 4\pi {r^2}\left( {{{dr} \over {dt}}} \right)$$
Now, it is given that $${{dV} \over {dt}} = 72\pi $$
$$\therefore$$ After $$49$$ min, Volume -
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {4500 - 49 \times 72} \right)\pi $$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {4500 - 3528} \right)\pi $$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 972\pi {m^3}$$
$$ \Rightarrow V = 972\,\,\pi {m^3}$$
$$\therefore$$ $$972\pi = {4 \over 3}\pi r{}^3$$
$$ \Rightarrow {r^3} = 3 \times 243 = 3 \times {3^5} = {3^6} = {\left( {{3^2}} \right)^3} \Rightarrow r = 9$$
Also, we have $${{dV} \over {dt}} = 72\pi $$
$$\therefore$$ $$72\pi = 4\pi \times 9 \times 9\left( {{{dr} \over {dt}}} \right) \Rightarrow {{dr} \over {dt}} = \left( {{2 \over 9}} \right)$$
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