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Question: A spherical balloon is filled with \[4500\pi \] cubic meters of helium gas. If a leak in the balloon...

A spherical balloon is filled with 4500π4500\pi cubic meters of helium gas. If a leak in the balloon causes the gas to escape at the rate of 72π72\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:
A.97\B.79C.29\D.92\begin{array}{l}A.\dfrac{9}{7}\\\B.\dfrac{7}{9}\\\C.\dfrac{2}{9}\\\D.\dfrac{9}{2}\end{array}

Explanation

Solution

At first, find the remaining value of gas after 49 minutes after subtracting (72π×49)\left( {72\pi \times 49} \right) cubic meters from 4500π4500\pi then, use the formula 43πR3\dfrac{4}{3}\pi {R^3} to find the radius of sphere as it represents formula for volume. Then, differentiate V=43πr3V = \dfrac{4}{3}\pi {r^3} with respect to time and then find the value of drdt.\dfrac{{dr}}{{dt}}.

Complete step-by-step answer:
In the question, we are told that a spherical balloon is given and is filled with 4500πcubicmeters4500\pi {\rm{ cubic meters}} of helium gas. Now, a leak is occurring in the balloon which causes the balloon to escape at the rate of 72πcubicmeters/minute72\pi {\rm{ cubic meters}}/{\rm{minute}}. Then, we have to find the rate at which the radius of the balloon decreases 49 minutes after the leakage began.
So, as the speed of escape of volume is 72πcubicmeters/minute72\pi {\rm{ cubic meters}}/{\rm{minute}}, hence, we will first find the volume left in the balloon after 49 minutes.
The volume escaped after 49 minutes will be 72πcubicmeters/minute×49minutes72\pi {\rm{ cubic meters/minute}} \times {\rm{49 minutes}} which can be calculated and equals 3528πcubicmeters.3528\pi {\rm{ cubic meters}}{\rm{.}}
Thus, volume left in the balloon is (4500π3528π)cubicmeters972πcubicmeters\left( {4500\pi - 3528\pi } \right){\rm{cubic meters}} \Rightarrow {\rm{972}}\pi {\rm{ cubic meters}}
The volume of the sphere can be found out by using formula 43πR3\dfrac{4}{3}\pi {R^3} where R is radius of sphere.
Thus, from this we can find out its radius of sphere when its volume is 972π.972\pi .
If the radius of sphere at that period of time is R, then, we can form the equation that,
43πR3=972π\dfrac{4}{3}\pi {R^3} = 972\pi
Now, on cross multiplying, we get,
R3=972π×34π R3=729\begin{array}{l}{R^3} = 972\pi \times \dfrac{3}{{4\pi }}\\\ \Rightarrow {R^3} = 729\end{array}
Hence, the value of R is (729)139m{\left( {729} \right)^{\dfrac{1}{3}}} \Rightarrow 9m. Thus, the length of the radius of the sphere after 49 minutes will be 9m.
Now, as we know that, the formula of volume of sphere is,
V=43πr3V = \dfrac{4}{3}\pi {r^3}
Now, we will differentiate with respect to t throughout the equation, so we get,
dvdt=4π×3r2drdt\dfrac{{dv}}{{dt}} = 4\pi \times 3{r^2}\dfrac{{dr}}{{dt}}
As 43π\dfrac{4}{3}\pi is a constant and r3{r^3} when differentiated with respect to t it gives 3r2drdt3{r^2}\dfrac{{dr}}{{dt}}.
Thus, after differentiating we get,
dvdt=4πr2drdt\dfrac{{dv}}{{dt}} = 4\pi {r^2}\dfrac{{dr}}{{dt}}
We know that, value of dvdt\dfrac{{dv}}{{dt}}is72π72\pi as given.
So, on substituting, we get,
72π=4π2drdt72\pi = 4{\pi ^2}\dfrac{{dr}}{{dt}}
At this time ‘t’ we know the value of r which is 9m, so, on substituting, we get,
72π=4π×9×9×drdt72\pi = 4\pi \times 9 \times 9 \times \dfrac{{dr}}{{dt}}
Or the value of drdt\dfrac{{dr}}{{dt}} is,
drdt=72π4π×9×9 drdt=29\begin{array}{l}\dfrac{{dr}}{{dt}} = \dfrac{{72\pi }}{{4\pi \times 9 \times 9}}\\\ \Rightarrow \dfrac{{dr}}{{dt}} = \dfrac{2}{9}\end{array}
Thus, the rate of decreasing radius is 29\dfrac{2}{9}.
Hence, the correct option is C.

Note: Students must know the correct formula of the volume or they will not be able to solve these kinds of problems. Also, they should be careful about calculations, otherwise, their answer might not be correct. If we were given the time in other units such as seconds or hours, then first we would have had to convert it to minutes because the rate is given to us in cubic meters/minute. Some students try to simplify the given rate by substituting the value of π\pi as 2273.14\dfrac{{22}}{7} \Rightarrow {\rm{3}}.{\rm{14}}, but this is not required because π\pi gets cancelled off during simplification.