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#### Please solve RD Sharma class 12 chapter Differential Equations exercise 21.11 question 1 maths textbook solution

Answer: $r=\sqrt{1+\frac{1}{3} t^{2}}$

Given: It is given that initial radius of balloon$=1$ unit. After 3 seconds, radius of balloon$=2$ units.

To find: We have to find the radius after time $t$.

Hint: Use surface area of a balloon $=4\pi r^{2}$. Then differentiate it to find the radius.

Solution: We have radius of the balloon$=1$ unit

After 3 seconds, radius of balloon$=2$ units

Let $r$ be the radius and $A$ be the surface area of the balloon at any time $t$.

\begin{aligned} &=\frac{d A}{d t} \propto t \\\\ &=\frac{d A}{d t}=k t \end{aligned}        where $k$ is constant.

\begin{aligned} &=\frac{d\left(4 \pi r^{2}\right)}{d t}=k t \\\\ &=8 \pi r \frac{d r}{d t}=k t \end{aligned}

Integrating on both sides

\begin{aligned} &=8 \pi \frac{r^{2}}{2}=k \frac{t^{2}}{2}+C \\\\ &=4 \pi r^{2}=k \frac{t^{2}}{2}+C \ldots(i) \end{aligned}

We are given that unit at $t=0, r=1$ unit

\begin{aligned} &=4 \pi(1)^{2}=k \times 0+C \\\\ &=C=4 \pi \end{aligned}

And $t=3, r=2$ units

\begin{aligned} &=4 \pi(2)^{2}=k \frac{(3)^{2}}{2}+C \quad \text { [From (i)] }\\ \\&=16 \pi=\frac{9}{2} k+4 \pi\\\\ &=\frac{9}{2} k=12 \pi \end{aligned}

By cross multiplication we get

\begin{aligned} &9 k=24 \pi \\\\ &k=\frac{24}{9} \pi \\\\ &k=\frac{8}{3} \pi \end{aligned}

Now substitute $C=4 \pi$ and   $k=\frac{8}{3} \pi$  in equation (i)

\begin{aligned} &4 \pi r^{2}=\frac{8}{3} \pi \frac{t^{2}}{2}+4 \pi \\\\ &4 \pi r^{2}-4 \pi=\frac{8}{6} \pi t^{2} \\\\ &4 \pi\left(r^{2}-1\right)=\frac{8}{6} \pi t^{2} \\\\ &\pi\left(r^{2}-1\right)=\frac{1}{4}\cdot \frac{8}{6} \pi t^{2} \end{aligned}

\begin{aligned} &\left(r^{2}-1\right)=\frac{1}{\pi} \frac{1}{3} \pi t^{2} \\\\ &r^{2}-1=\frac{1}{3} t^{2} \\\\ &r^{2}=1+\frac{1}{3} t^{2}=\sqrt{1+\frac{1}{3} t^{2}} \end{aligned}

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