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  • Please solve RD Sharma Class 12 Chapter 21 Differential Equations Exercise Case Study Based Question CSBQ Question 1 Subquestion (iv) maths textbook solution.

Please solve RD Sharma Class 12 Chapter 21 Differential Equations Exercise Case Study Based Question (CSBQ) Question 1 Subquestion (iv) maths textbook solution.

Answers (1)

Answer: option(a) \therefore r=\left(27-\frac{3 \lambda t}{4 \pi}\right)^{\frac{1}{3}}

Hint: Solve the differential equation & find value of radius at time t.

Given: radius of balloon is 3 units, time is t and r is the radius.

Solution: 4 \pi r^{2} \frac{d r}{d t}=-\lambda

 4 \pi r^{2} d r=-\lambda d t

Integrating on both sides,

\begin{aligned} &\int 4 \pi r^{2} d r=\int-\lambda d t \\ &4 \pi \int r^{2} d r=-\lambda \int d t \\ &4 \pi \frac{r^{3}}{3}=-\lambda t+c\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; ....(i) \end{aligned}

 

Now given radius is 3 and & initially i.e., time, t= 0

 

Integrating on both sides,

\begin{aligned} 4 \pi \frac{(3)^{3}}{3} &=-\lambda(0)+c \\ 4 \pi \frac{9 \times 3}{3} &=0+c \\ 36 \pi &=c\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; (ii) \end{aligned}

 

Put (ii) in (i)

\begin{aligned} 4 \pi \frac{r^{3}}{3} &=-\lambda t+36 \pi \\ 4 \pi r^{3} &=-3 \lambda t+108 \pi \\ r^{3} &=\frac{-3 \lambda t+108 \pi}{4 \pi} \\ r^{3} &=\frac{108}{4}-\frac{3 \lambda t}{4 \pi} \\ \therefore \quad r &=\left(27-\frac{3 \lambda t}{4 \pi}\right)^{\frac{1}{3}} \end{aligned}

 

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