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

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Answer: option(a) 4 \pi r^{2} \frac{d r}{d t}=-\lambda, \lambda>0

Hint: Use formula of volume of \text { sphere }=\frac{4}{3} \pi r^{3}

Given: r is the radius of balloon.

Solution: \frac{d V}{d t}=-\lambda as framed above

So, using formula ,\mathrm{V}=\frac{4}{3} \pi r^{3}

Where, r = radius of spherical balloon.

\begin{aligned} &\frac{d}{d t}\left(\frac{4}{3} \pi r^{3}\right)=-\lambda \\ &\frac{4}{3} \pi \frac{d}{d t}\left(r^{3}\right)=-\lambda \\ &\frac{4 \pi}{3} 3 r^{2} \frac{d r}{d t}=-\lambda \\ &4 \pi r^{2} \frac{d r}{d t}=-\lambda \end{aligned}


Thus the radius at instant t is


4 \pi r^{2} \frac{d r}{d t}=-\lambda, \lambda>0

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