JEE Mains · Maths · STD 12 - 6. Application of derivatives
The surface area of a balloon of spherical shape being inflated, increases at a constant rate. If initially, the radius of balloon is \(3\) units and after \(5\) seconds, it becomes \(7\) units, then its radius after \(9\) seconds is
- A \(9\)
- B \(10\)
- C \(11\)
- D \(12\)
Answer & Solution
Correct Answer
(A) \(9\)
Step-by-step Solution
Detailed explanation
Let \(r\) be the radius of spherical balloon \(S =\) Surface area \(S =4 \pi r ^{2}\) \(\frac{ dS }{ dt }=8 \pi r \times \frac{ dr }{ dt }= k \text { (constant) }\) \(4 \pi r ^{2}= kt + C\) (C is constant of integration) For \(t =0, r =3 \Rightarrow 36 \pi= C\) For…
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