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JEE Mains · Maths · STD 12 - 6. Application of derivatives
A spherical balloon is being inflated at the rate of \(35\,cc/min\). The rate of increase in the surface area (in \(cm^2/min.\)) of the balloon when its diameter is \(14\, cm\), is
- A \(10\)
- B \(\sqrt {10} \)
- C \(100\)
- D \(10\sqrt {10} \)
Answer & Solution
Correct Answer
(A) \(10\)
Step-by-step Solution
Detailed explanation
Vloum of sphere \(V = \frac{4}{3}\pi {r^3}\) \(\frac{{dV}}{{dt}} = \frac{4}{3}.\pi .3{r^2}.\frac{{dr}}{{dt}}\) \(35 = 4\pi {r^2}.\frac{{dr}}{{dt}}\) or \(\frac{{dr}}{{dt}} = \frac{{35}}{{4\pi {r^2}}}\,\,\,\,\,\,\,\,\,\,......\left( 1 \right)\) Surface area of sphere…
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