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Question 55 of 58

Q.The volume of a spherical balloon increases at the rate of 10 cm3/sec10\ \text{cm}^3/\text{sec}. The rate of change of its surface area when its radius is 16 cm, is

(a) 1⋅5 cm2/sec1{\cdot}5\ \text{cm}^2/\text{sec}
(b) 1⋅8 cm2/sec1{\cdot}8\ \text{cm}^2/\text{sec}
(c) 2 cm2/sec2\ \text{cm}^2/\text{sec}
(d) 1⋅25 cm2/sec1{\cdot}25\ \text{cm}^2/\text{sec}
West Bengal WbchseWest Bengal HS (WBCHSE) Board 2026MCQ· 1mImportance★★★★★
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Relate dSdt\tfrac{dS}{dt} to dVdt\tfrac{dV}{dt} through drdt\tfrac{dr}{dt}; the surface area changes at 20r=1⋅25 cm2/sec\tfrac{20}{r}=1{\cdot}25\ \text{cm}^2/\text{sec}.

This is a classic related-rates problem from the NCERT/CBSE Class 12 application of derivatives chapter.

Volume V=43πr3V=\tfrac43\pi r^3, so

dVdt=4πr2drdt=10⇒drdt=104πr2.\frac{dV}{dt} = 4\pi r^2\frac{dr}{dt} = 10 \Rightarrow \frac{dr}{dt} = \frac{10}{4\pi r^2}.

Surface area S=4πr2S = 4\pi r^2, so …

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