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Question 142 of 160

Q.The surface area of a spherical balloon is increasing at the rate of 2 cm2^2/sec. At what rate the volume of the balloon is increasing when the radius of the balloon is 6 cm?

Maharashtra MsbshseMaharashtra HSC (MSBSHSE) Board 2020Subjective· 3mImportance★★★★★
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Relate dSdt\dfrac{dS}{dt} to drdt\dfrac{dr}{dt}, then find dVdt\dfrac{dV}{dt}.

S=4πr2  ⟹  dSdt=8πrdrdtS=4\pi r^2 \implies \dfrac{dS}{dt}=8\pi r\dfrac{dr}{dt}

Given dSdt=2\dfrac{dS}{dt}=2:  2=8πrdrdt  ⟹  drdt=14πr\ 2 = 8\pi r\dfrac{dr}{dt} \implies \dfrac{dr}{dt} = \dfrac{1}{4\pi r}

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