Question

A spherical balloon is pumped, at the constant rate of 3m3/min. The rate of increase of it's surface area at certain instant is found to be 5m2/min. At this instant it's radius is equal to:

Answer 1

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evaluate the density change of water. ..... The film contact area is A and its thickness h. .... P1.53 the cone surface and instantaneous rate ω, æ rω öæ dr ö d

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Answer 2

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$V = (\frac{4}{3}) × πr³$

$\frac{dV}{dt} = (\frac{4}{3}) × π3r² (\frac{dr}{dt})$

$\frac{dr}{dt}$ = $\frac{3}{(4πr²)}$ — (1)

Surface area = $4πr²$

$\frac{d (surface\:\:area)}{dt} = 4π (2R) (\frac{dr}{dt})$

$5 = 4π × (2R) × (\frac{3}{4πr²}$

$r = \frac{6}{5}\:\:m$

$r = 1.2 \:\:m$

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