Water is being filled at the rate of 100 litre per minute in a conical vessel (vertex downward) with axis vertical. Its semi vertical angle is tan−1(512). The rate in square meter per minute, at which the wet curved surface area of the tank is increasing, when the depth of the water tank is 3 meters is mn(m,n∈N, G.C.D=1) so that m+n equals

tanθ=512=rn⇒ r=5h12, Given dvdt=0.1,V=13πr2h=25π432h3Diff dhdt=16250π Now 5=πrl=65πh2144 ⇒dδdt=1375⇒m+n=88

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Water is being filled at the rate of 100 litre per minute in a conical vessel (vertex downward) with axis vertical. Its semi vertical angle is $\tan^{- 1} (\frac{5}{12}) .$ The rate in square meter per minute, at which the wet curved surface area of the tank is increasing, when the depth of the water tank is 3 meters is $\frac{m}{n} (m, n \in N, G . C . D = 1)$ so that $m + n$ equals

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answer is 88.

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Detailed Solution

$\tan θ = \frac{5}{12} = \frac{r}{n} \Rightarrow r = \frac{5 h}{12},$ Given $\frac{d v}{d t} = 0.1, V = \frac{1}{3} π r^{2} h = \frac{25 π}{432} h^{3}$
Diff $\frac{d h}{d t} = \frac{16}{250} π$ Now $5 = π r l = \frac{65 π h^{2}}{144}$

$\Rightarrow \frac{d δ}{d t} = \frac{13}{75} \Rightarrow m + n = 88$