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QJEEM19Ma0155: JEE Main 2019 Mathematics - Class 12 JEEM19Ma0155 Solution: A water tank has the shape of an inverted right circular cone,…

A water tank has the shape of an inverted right circular cone, whose semi-vertical angle is tan⁡−1(12){\tan ^{ - 1}}\left( {{1 \over 2}} \right) . Water is poured into it at a constant rate of 5 cubic meter per minute. The the rate (in m/min.), at which the level of water is rising at the instant when the depth of water in the tank is 10m; is :-

Multiple Choice Options

  • A 115π{1 \over {15\pi }}
  • B 15π{1 \over {5\pi }}
  • C 110π{1 \over {10\pi }}
  • D 2π{2 \over \pi }

Step-by-Step Verified Solution

JEE Main 2019 (Online) 9th April Evening Slot Mathematics - Application of Derivatives Question 147 English Explanation

Given dvdt{{dv} \over {dt}} = 5 cm3/min

and θ\theta = tan⁡−1(12){\tan ^{ - 1}}\left( {{1 \over 2}} \right)

⇒\Rightarrow tan θ\theta = 12{1 \over 2} = rh{r \over h}

⇒\Rightarrow h = 2r

Volume of the cone,

v = 13πr2h{1 \over 3}\pi {r^2}h = 13π(h2)2h{1 \over 3}\pi {\left( {{h \over 2}} \right)^2}h = πh312{{\pi {h^3}} \over {12}}

∴\therefore dvdt=π12(3h2)dhdt{{dv} \over {dt}} = {\pi \over {12}}\left( {3{h^2}} \right){{dh} \over {dt}}

⇒\Rightarrow 5 = π4h2dhdt{\pi \over 4}{h^2}{{dh} \over {dt}}

⇒\Rightarrow 5 = π4(10)2dhdt{\pi \over 4}{\left( {10} \right)^2}{{dh} \over {dt}}

⇒\Rightarrow dhdt{{dh} \over {dt}} = 15π{1 \over {5\pi }}

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