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Application of Derivatives question

2019 · 9 Apr · Shift 2 · Q40
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  5. /2019 · 9 Apr · Shift 2 · Q40

Application of Derivatives question

2019 · 9 Apr · Shift 2 · Q40

JEE MainMathematicsApplication of DerivativesMCQ+4 / −1
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)tan−1(21​). 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 :-
  1. A
    115π{1 \over {15\pi }}15π1​
  2. B
    15π{1 \over {5\pi }}5π1​
  3. C
    110π{1 \over {10\pi }}10π1​
  4. D
    2π{2 \over \pi }π2​
View written solutionFree

Correct answer: B

  1. Relate radius and height using the semi-vertical angle

For the cone, tan⁡θ=rh\tan \theta = \frac{r}{h}tanθ=hr​ where θ=tan⁡−1(12)\theta = \tan^{-1}\left(\frac12\right)θ=tan−1(21​).

So, rh=12  ⟹  r=h2\frac{r}{h} = \frac12 \implies r = \frac{h}{2}hr​=21​⟹r=2h​

  1. Volume of water in the cone

Volume of a cone: V=13πr2hV = \frac13 \pi r^2 hV=31​πr2h

Substitute r=h2r = \frac{h}{2}r=2h​:

= \frac13 \pi \cdot \frac{h^2}{4} \cdot h = \frac{\pi}{12} h^3$$ 3. **Differentiate with respect to time** Given water is poured at the constant rate $$\frac{dV}{dt} = 5 \text{ m}^3/\text{min}$$ Differentiate: $$\frac{dV}{dt} = \frac{\pi}{12} \cdot 3h^2 \frac{dh}{dt} = \frac{\pi}{4} h^2 \frac{dh}{dt}$$ Thus, $$5 = \frac{\pi}{4} h^2 \frac{dh}{dt}$$ 4. **At the instant when depth is 10 m** Put $h=10$: $$5 = \frac{\pi}{4}(10)^2 \frac{dh}{dt} = \frac{\pi}{4} \cdot 100 \frac{dh}{dt} = 25\pi \frac{dh}{dt}$$ Hence, $$\frac{dh}{dt} = \frac{5}{25\pi} = \frac{1}{5\pi}$$ 5. **Match with the options** $$\frac{dh}{dt} = \frac{1}{5\pi} \text{ m/min}$$ So the correct option is **B**.
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