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Application of Derivatives question

2022 · 25 Jun · Shift 2 · Q30
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  5. /2022 · 25 Jun · Shift 2 · Q30

Application of Derivatives question

2022 · 25 Jun · Shift 2 · Q30

JEE MainMathematicsApplication of DerivativesMCQ+4 / −1
Water is being filled at the rate of 1 cm3 / sec in a right circular conical vessel (vertex downwards) of height 35 cm and diameter 14 cm. When the height of the water level is 10 cm, the rate (in cm2 / sec) at which the wet conical surface area of the vessel increases is
  1. A
    5
  2. B
    215{{\sqrt {21} } \over 5}521​​
  3. C
    265{{\sqrt {26} } \over 5}526​​
  4. D
    2610{{\sqrt {26} } \over {10}}1026​​
View written solutionFree

Correct answer: C

  1. Relate radius and height using similarity

For the conical vessel:

  • Height H=35H = 35H=35 cm
  • Radius R=7R = 7R=7 cm

At any instant, if water has height hhh and radius rrr, then by similarity, rh=RH=735=15\frac{r}{h} = \frac{R}{H} = \frac{7}{35} = \frac{1}{5}hr​=HR​=357​=51​ so r=h5.r = \frac{h}{5}.r=5h​.

  1. Volume of water in the cone

Volume of water: V=13πr2h.V = \frac{1}{3}\pi r^2 h.V=31​πr2h. Substitute r=h5r = \frac{h}{5}r=5h​:

= \frac{1}{3}\pi \cdot \frac{h^2}{25} \cdot h = \frac{\pi h^3}{75}.$$ Differentiate with respect to time $t$: $$\frac{dV}{dt} = \frac{\pi}{75} \cdot 3h^2 \frac{dh}{dt} = \frac{\pi h^2}{25}\frac{dh}{dt}.$$ Given $\frac{dV}{dt} = 1$ cm$^3$/sec. At $h=10$ cm, $$1 = \frac{\pi (10)^2}{25}\frac{dh}{dt} = 4\pi \frac{dh}{dt}.$$ Hence, $$\frac{dh}{dt} = \frac{1}{4\pi}. $$ 3. **Wet conical surface area of water** The wet curved surface area is $$S = \pi r l,$$ where $l$ is the slant height of the water cone. Now, $$l = \sqrt{r^2 + h^2}.$$ Using $r=\frac{h}{5}$, $$l = \sqrt{\left(\frac{h}{5}\right)^2 + h^2} = h\sqrt{\frac{1}{25}+1} = h\sqrt{\frac{26}{25}} = \frac{h\sqrt{26}}{5}.$$ Therefore, $$S = \pi \cdot \frac{h}{5} \cdot \frac{h\sqrt{26}}{5} = \frac{\pi \sqrt{26}}{25}h^2.$$ Differentiate: $$\frac{dS}{dt} = \frac{\pi \sqrt{26}}{25} \cdot 2h \frac{dh}{dt}.$$ At $h=10$, $$\frac{dS}{dt} = \frac{2\pi\sqrt{26}}{25} \cdot 10 \cdot \frac{1}{4\pi}.$$ Simplify: $$\frac{dS}{dt} = \frac{20\pi\sqrt{26}}{25} \cdot \frac{1}{4\pi} = \frac{4\sqrt{26}}{5} \cdot \frac{1}{4} = \frac{\sqrt{26}}{5}.$$ 4. **Match with the options** Thus, the required rate is $$\boxed{\frac{\sqrt{26}}{5}}$$ which corresponds to **Option C**.
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