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← JEE Main 2022 · 25 Jun · Shift 2
MathematicsSingle correct+4 / -1Question 11 of 90

JEE Main 2022 · 25 Jun · Shift 2, Mathematics question 11

Water is being filled at the rate of 1 cm3^3/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 cm2^2/sec) at which the wet conical surface area of the vessel increases is

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Answer and worked solution

Official answer: C

Shape. r=h5r=\frac h5 (radius 7 at height 35), so V=πh375V=\frac{\pi h^3}{75} and dVdt=πh225dhdt=1\dfrac{dV}{dt}=\dfrac{\pi h^2}{25}\dfrac{dh}{dt}=1: at h=10h=10, dhdt=14π\dfrac{dh}{dt}=\dfrac1{4\pi}.

Wet area. S=πrl=π⋅h5⋅h265=π2625h2S=\pi rl=\pi\cdot\frac h5\cdot\frac{h\sqrt{26}}5=\frac{\pi\sqrt{26}}{25}h^2.

Rate. dSdt=2π2625(10)⋅14π=265\dfrac{dS}{dt}=\dfrac{2\pi\sqrt{26}}{25}(10)\cdot\dfrac1{4\pi}=\dfrac{\sqrt{26}}5.

(C)

Topics:Rate of change
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