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QJEEM22Ma5539: JEE Main 2022 Mathematics - Class 12 JEEM22Ma5539 Solution: Water is being filled at the rate of 1 cm 3 / sec in a right…

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

Multiple Choice Options

  • A 5
  • B 215{{\sqrt {21} } \over 5}
  • C 265{{\sqrt {26} } \over 5}
  • D 2610{{\sqrt {26} } \over {10}}

Step-by-Step Verified Solution

∵\because V=13πr2hV = {1 \over 3}\pi {r^2}h and rh=735=15{r \over h} = {7 \over {35}} = {1 \over 5}

⇒V=175πh3\Rightarrow V = {1 \over {75}}\pi {h^3}

JEE Main 2022 (Online) 25th June Evening Shift Mathematics - Application of Derivatives Question 74 English Explanation

dVdt=125πh2dhdt=1{{dV} \over {dt}} = {1 \over {25}}\pi {h^2}{{dh} \over {dt}} = 1

⇒dhdt=25πh2\Rightarrow {{dh} \over {dt}} = {{25} \over {\pi {h^2}}}

Now, S=πrl=π(h5)h2+h225=π2526h2S = \pi rl = \pi \left( {{h \over 5}} \right)\sqrt {{h^2} + {{{h^2}} \over {25}}} = {\pi \over {25}}\sqrt {26} {h^2}

⇒dSdt=226πh25 . dhdt=226h\Rightarrow {{dS} \over {dt}} = {{2\sqrt {26} \pi h} \over {25}}\,.\,{{dh} \over {dt}} = {{2\sqrt {26} } \over h}

⇒dSdt(h=10)=265\Rightarrow {{dS} \over {d{t_{(h = 10)}}}} = {{\sqrt {26} } \over 5}

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