JEE MAIN - Mathematics (2022 - 25th June Evening Shift - No. 8)
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
5
$${{\sqrt {21} } \over 5}$$
$${{\sqrt {26} } \over 5}$$
$${{\sqrt {26} } \over {10}}$$
Explanation
$$\because$$ $$V = {1 \over 3}\pi {r^2}h$$ and $${r \over h} = {7 \over {35}} = {1 \over 5}$$
$$ \Rightarrow V = {1 \over {75}}\pi {h^3}$$
_25th_June_Evening_Shift_en_8_1.png)
$${{dV} \over {dt}} = {1 \over {25}}\pi {h^2}{{dh} \over {dt}} = 1$$
$$ \Rightarrow {{dh} \over {dt}} = {{25} \over {\pi {h^2}}}$$
Now, $$S = \pi rl = \pi \left( {{h \over 5}} \right)\sqrt {{h^2} + {{{h^2}} \over {25}}} = {\pi \over {25}}\sqrt {26} {h^2}$$
$$ \Rightarrow {{dS} \over {dt}} = {{2\sqrt {26} \pi h} \over {25}}\,.\,{{dh} \over {dt}} = {{2\sqrt {26} } \over h}$$
$$ \Rightarrow {{dS} \over {d{t_{(h = 10)}}}} = {{\sqrt {26} } \over 5}$$
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