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Mathematics Question on Surface Areas and Volumes

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 increase, is

A

5

B

215\frac{\sqrt21}{5}

C

265\frac{\sqrt26}{5}

D

2610\frac{\sqrt26}{10}

Answer

5

Explanation

Solution

The correct option is(C): 265\frac{\sqrt26}{5}.

s=πl=π(h5h2+h225=π2526h2)s=\pi{l}=\pi(\frac{h}{5}\sqrt{h^2+\frac{h^2}{25}}=\frac{\pi}{25}\sqrt{26h^2})

⇒dsdt(h=10)=265⇒\frac{ds}{dt}(h=10)=\frac{\sqrt26}{5}