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Water is being filled at the rate of $\mathrm{1cm^{3}/sec}$ in a right circular conical vessel ( vertex downwards ) of height $\mathrm{35cm}$ and diameter $\mathrm{14cm}$ . Whe the height of the water level is $\mathrm{10cm}$ , the rate ( in $\mathrm{cm^{2}/sec}$ ) at which the wet coical surface area of the vessel increases is
Option: 1
$\mathrm {5}$

Option: 2
$\mathrm{\frac{\sqrt{21}}{5}}$

Option: 3
$\mathrm{\frac{\sqrt{26}}{5}}$

Option: 4
$\mathrm{\frac{\sqrt{26}}{10}}$

Answers (1)

best_answer

Let the volume of the cone be $\mathrm{V \: cm^{3}}$

Given $\mathrm{\frac{dV}{dt}=1cm^{3}/sec,h=35cm,r= 7cm}\\$

i.e $\mathrm{\frac{h}{r}=5}\\$

We know for a cone $\mathrm{l^{2}=r^{2}+h}\\$

Lateral Surface area,

$\mathrm{S=\pi r\sqrt{r^{2}+h^{2}}}\\$

$\mathrm{S=\pi\frac{h}{5}\sqrt{\frac{h^{2}}{25}+h^{2}}}=\mathrm{\pi\frac{\sqrt{26}}{25}h^{2}}\\$

$\mathrm{V=\frac{1}{3}\pi r^{2}h\frac{1}{3}\pi \left ( \frac{h}{5} \right )^{2}h =\frac{\pi}{75}h^{3}}\\$

$\Rightarrow\mathrm{\frac{dV}{dt}=\frac{\pi}{25}h^{2}\frac{dh}{dt}}\Rightarrow\mathrm{\frac{\pi}{25}h^{2}\frac{dh}{dt}=1}\\$

$\Rightarrow\mathrm{\frac{dh}{dt}=\frac{25}{\pi h^{2}}}\\$ .........(1)

$\mathrm{\frac{ds}{dt}=\frac{\pi\sqrt{26}}{25}\times 2h\frac{dh}{dt}=\frac{2\sqrt{26}}{h}}\\$ ..........from eqn(1)

$\mathrm{\left ( \frac{ds}{dt} \right )_{h=10}=\frac{\sqrt{26}}{5}}$

Hence the correct answer is option 3

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