Water is being filled at the rate of 1 cm^3 sec^-1 in a right circular conical vessel (vertex downwards) of height 35 cm and diameter 14 cm. When the…

Let the volume of the cone be $Vc.c.$Vc.c.

Given $\frac{dv}{dt}=1\mathrm{cc}/\mathrm{sec},h=35\mathrm{cm},r=7\mathrm{cm}$dtdv​=1cc/sec,h=35cm,r=7cm

i.e. $\frac{h}{r}=5$rh​=5

We know for a cone ${l}^{2}={r}^{2}+{h}^{2}$l2=r2+h2

Lateral surface area, $S=\pi r\sqrt{{r}^{2}+{h}^{2}}$S=πrr2+h2​

$$S=\pi \frac{h}{5}\sqrt{\frac{{h}^{2}}{25}+{h}^{2}}=\pi \frac{\sqrt{26}}{25}{h}^{2}$$S=π5h​25h2​+h2​=π2526​​h2

Also $V=\frac{1}{3}\pi {r}^{2}h=\frac{1}{3}\pi {(\frac{h}{5})}^{2}h=\frac{\pi }{75}{h}^{3}$V=31​πr2h=31​π(5h​)2h=75π​h3

$$\Rightarrow \frac{dV}{dt}=\frac{\pi }{25}{h}^{2}\frac{dh}{dt}\Rightarrow \frac{\pi }{25}{h}^{2}\frac{dh}{dt}=1$$⇒dtdV​=25π​h2dtdh​⇒25π​h2dtdh​=1

$$\Rightarrow \frac{dh}{dt}=\frac{25}{\pi {h}^{2}}\ldots (i)$$⇒dtdh​=πh225​…(i)

Now $\frac{dS}{dt}=\frac{\pi \sqrt{26}}{25}\times 2h\frac{dh}{dt}=\frac{2\sqrt{26}}{h}$dtdS​=25π26​​×2hdtdh​=h226​​ (from $(i)$(i))

$${(\frac{dS}{dt})}_{h=10}=\frac{\sqrt{26}}{5}$$(dtdS​)h=10​=526​​