Water is being filled at the rate of 1 , cm^3 | vedclass

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(C) Let the radius of the cone be $R = 7 \, cm$ and height be $H = 35 \, cm$. By similar triangles,$\frac{r}{h} = \frac{R}{H} = \frac{7}{35} = \frac{1

Vedclass · Accepted Answer

$\frac{\sqrt{26}}{5}$

Vedclass · Answer

(C) Let the radius of the cone be $R = 7 \, cm$ and height be $H = 35 \, cm$. By similar triangles,$\frac{r}{h} = \frac{R}{H} = \frac{7}{35} = \frac{1}{5}$,so $h = 5r$.The volume of water is $V = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi r^2 (5r) = \frac{5}{3} \pi r^3$.Given $\frac{dV}{dt} = 1 \, cm^3/sec$,we have $\frac{d}{dt} (\frac{5}{3} \pi r^3) = 5 \pi r^2 \frac{dr}{dt} = 1$,so $\frac{dr}{dt} = \frac{1}{5 \pi r^2}$.The wet conical surface area is $S = \pi r l = \pi r \sqrt{r^2 + h^2} = \pi r \sqrt{r^2 + (5r)^2} = \pi r \sqrt{26r^2} = \sqrt{26} \pi r^2$.Differentiating with respect to $t$,$\frac{dS}{dt} = 2 \sqrt{26} \pi r \frac{dr}{dt}$.Substituting $\frac{dr}{dt} = \frac{1}{5 \pi r^2}$,we get $\frac{dS}{dt} = 2 \sqrt{26} \pi r (\frac{1}{5 \pi r^2}) = \frac{2 \sqrt{26}}{5r}$.When $h = 10 \, cm$,$r = \frac{h}{5} = \frac{10}{5} = 2 \, cm$.Thus,$\frac{dS}{dt} = \frac{2 \sqrt{26}}{5(2)} = \frac{\sqrt{26}}{5} \, cm^2/sec$.

Water is being filled at the rate of $1 \, cm^3/sec$ 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 $cm^2/sec$) at which the wet conical surface area of the vessel increases is

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