Question

Water is dripping out from a conical funnel of semi-vertical angle π/4 at the uniform rate of 2 square cms /sec in the surface area, through a tiny hole at the vertex of the bottom. When the slant height of cone is 4 cm, then the rate of decrease of the slant height of water is

• A. $\frac{\sqrt{2}}{4\pi }\mathrm{cm}/\sec .$
• B. $\frac{\sqrt{2}}{2\pi }\mathrm{cm}/\sec .$
• C. $\frac{3\sqrt{2}}{4\pi }\mathrm{cm}/\sec .$
• D. $\frac{1}{4\sqrt{2}\pi }\mathrm{cm}/\sec .$

Answer 1

The correct option is A $\frac{\sqrt{2}}{4\pi }\mathrm{cm}/\sec .$

If S be the surface area of the connical funnel, then
$\frac{\mathrm{dS}}{\mathrm{dt}}=2{\mathrm{cm}}^2/\sec .S=\mathrm{\pi rl}=\pi \left(l\sin \frac{\pi }{4}\right)l=\pi \frac{l^2}{\sqrt{2}}\left(\mathrm{As}\theta =\frac{\pi }{4}.\mathrm{By}\mathrm{Trigonometry},\sin \theta =\frac{r}{l}\Rightarrow r=l\sin \frac{\pi }{4}\right]\therefore \frac{\mathrm{dS}}{\mathrm{dt}}=\frac{\pi }{\sqrt{2}}\times 2l\frac{\mathrm{dl}}{\mathrm{dt}}=\sqrt{2}\mathrm{\pi l}\frac{\mathrm{dl}}{\mathrm{dt}}\mathrm{When}l=4\mathrm{cm},\frac{\mathrm{dl}}{\mathrm{dt}}=\frac{2}{\sqrt{2}\pi \times 4}=\frac{\sqrt{2}}{4\pi }\mathrm{cm}/\sec .$