Question

A cylindrical vessel, whose diameter and height both are equal to 30 cm, is placed on a horizontal surface and a small particle P is placed in it at a distance of 5.0 cm from the centre. An eye is placed at a position such that the edge of the bottom is just visible (see Figure). The particle P is in the plane of drawing. Up to what minimum height should water be poured in the vessel to make the particle P visible ?

Answer 1

$tan{\theta }_1=1=\frac{30}{30}$

${\theta }_1={45}^{\circ }$

In $\Delta Tus$

$tan{45}^{\circ }=\frac{30-4}{TS}$

TS = 30 - 4

TS = VR

$P_v=20-vR=h-10$

$sin{\theta }_2=\frac{h-10}{\sqrt{(h-10{)}^2+h^2}}$

$sin{\theta }_2\times {\mu }_2={\mu }_{1}sin{\theta }_1$

or, $\frac{h-10}{\sqrt{(h-10{)}^2+h^2}}\times \frac{4}{3}=1\times \frac{1}{v^2}$

or, $\left(\right(u{v}_2\left)\right(n-10){)}^2={\left(\sqrt{(h-10{)}^2+h^2}3\right)}^2$

or, $32(h-10{)}^2=9(h-10{)}^2+9h^2$

or, $23(h-10{)}^2=9h^2$

or, ${\left(\frac{h-10}{h}\right)}^2=\frac{9}{23}$

or, $\frac{h-10}{h}=\frac{3}{4.8}$

or, $1-\frac{10}{h}=\frac{3}{4.8}$

or, $\frac{10}{h}=\frac{1.8}{4.8}$

$\therefore h=\frac{48\times 10}{1.8}=26.6\text{cm}$