A bulb is placed at a depth of$2\sqrt{7 }m$ in water and floating opaque disc is placed over the bulb so that the bulb is not visible from the surface. The minimum diameter of the disc is

given that

A bulb is placed at the bottom of water tank of depth $2\sqrt{7}m$

A floating opaque disc of radius r is kept on the surface to make the bulb not visible

From $∆BOQ$

$\angle OBQ=C$  critical angle

$OB=2\sqrt{7} OQ=r$

We know

${}_{air}\mu _{water}=\frac{{\mu }_{water}}{{\mu }_{air}}=\frac{1}{\sin C}$

${\mu }_w=4/3$

$\Rightarrow \frac{4}{3}=\frac{1}{\sin C}$

$\sin C=\frac{3}{4}$

We can have from $△\mathrm{BOQ}$

$\tan C=\frac{OQ}{OB}$

$\frac{3}{\sqrt{7}}=\frac{r}{2\sqrt{7}}$

$\therefore \mathrm{r}=6 \mathrm{m}$

Diameter of disc $d=2r=12 m$