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An observer can see through a pin hole the top end of a thin rod of height h, placed as shown. The beaker height is 3 h and its radius is h. When the beaker is filled with a liquid up to a height 2h ,he can see the lower end of the rod. Then RI of liquid is .

$(a)\;\frac{5}{2} \\ (b)\;\sqrt {\frac{5}{2}} \\ (c)\;\sqrt {\frac{3}{2}} \\ (d)\;\frac{3}{2} $

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$\large\frac{\sin i}{\sin r}=\frac{1}{\mu}$
$\large\frac{\Large\frac{x}{\sqrt{x^2+4h^2}}}{\Large\frac{2h-x}{\sqrt {(2h-x)^2+h^2}}}=\frac{1}{\mu}$----(1)
$\sin r=\large\frac{2h-x}{\sqrt {(2h-x)^2+h^2}}=\frac{x}{h}$---(2)
using (1) and (2)
$\mu =\sqrt {\large\frac{5}{2}}$
Hence b is the correct answer.
answered Feb 19, 2014 by meena.p

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