# An observer can see through a small hole on the side of the jar ( radius $15\; cm$) at a point of height of $45\; cm$. When the jar is filled with a liquid up to a height of $30\; cm$ the same observer can see the edge at the bottom of the jar. If the R.I. of the liquid is $\frac{N}{100}$, where $N$  is an integer the value of $N$ is  Option: 1 5 Option: 2 46 Option: 3 158 Option: 4 96

$\\ tan \theta_1=\frac{15}{15}=1\\ \Rightarrow \theta_1=45^o\\ tan \theta_2=\frac{15}{30}\Rightarrow tan \theta_2=\frac{1}{2}\Rightarrow \theta_2=\tan^{-1}\frac{1}{2}\\ Sin \theta_2=\frac{1}{\sqrt5}\\ \frac{Sin \theta_1}{Sin \theta_2}=N\Rightarrow \frac{\frac{1}{\sqrt2}}{\frac{1}{\sqrt5}}=N\\ \Rightarrow N=\frac{\sqrt5}{\sqrt2}=1.58\\ N=\frac{158}{100}\\ So, N=158$

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