An observer can see through a small hole on the side of a jar (radius $15\, cm$) at a point at a height of $15\, cm$ from the bottom (see figure). The hole is at a 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 refractive index of the liquid is $N/100$,where $N$ is an integer,the value of $N$ is $.....$

(A) Let the radius of the jar be $R = 15\, cm$. The height of the liquid is $h = 30\, cm$. The hole is at a height of $45\, cm$ from the bottom,so the distance from the liquid surface to the hole is $45 - 30 = 15\, cm$.
When the observer looks at the bottom edge,the light ray travels from the bottom edge to the liquid surface and then refracts towards the hole.
Let $r$ be the angle of refraction in the liquid with respect to the normal. From the geometry,the horizontal distance from the edge to the point where the ray hits the surface is $15\, cm$,and the vertical depth is $30\, cm$. Thus,$\tan r = \frac{15}{30} = \frac{1}{2}$.
This implies $\sin r = \frac{1}{\sqrt{1^2 + 2^2}} = \frac{1}{\sqrt{5}}$.
The angle of incidence $i$ at the liquid-air interface is the angle the ray makes with the normal. Since the horizontal distance from the hole to the point on the surface is $15\, cm$ and the vertical distance is $15\, cm$,$\tan i = \frac{15}{15} = 1$,so $i = 45^{\circ}$.
Applying Snell's Law at the interface: $1 \cdot \sin 45^{\circ} = \mu \cdot \sin r$.
$\frac{1}{\sqrt{2}} = \mu \cdot \frac{1}{\sqrt{5}}$.
$\mu = \sqrt{\frac{5}{2}} = \sqrt{2.5} \approx 1.5811$.
Given $\mu = \frac{N}{100}$,we have $N = 100 \mu = 100 \times 1.5811 = 158.11$.
Since $N$ is an integer,the value is $N = 158$.