

Step 1: Relation in Prism
From the prism relation:
$$ r_1 + c = A $$
Rearranging for \( r_1 \):
$$ r_1 = 90^\circ - c \quad \text{...(1)} $$
Step 2: Expression for \( \cos c \)
We know:
$$ \sin c = \frac{1}{\mu} $$
Using trigonometric identity:
$$ \cos c = \frac{\sqrt{\mu^2 - 1}}{\mu} $$
Step 3: Apply Snell's Law on Incidence Surface
Using Snell’s law at the first surface:
$$ \sin 30^\circ = \mu \sin (r_1) $$
Substituting \( r_1 = 90^\circ - c \):
$$ \frac{1}{2} = \mu \sin (90^\circ - c) $$
Since \( \sin (90^\circ - c) = \cos c \), we get:
$$ \frac{1}{2} = \mu \times \frac{\sqrt{\mu^2 - 1}}{\mu} $$
Simplifying:
$$ \frac{1}{2} = \frac{\sqrt{\mu^2 - 1}}{1} $$
Step 4: Solve for \( \mu \)
Squaring both sides:
$$ \frac{1}{4} = \mu^2 - 1 $$
Rearranging:
$$ \mu^2 = \frac{5}{4} $$
Taking square root:
$$ \mu = \frac{\sqrt{5}}{2} $$
Conclusion
The refractive index \( \mu \) of the prism is \(\frac{\sqrt{5}}{2}\).
A sphere of radius R is cut from a larger solid sphere of radius 2R as shown in the figure. The ratio of the moment of inertia of the smaller sphere to that of the rest part of the sphere about the Y-axis is : 
AB is a part of an electrical circuit (see figure). The potential difference \(V_A - V_B\), at the instant when current \(i = 2\) A and is increasing at a rate of 1 amp/second is: