On a hypotenuse of a right prism (30^∘-60^∘-90^∘) of refractive index 1.50, a drop of liquid is placed as shown in figure. Light is allowed to fall normally on the short face of the prism. In order that the ray of light may get totally reflected, the maximum value of refractive index of the liquid is 
Step 1: Light enters the prism normally, so it travels undeviated inside.
Step 2: Angle of incidence at the hypotenuse: i = 60^∘
Step 3: For total internal reflection at prism–liquid interface: sin c = fracnlᵢqᵤᵢdnprism and i ≥ c
Step 4: Maximum nlᵢqᵤᵢd occurs when i=c: nlᵢqᵤᵢd = nprism sin i = 1.5 × sin 60^∘ nlᵢqᵤᵢd = 1.5 × frac√(3)2 ≈ 1.47


Inductance of a coil with \(10^4\) turns is \(10\,\text{mH}\) and it is connected to a DC source of \(10\,\text{V}\) with internal resistance \(10\,\Omega\). The energy density in the inductor when the current reaches \( \left(\frac{1}{e}\right) \) of its maximum value is \[ \alpha \pi \times \frac{1}{e^2}\ \text{J m}^{-3}. \] The value of \( \alpha \) is _________.
\[ (\mu_0 = 4\pi \times 10^{-7}\ \text{TmA}^{-1}) \]