1. Find the Refractive Index ($\mu$)
The relationship between the critical angle ($\theta_c$) and the refractive index ($\mu$) is given by:
$$\mu = \frac{1}{\sin \theta_c}$$
Given $\theta_c = 30^\circ$:
$$\mu = \frac{1}{\sin 30^\circ} = \frac{1}{0.5} = 2$$
2. Find the Brewster Angle ($\theta_B$)
According to Brewster's Law, the tangent of the Brewster angle (polarizing angle) equals the refractive index of the material:
$$\tan \theta_B = \mu$$
Substitute $\mu = 2$:
$$\tan \theta_B = 2$$
$$\theta_B = \tan^{-1}(2)$$
3. Calculate the Value
Using the inverse tangent function:
$$\theta_B \approx 63.43^\circ$$
4. Round Off
The problem asks to round off to the nearest integer.
$$63.43^\circ \approx 63^\circ$$
Final Answer:
The Brewster angle is 63 degrees.
In order to achieve the static equilibrium of the see-saw about the fulcrum \( P \), shown in the figure, the weight of Box B should be _________ kg, if the weight of Box A is 50 kg.

A particle of mass 1kg, initially at rest, starts sliding down from the top of a frictionless inclined plane of angle \(\frac{𝜋}{6}\)\(\frac{\pi}{6}\) (as schematically shown in the figure). The magnitude of the torque on the particle about the point O after a time 2seconds is ______N-m. (Rounded off to nearest integer) 
(Take g = 10m/s2)


At a particular temperature T, Planck's energy density of black body radiation in terms of frequency is \(\rho_T(\nu) = 8 \times 10^{-18} \text{ J/m}^3 \text{ Hz}^{-1}\) at \(\nu = 3 \times 10^{14}\) Hz. Then Planck's energy density \(\rho_T(\lambda)\) at the corresponding wavelength (\(\lambda\)) has the value \rule{1cm}{0.15mm} \(\times 10^2 \text{ J/m}^4\). (in integer)
[Speed of light \(c = 3 \times 10^8\) m/s]
(Note: The unit for \(\rho_T(\nu)\) in the original problem was given as J/m³, which is dimensionally incorrect for a spectral density. The correct unit J/(m³·Hz) or J·s/m³ is used here for the solution.)