VITEEE 2018 :: Physics :: General questions

• Physics - General questions

A ray PQ incident on the refracting face BA is refracted in the prism BAC as shown in the figure and emerges from the other refracting face AC as RS such that AQ = AR. If the angle of prism A = 60º and the refractive index of the material of prism is √3 then the angle of deviation of the ray is

Answer:

Explanation:

Given AQ = AR and $\angle A = 60^{0}$

$\therefore \angle AQR =\angle ARQ =  60^{0}$

$\therefore r_{1} = r_{2} = 30^{o}$

Applying Snell's law on face AB

$\sin i_{1} = \mu\sin r_{1}$

$\sin i_{1} = \sqrt{3}\sin 30^{o} = \sqrt{3}\times\frac{1}{2}$

= $\frac{\sqrt{3}}{2}$

$\therefore i_{1} = 60^{o} = i_{2}$

In a prism, deviation

$\delta =i_{1} + i_{2}-A$ = 60º + 60º - 60º = 60º

A large number of liquid drops each of radius r coalesce to form a single drop of radius R. The energy released in the process is converted into kinetic energy of the big drop so formed. The speed of the big drop is (given, the surface tension of liquid T, density ρ )

Answer:

Explanation:

When drops combine to form a single drop of radius R

Then energy released,

$E = 4\pi TR^{3}\left[\frac{1}{r}-\frac{1}{R}\right]$

If this energy is converted into kinetic energy then

$\frac{1}{2}mv^{2} = 4\pi TR^{3}\left[\frac{1}{r}-\frac{1}{R}\right]$

$\frac{1}{2}\left[\frac{4}{3}\pi R^{3}\rho\right] v^{2} = 4\pi TR^{3}\left[\frac{1}{r}-\frac{1}{R}\right]$

$v = \sqrt{\frac{6T}{\rho}\left[\frac{1}{r}-\frac{1}{R}\right]}$

The given electrical network is equivalent to :

Answer:

Explanation:

The ratio of radii of the first three Bohr orbits is

Answer:

Explanation:

Atomic number Z is equal to 1.

The radius of n^th orbit r_n=0.529n²

The first three orbits values are 1,2,3

Ratio of radius of first three orbit  r₁: r₂: r₃ = $n_{1}^{2} : n_{2}^{2} : n_{3}^{2}$

$\therefore$ 1²:2²:3²

$\therefore$ 1:4:9

An alternating voltage of 220 V, 50 Hz frequency is applied across a capacitor of capacitance 2 μF. The impedance of the circuit is

Answer:

Explanation:

Impedance of a capacitor is X_C = 1/ωC

X_C = $\frac{1}{2\pi fC}$ = $\frac{1}{2\pi\times50\times2\times10^{-6}}$ = $\frac{5000}{\pi}$

• Physics - General questions