VITEEE 2018 :: Physics :: General questions

• Physics - General questions

A running man has half the kinetic energy of that of a boy of half of his mass. The man speeds up by 1m/s so as to have same K.E. as that of the boy. The original speed of the man will be

Answer:

Explanation:

Let m = mass of boy, M = mass of man v = velocity of boy, V = velocity of man,

$\frac{1}{2}MV^{2} = \frac{1}{2}\left[\frac{1}{2}mv^{2}\right]$ ...(i)

$\frac{1}{2}M(V+1)^{2} = 1\left[\frac{1}{2}mv^{2}\right]$ ....(ii)

Putting m = M/2 and  solving V= $\frac{1}{\sqrt{2}-1}$

A metal disc of a radius of 100 cm is rotated at a constant angular speed of 60 rad/s in a plane at right angles to an external field of magnetic induction 0.05 Wb/m². The emf induced between the centre and a point on the rim will be

Answer:

Explanation:

Induced emf produced between the centre and a point on the disc is given by

e = $\frac{1}{2}\omega BR^{2}$

Putting the values

ω = 60 rad/s, $B=0.05 Wb/m^{2}$

and R = 100cm = 1m

e = 1/2 × 60 × 0.05 × (1)² = 1.5 V

Consider the junction diode as ideal. The value of current fiowing through AB is

Answer:

Explanation:

Since diode is in forward bias, so the value of current flowing through AB,

$i =\frac{ΔV}{R} = \frac{4-(-6)}{1\times10^{3}}=\frac{10}{10^{3}}$

$10^{-2}$ A

A body cools from 50.0ºC to 49.9ºC in 5s. How long will it take to cool from 40.0ºC to 39.9ºC? Assume the temperature of surroundings to be 30.0ºC and Newton's law of cooling to be valid.

Answer:

Explanation:

$\frac{50-49.9}{5} = K\left(\frac{50+49.9}{2}-30\right)$ ....(i)

$\frac{40-39.9}{t} = K\left(\frac{40+39.9}{2}-30\right)$ .....(ii)

From equations (i) and (ii), we get t ≈ 10 s

In the given circuit the reading of voltmeter V₁ and V₂ are 300 volt each. The reading of the voltmeter V₃ , and ammeter A are respectively

Answer:

Explanation:

As V_L = V_C = 300 V, resonance will take place.

$\therefore V_{R}= 220 V;I = \frac{220}{100}=2.2A$

Reading of V₃ = 220 V and reading of A =2.2 A

• Physics - General questions