Here are two questions on self inductance:

(1) The inductance between A and D is
(a) 3.66 H (b) 9 H (c) 0.66 H (d) 1 H

This question appeared in AIEEE 2002 question paper. If you examine the arrangement carefully, you will realise that the three iductors, each having value 3 henry are connected in parallel, between the points A and D.
Inductors connected in parallel will produce an effective value given by the reciprocal relation,
1/L = 1/L₁ + 1/L₂ + 1/L₃.
Since all the three inductors have the same value 3H, the effective value of the inductance is 3H/3 = 1 H
[If the three components were capacitors of value 3 μF each, the effective value would be 9 μF because capacitances in parallel get added as C = C₁ + C₂ + C₃ + etc. It will be convenient to remember that Resistors, reactances ( inductive as well as capacitive) and impedances get added when connected in series ( Z = Z₁ + Z₂ + Z₃ + etc.) while their effective value is given by the reciprocal relation when they are connected in parallel (1/Z = 1/Z₁ + 1/Z₂ + 1/Z₃ +…etc.)].

(2) An air cored coil of 600 turns has a self inductance of 90 mH. If the number of turns is reduced to 200 without changing the area of the coil, the self inductance will be
(a) 30 mH (b) 270 mH (c) 180 mH (d) 90 mH (e) 10 mH

The answer to this question is 10 mH since the self inductance of a coil is directly proportional to the square of the numbr of turns. When the number of turns becomes one third, the inductance becomes one nineth.
[The self inductance is equal to the total magnetic flux linked with the coil (because of its own magnetic field) when unit current flows in it. But this magnetic flux is directly proportional to the magnetic field produced by the coil and the total number of turns in the coil. Since the magnetic field produced by the coil is directly proportional to the total number of turns in the coil, the total magnetic flux linked (and hence the self inductance) is directly proportional to the square of the number of turns in the coil]

Two Questions Involving Gravitational Potential Energy & Escape Velocity

The following question will be simple if you have a clear idea of the potential energy in a gravitational field:
Two bodies of masses m₁ and m₂ are initially at rest and and at infinite distance apart. When a very feeble momentary push is given to one of them, they move towards each other because of the gravitational force between them. When the separation between them is ‘r’, their relative velocity of approach is (if G is the gravitational constant)
(a) [2Gr(m₁+m₂)]^½ (b) [2Gr(m₁–m₂)]^½ (c) [1/2Gr(m₁+m₂)]^½

(d) [2G (m₁+m₂)/r]^½ (e) [2G(m₁ – m₂)/r]^½
Imagine that initially the masses are at a distance ‘r’ apart. The gravitational potential energy of the system then is –Gm₁m₂/r. If the separation between the masses is to be increased to infinity, kinetic energy equal to Gm₁m₂/r is to be supplied to the system. (At infinite separation, kinetic energy as well as potential energy is zero). When the masses are allowed to approach, they lose potential energy (which becomes negative) and gain kinetic energy and at a separation ‘r’, the kinetic energy is +Gm₁m₂/r.
If v₁ and v₂ are the velocities of the masses, we have
½ m₁v₁ ² + ½ m₂v₂ ² = Gm₁m₂/r.
Since the momenta are equal (in magnitude), m₁v₁ = m₂v₂ so that v₂ = m₁v₁/m₂.
Substituting this in the energy equation above, v₁ = [2G/(m₁ + m₂)r]^½ ×m₂.
Similarly, v₂ = [2G/(m₁ + m₂)r] ^½ ×m1.
Since the masses are moving in opposite directions, their relative velocity is
v₁ + v₂ =[2G/(m₁ + m₂)r]^½ × (m₂ + m₁) = [2G(m₁+m₂)/r]^½
[The solution to this problem has been edited. Thank you Dr. Ravikrisnan, for pointing out the mistake in the solution which I had posted yesterday].

Now, consider the following question, which will be simple if you have a clear idea of the velocity of escape in a gravitational field:
The surface value of acceleration due to gravity on a planet of radius ‘R’ is ‘g’. If a body of mass ‘m’ at infinite distance from the planet is given a feeble momentary push so that it moves towards the planet, what will be its kinetic energy when it strikes the surface of the planet? (Assume that the gravitational field of the planet only is significant and the effect of atmosphere on the motion of the body is negligible).
(a) Infinite (b) mgR² (c) ) mgR³ (d) 2mgR (e) mgR
The body will strike the surface of the planet with a velocity equal to the escape velocity which is √(2gR). The kinetic energy with which it will strike the surface is therefore ½ m[√(2gR)]² = mgR.

Power in A.C. Circuits

The power(P) dissipated in an alternating current circuit is given by
P = V_rms I_rms cosΦ where cosΦ is the power factor. [You should remember that Φ is the phase difference between current and voltage].
Power in an A.C. circuit is also given by P = (I_rms)²R. Note that power is dissipated in resistance only.
Another form of the power relation is P = (I_rms)²Z cosΦ, which follows from cosΦ = R/Z

Some of you may be wondering which relation is to be used for calculating power. Well, you may use any one of them which is convenient to you in the context of the problem. Now, consider the following MCQ:
A voltage v = V₀ cosωt is applied in series with an inductance L and a resistance R. The average power dissipated per cycle in the circuit is
(a) V₀² /R (b) V₀²/2R (c) V₀²/(R+L) (d) V₀²R/[(R²+ L²ω ²)] (e) V₀²R/[2(R²+ L²ω ²)]
We shall use the relation, P = (I_rms)²R. Since I_rms = V_rms/Z = (V₀/√2)/[√(R²+ L²ω ²)], we obtain P = V₀²R/[2(R² + L²ω ²)].
You could have used the other relations (for power) also to obtain the result.
Consider the folowing MCQ which appeared in C.E.T. J&K 2000 question paper:
In an A.C. circuit V and I are given by
V = 100 sin (1000t) volt
I = 1000 sin (1000t + π/3 ) mA.
The power dissipated in the circuit is
(a) 10⁴ W (b) 25 W (c) 10 W (d) 250 W
Here current, voltage and the phase difference (between current and voltage) are given so that it is convenient to use the relation P = V_rms I_rms cosΦ. [Note that the current has a phase lead of π/3 and is given in mA].
Therefore, P = (100/√2) × (1000×10 –3/√2) ×cos (π/3) = 25 W

Magnetic Fields Produced by Current Carrying Conductors

Let us consider the following MCQ which appeared in AIEEE 2004 question paper:
The magnetic field due to a current carrying circular loop of radius 3 cm at a point on the axis at a distance of 4 cm from the centre is 54 μT. What will be its value at the centro of the loop?
(a) 250 μT (b) 150 μT (c) 125 μT (d) 75 μT
The magnetic flux density(B) at a point on the axis at distance ‘x’ from the centre for a single turn loop is given by
B = μ₀ r²I / [2(r²+ x²)^3/2] where μ₀ is the permeabiloty of free space, ‘r’ is the radius of the loop and ‘I’ is the current.
The magnetic field(B') at the centre of the loop is given by
B' = μ₀ I/2r
Dividing, B/B' = r³/(r²+ x²)^3/2 from which B' = B[(r²+ x²)^3/2] /r³.
You may substitute distances in cm itself since their units will get cancelled in the ratio. On substituting B in μT itself, you will get the answer in μT.
Therefore, B' = 54×[(9+16)^3/2]/27 = 54×125/27 = 250 μT.
The following multiple choice question appeared in Kerala Medical Entrance 2003 test paper:

A wire of certain length carries a steady current. It is first bent to form a circular coil of one turn. The same wire is next bent to form a circular coil of three turns. The ratio of magnetic induction at the centre of the coil in the two cases is
(a) 9:1 (b) 1:9 (c) 1:3 (d) 3:1 (e) 1:1

Since the magnetic induction (flux density) at the centre of a circular coil is given by
B = μ₀ nI/(2r) where μ₀ is the permeability of free space, ‘n’ is the number of turns in the coil, I is the current and ‘r’ is the radius of the coil, we have, B α n/r for a given current.

The ratio of magnetic inductions at the centre is therefore B/B' = (n/n') ×(r'/r) = (1/3) ×(1/3) = 1/9, since 2πr = 3×2πr' so that r'/r = 1/3. The correct option therefore is (b).
Here is a simple question (but be careful):

An infinitely long aluminium pipe of radius ‘r’ carries a current. The magnetic flux density out side the pipe at a distance 3r/2 from the axis is 0.04 T. The magnetic flux density inside the pipe at a distance r/2 from the axis will be
(a) 0.01 T (b) 0.02 T (c) 0.04 T (d) zero (e) infinite
Don’t be distracted by the value of the field out side. You should remember that the magnetic flux density at any point inside a long current carrying pipe is zero. The correct option therefore is (d).

Questions involving Force of Buoyancy

A wooden cylinder is floating (with its axis horizontal) at the interface of water and oil, with a quarter of its volume in water. If the density of oil is 800 kgm^–3 , the density of wood is
(a) 820 kgm^–3 (b) 830 kgm^–3 (c) 840 kgm^–3 (d) 850 kgm^–3 (e) 860 kgm^–3
Even if the cylinder is floating upright, a quarter of its volume will be in water. The cylinder is in equilibrium because the real weight of the cylinder acting vertically downwards is balanced by the forces of buoyancy exerted by water and oil, acting vertically upwards. Therefore we can write
vρg = v₁ ρ₁g + v₂ρ₂g where ‘v’ and ‘ρ’ are the volume and the density of the wooden cylinder, ρ₂ and ρ₂ are the densities of water and oil and v₁ and v₂ are the volumes of the cylinder within water and oil respectively. [Note that the force of buoyancy is the weight of the displaced fluid]. Thus, we have
vρ = (v/4) ×1000 + (3v/4) ×800 from which ρ = 850 kgm^–3
The same question could have been asked as follows:
A wooden cylinder of height 6 cm and radius 2 cm is floating upright at the interface of water and oil, with 1.5 cm of its length in water. If the density of oil is 800 kgm^–3, the density of wood is
(a) 820 kgm^–3 (b) 830 kgm^–3 (c) 840 kgm^–3 (d) 850 kgm^–3 (e) 860 kgm^–3
Since the cylinder is upright and a quarter of its height is within water, a quarter of its volume is within water as in the above case and you will obtain the same answer.
Now consider the following MCQ:
A small piece of wood of density 900 kgm^–3 is released from a depth of 20 cm inside water. If the viscous forces are ignored, the height above the surface of water up to which the ball will jump is
(a) 7.5 cm (b)12.5 cm (c) 16.5 cm (d)20.5 cm (e) 25 cm
The apparent weight of the wooden piece inside water is v(ρ-σ)g where v is its volume, ρ is its density and σ is the density of water. Since ρ is less than σ the apparent weight of the ball is negative which means it is directed upwards. It is this force which drives the wooden piece upwards. The magnitude of this force is v(σ- ρ)g.
Since the mass of the wooden piece is vρ, its upward acceleration inside water is, a = v(σ- ρ)g/vρ = (σ- ρ)g/ρ. The velocity ‘v’ of the ball on reaching the water surface is given by the usual equation of uniformly accelerated motion, v²=u²+2as so that
v² = 0+2[(σ- ρ)g/ρ]s, where ‘s’ is the distance traveled within water (0.2 m ).
Let us use the energy equation for finding the height ‘h’ up to which the ball will jump outside water:
½ mv² = mgh, from which h=v²/2g=[(σ-ρ)/ρ]s = [(1000-900)/900]×0.2 =0.022 m = 2.2 cm [Option (e)].