Circular motion of electrons- Angular Momentum and magnetic dipole moment:
If an electron (charge ‘e’, mass ‘m’) is in uniform circular motion with angular velocity ‘ω’ the orbital angular momentum of the electron about a perpendicular axis passing through the centre of the circle is mωr^2 (=mvr). Often you will encounter questions regarding the direction of the angular momentum vector. You should note that the orbital angular momentum of the electron is directed perpendicular to the plane of the orbit.
The magnetic dipole moment of an orbiting electron is eωr^2/2 (= evr/2). [Magnetic dipole moment = IA = (e/T)×πr^2 = (eω/2π)×πr^2 = eωr^2/2=evr/2. Here I is the equivalent current, T is the orbital period and r is the orbital radius].
Now consider the following M.C.Q. which appeared in Kerala Engineering Entrance Test paper of 2006:
Two electrons (each of charge = e and mass = m) are attached one at each end of a light rigid rod of length 2r. The rod is rotated at constant angular speed about a perpendicular axis passing through its centre. The ratio of the angular momentum about the axis of rotation to the magnetic dipole moment of the system is
(a) 2me/3 (b) e^2/2m (c) 2×specific charge of electron (d) 5m/2e (e) 2m/e
The ‘light rigid rod’ in the problem could have been dispensed with.
Angular momentum of two electrons = 2mωr^2. As shown above, the magnetic dipole moment of two orbiting electrons = 2 eωr^2/2. The ratio of angular momentum to the magnetic dipole moment = 2m/e. [Option(e)].
Note that this result is independent of the number of electrons.
You should also note that the ratio of the magnetic moment to the angular momentum of a charged particle in angular motion is called gyromagnetic ratio.

Two Questions from Sound

(1) When two tuning forks are sounded together, 4 beats per second are produced. The frequency of one fork is 512Hz. When the other fork is loaded with a little wax, the beat frequency is 6Hz. The frequency of the other tuning fork is (in Hz)
(a) 516 (b) 514 (c) 512 (d) 508 (e) 502
This is a very simple question and you know that the beat frequency is the difference between the frequencies of the tuning forks. Since 4 beats are produced, the frequency of the other fork is either 516 or 508.
Loading of the other fork will reduce its frequency. So, if its actual frequncy is 516, the beat frequency is reduced. Since the beat frequency is increased, the actual frequency of the other fork must be 508 [option(d)].
Note that if the other fork is filed slightly, its frequency will be increased. If the question is modified this way and the beat frequency after filing is given as 6Hz itself, the answer will be option (a).
(2) The tension in a sonometer wire is increased by 96%. In order to keep the frequency of vibration of the wire unchanged, the length of the wire must be increased by
(a) 38% (b) 40% (c) 42% (d) 44% (e) 48%
The frequency of vibration of the wire is given by n = (1/2L)×√(T/m) where ‘L’ is the length, ‘T’ is the tension and and ‘m’ is the linear density of the wire. When T becomes 1.96T, the length L must become 1.4 L. The increment in L therefore is 40% [option (b)].

Elastic and Inelastic Collisions

As you know, momentum and kinetic energy are conserved in elastic collisions. But kinetic energy is not conserved in inelastic collisions, eventhough momentum is conserved. Consider the following M.C.Q.
A steel sphere of mass 20 gram moving horizontally with a velocity of 2m/s collides elastically with the bob of a long simple pendulum at rest. If the mass of the pendulum bob is 20 gram what is the height up to which the bob is raised?
(a) 10cm (b) 20cm (c) 40cm (d) 50cm (e) 60cm
As the collision is elastic and the masses of the sphere and the bob are the same, the sphere will transfer the entire kinetic energy to the bob. (This can be easily proved by equating the total initial momentum and the total initial kinetic energy to the total final momentum and the total final kinetic energy respectively). After receiving the kinetic energy from the sphere, the pendulum bob rises to a height ‘h’ given by,
½ mv^2 = mgh, from which h = v^2/2g = 4/20 =0.2m = 20cm.[Option(b)]
This simple problem changes notably if the collision is inelastic as in the modified question below:
A small metallic dart of mass 20gram moving horizontally with a velocity of 2m/s strikes the wooden bob of a long simple pendulum at rest and gets stuck to it. If the mass of the pendulum bob is 20 gram what is the approximate height up to which the bob is raised?

(a) 2.5cm (b) 5cm (c) 7.5cm (d) 10cm (12.5cm
This is a case of inelastic collision since the dart sticks to the bob and moves with the bob after the impact (Part of the kinetic energy of the dart will be lost in piercing the bob). You should not equate the initial kinetic energy of the dart to the gravitational potential energy of the bob-dart system. The common velocity of the bob-dart system after the impact is to be found from the law of conservation of momentum and the kinetic energy of the bob-dart combination is to be found and equated to the gravitational potential energy.
Equating the initial momentum to the final momentum,we have
0.02×2 + 0=(0.02+0.02)v, from which the common velocity ‘v’of the bob-dart combination works out to be 1m/s. Equating the K.E.and P.E.,
½ (0.02+0.02)×1^2 = (0.02+0.02)×gh.
Taking ‘g’ to be nearly 10, h = 0.05m = 5cm [option(b)]

Radioactive Decay Law

The radioactive decay law as you might be remembering well is expressed mathematically as
N = N0e^-λt with usual notations.
In most entrance examinations such as Medical and Engineering entrance examination, you wont be allowed to use calculators or logarithm tables. The above equation, modified in terms of half life will be very useful in this context. If N is the number of nuclei remaining undecayed after ‘n’ half life periods, it is related to the initial number N0 as,
N = N0/2ⁿ.
Now let us discus the following M.C.Q.:
Out of 1.414×10²⁴ nuclei, only 10²⁴ nuclei remain undecayed after 15 minutes in a radioactive sample. The half life period of the sample in minutes is
(a) 64 (b) 55 (c) 40 (d)30 (e) 24
We have, 10²⁴ = (1.414×10²⁴)/2ⁿ from which 2ⁿ = 1.414 so that n= ½. This means that 15 minutes is half of the half life period. The half life of the sample therefore is 30 minutes.
The above question can be asked in a modified manner, involving the activity of the sample as follows:
The activity of a radioactive sample drops from 1.414×10⁸ disintegrations per second to 10⁸ disintegrations per second in 15 minutes. The half life period of the sample in minutes is
(a) 64 (b) 55 (c) 40 (d)30 (e) 24

Since the activity of a sample is directly proportional to the number of nuclei present at the instant, we can express the activity ‘A’ after ‘n’ half lives in terms of the initial activity ‘A0’ as,
A = A0/2ⁿ
Substituting the values of A and A0, we have 10⁸ = (1.414×10⁸)/2ⁿ from which n=½. So 15 minutes is half of the half life period of the sample and the answer to the question is 30 minutes [option (d)].

A Question on TV transmitter Height

Unlike A.M.radio transmitters, TV transmitters can cover a distance of 60km to 65km only. You might have noted that this limitation in the range of a TV transmitter is due to the fact that the carrier used for TV transmission is in the V.H.F and U.H.F. range. They are not reflected by the ionosphere and the curvature of the earth places the above mentioned limit in the range of terrestrial TV transmitters.
You might be remembering the expression for the range ‘d’ of a TV transmitter:
d = √(2Rh) where R is the radius of the earth and h is the height of the TV transmitter tower.
The above expression is the approximation of the expression, d = √(2Rh+h^2), obtained by ignoring h^2 which is small compared to 2Rh.
Now consider the following M.C.Q.:
To cover a population of 3 million, what should be the height of a TV transmitter tower? (Population per square km = 1000)
(a) 25m (b) 45m (c) 55m (d) 65m (e) 75m
If ‘d’ is the required range of the TV transmitter, the coverage area = πd^2 = π×2Rh so that we have, 1000 π×2Rh = 3×10^6.
The radius of the earth is 6400km which you are expected to remember. You may substitute for R in km itself in the above equation to obtain h to be approximately 0.075km = 75m. The correct option therefore is (e).