Two Questions involving Energy and Power

(1) A bullet of mass 100g penetrates a distance of 10cm through a fixed block of mass 2kg. If the block is free to move, the distance penetrated will be nearly
(a) 3.5cm (b) 4.5cm (c) 5.5cm (d) 7.5cm (e) 9.5cm
If ‘F’ is the retarding force offered by the block and ‘s’ is the distance of penetration when the block is fixed, we have
Fs = ½ mu^2 ……………..(1)
where ‘m’ is the mass and ‘u’ is the initial velocity of the bullet. If ‘x’ is the distance of penetration when the block is free to move, we have
Fx = ½ mu^2 - ½ (M+m) v^2 where ‘M’ is the mass of the block and ‘v’ is the common velocity of the bullet and the block after the impact. Evidently, v = mu/(M+m) from the momentum conservation law so that we have
Fx = ½ mu^2 - ½ (M+m) [mu/(M+m)]^2
Simplifying, Fx = ½ Mmu^2 /(M+m) ………….(2)
Dividing eq(2) by eq(1) we get x/s = M/(M+m) so that x = Ms/(M+m) = 2×0.1/(2+0.1) = 0.095m = 9.5cm.
(2) An engine delivering constant power displaces a large block through 40cm in 4 seconds. What will be the displacement of the block in 16seconds?
(a) 1.6m (b) 3.2m (c) 4.8m (d) 6.4m (e) 8m
If the power is constant, the displacement will be proportional to t^(3/2) as shown below:
Power P = Fv = m(dv/dt)×v with usual notations. Since the power is constant (k) we have m(dv/dt)×v = k so that v(dv) = k(dt)/m. Integrating, we get
½ v^2 = kt/m. Therefore, v α√t. Since s=vt the displacement ‘s’ is proportional to t√t. Remember this: If power is constant, displacement is directly proportional to t^(3/2)
In the above problem therefore we can write
0.4/x = (4√4)/(16√16) =8/64, from which x = 3.2m

Multiple Choice Questions on Friction:

The following question appeared in the Physics paper of A.I.E.E.E.-2005:

The upper half of an inclined plane with inclination Ф is perfectly smooth while the lower half is rough. A body starting from rest at the top will again come to rest at the bottom if the coefficient of friction for lower half is
(a) tanФ (b) 2tanФ (c) 2cosФ (d) 2sinФ

You have to use the equation of linear motion, v² = u² + 2as for both parts of the motion. For the smooth half of the incline, we have v² = 0 + 2gsinФ×s and for the rough half we have 0 = v² +2(gsinФ - μgcosФ)×s. Combining these two equations, we get
0 = 2gsinФ×s + 2(gsinФ - μgcosФ)×s, from which 2sinФ = μcosФ
Therefore, μ = 2tanФ [Option (b)]
Consider the following MCQ:

A car of mass 1000kg moves on a circular track of radius 20m. If the coefficient of friction is 0.64, then the maximum velocity with which the car can move is
(a) 15m/s (b) 11.2m/s (c) 20m/s (d) 18m/s (e) 22.4m/s
This simple question appeared in the Kerala Engineering Entrance Test paper of 2006. As the frictional force supplies the centripetal force required for the circular motion, we have μmg = mv² /r so that v = √(μrg) = √(0.64×20×9.8) = 11.2m/s

Let us discuss two more questions involving friction:
A coin of mass ‘m’ is placed on a rough horizontal turn table with coefficient of limiting friction ‘μ’. The distance of the coin from the axis of the turn table is ‘r’ and the turn table is turning with angular velocity ‘ω’. The maximum value of ω so as to keep the coin intact on the turn table is
(a) √(μg/r) (b) √(μgr) (c) μgr (d) √(μr/g) (e) μgr²
It is the frictional force that prevents the coin from sliding outwards under the action of the centrifugal force. Therefore we equate the limiting value of the frictional force to the centrifugal force (magnitudes): μmg = mrω² from which ω = √(μg/r). The correct option is therefore (a).
Suppose you are not asked to find the maximum value of ω in the above problem. Instead, you are asked to find the frictional force at the angular velocity ω, as in the modified question below:
A coin of mass ‘m’ is placed on a rough horizontal turn table with coefficient of limiting friction ‘μ’. The distance of the coin from the axis of the turn table is ‘r’ and the turn table is turning with angular velocity ‘ω’. If the coin remains intact on the turn table, the frictional force between the coin and the surface of the turn table is
(a) μmg (b) zero (c) mg (d) mrω² (e) mrω
Nothing is mentioned about the limiting equilibrium of the coin and therefore, the self adjusting frictional force just balances the centrifugal force mrω²
. The correct option therefore is (d).

Two questions from Communication Systems

Questions on topics in Communication Systems are generally simple and not at all time consuming. You should never omit them. Consider the following:
(1) Photo diodes used as light detectors in optical communication systems are operated in the reverse biased manner because
(a) forward biasing will damage the photo diode (b) power requirement is small in reverse biased mode (c) it is more convenient to measure reverse current (d) reverse current alone is affected by illumination (e) the fractional change in reverse current is much greater than the fractional change in forward current.
You know that the forward current is typically of the order of milliamperes while the reverse current is of the order of micro amperes. When light is incident on the photodiode, electrons and holes are produced in equal number, contributing equally to forward current and reverse current. Since the existing reverse current is a small current compared to the forward current, the change in reverse current will be significant compared to the reverse current itself. The change in reverse current is therefore much easier to detect compared to the change in forward current. The correct option therefore is (e).
(2) A TV transmitter tower has a height of 80m. If the range is to be doubled, the height of the tower is to be increased by
(a) 160m (b) 80m (c) 240m (d) 320m (e) 33m
With usual notations, range = √(2Rh). Therefore, to double the range, the height ‘h’ is to be made 4 times (320m). The increment in height is 240m [Option (c)]. In your hurry to answer this simple question, don’t fail to note that it is the increase in height that is asked for.

Rotation of Rigid Bodies:
Let us consider a couple of questions involving rigid body rotation.
(1) A meter scale is held vertically with one end on the floor and is allowed to fall. Assuming that the end on the floor does not slip, what will be the linear velocity of the other end when it strikes the floor?
(a) 2.7m/s (b) 3.1m/s (c) 5.4m/s (d) 9.8m/s (e) 11.2m/s
When the meter scale is allowed to fall, its gravitational potential energy gets converted into rotational kinetic energy so that we have
mgl/2 = ½ I ω ² where ‘l’ is the length(1metre for a metre scale), ‘m’ is the mass, ‘ω’ is the angular velocity and ‘I’ is the moment of inertia of the scale. l/2 appears in the potential energy expression since the centre of gravity of the scale is initially at a height l/2. You should note that the moment of inertia of the scale is about the end in contact with the floor and is equal to ml²/3.
From the above equation, ω =√(3g/l).
The linear velocity of the feree end of the scale = ωl = √(3gl). On substituting for l(=1) and g(=9.8) the linear velocity is 5.4m/s [Option (c)].
(2) An impulsive force F acting for a short time interval ∆t is applied at one end of a thin uniform bar of mass M and length L, in a direction perpendicular to the lengthof the bar. The angular velocity with which the bar will rotate is
(a) F∆t/4ML (b) F∆t/2ML (c)2F∆t/ML (d) 4F∆t/ML (e) 6F∆t/ML
The impulse received by the bar is F∆t which is equal to the linear momentum supplied. The bar will rotate about its centre of mass. The ‘lever arm’ for the angular momentum is L/2 so that we have, (L/2)F∆t = Iω = ML²ω /12. So, ω =6F∆t/ML

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).