Entrance Physics


AIPMT 2009 questions:

The electric field part of of an electromagnetic wave in a medium is represented by
Ex = 0;
Ey = 2.5 N/C cos[(2π×106 rad/s)t – (π×10–2 rad/m)x];
Ez = 0.
The wave is:
(1) moving along x-direction with frequency 106 Hz and wave length 100 m.
(2) moving along x-direction with frequency 106 Hz and wave length 200 m.
(3) moving along –x-direction with frequency 106 Hz and wave length 200 m.
(4) moving along y-direction with frequency 2π×106 Hz and wave length 200 m.

The electric field variation is in accordance with the equation y = A sin (ωt – kx) which represents a progressive wave proceeding along the positive x-direction. Instead of the usual displacement y we have the electric field E. In place of the angular frequency ω we have 2π×106 (remember ω = n) which means that the linear frequency n is 106 Hz.
Since the propagation constant k = 2π/λ where λ is the wave length, we have
π×10–2 = 2π/λ from which λ = 200 ms–1.
So the correct option is (2). 

[The units of electric field E (N/C), angular frequency ω (rad/s) and the propagation constant k (rad/m) given in the wave equation in the question should not distract you.
You should note that the negative sign in the wave equation y = A sin (ωt – kx) or y = A sin (kx – ωt) indicates that the wave is propagating along the positive x-direction. A wave propagating along the negative x-direction is represented by y = A sin (ωt + kx).
You can make use of any other form of the wave equation, for instance, y = A sin [2π(t/T – x/ λ)], also to solve the problem].

A wave in a string has an amplitude of 2 cm. The wave travels in the + ve direction of x axis with a speed of 128 m/sec. and it is noted that 5 complete waves fit in 4 m length of the string. The equation describing the wave is:
(1) y = (0.02) m sin (15.7x − 2010t)
(2) y = (0.02) m sin (15.7x + 2010t)
(3) y = (0.02) m sin (7.85x − 1005t)
(4) y = (0.02) m sin (7.85x + 1005t)

The unit of amplitude is metre, shown by its symbol ‘m’ in the equation (which should not distract you).
The equation describing the wave propagating in the +ve x-direction is
y = A sin (kx – ωt)
The amplitude A as given in the question is 2 cm = 0.02 m.
The velocity of the wave, v = ω/k = 128 ms–1.
Wave length λ = 4/5 m.
But k = 2π/λ = 2π×5/4 = 7.85 m and so ω = kv = 7.85×128 = 1005
So the equation of the wave is
y = (0.02) m sin (7.85x − 1005t)
The correct option is (3).
AIPMT 2008 question:

The wave described by y = 0.25 sin(10πx – 2πt) where x and y are in metres and t in seconds is a wave traveling along the
(1) negative x-direction with amplitude 0.25 m and wavelength λ = 0.2 m.
(2) negative x-direction with frequency 1 Hz.
(3) positive x-direction with frequency π Hz and wavelength λ = 0.2 m.
(4) positive x-direction with frequency 1 Hz and wavelength λ = 0.2 m.

The given wave equation is in the form y = A sin [2π(x/λ t/T)].
The negative sign in the equation shows that the wave is propagating along the positive x-direction.
By comparison we obtain 2πx/λ = 10πx from which λ = 0.2 m. Also, 2πt/T =2πt from which the frequency 1/T = 1 Hz. [Option (4)].

Multiple Choice Questions on Dimensions of Physical Quantities :

Van der Waals equation of state for real gases is [P + (a/V2)](Vb) = RT where P is the pressure, V is the volume of one mole of gas, R is the universal gas constant, T is the temperature and a and b are constants. The dimensional formula of a is
(a) [M0L6T0]
(b) [ML–1T–2]
(c) [ML5T–2]
(d) [ML–2T–2]
(e) [ML3T–2]

This is a question popular among question setters and has appeared many times in various entrance test papers. Since a/V2 appears as added to P, the dimensions of a/V2 must be the same as those of P. Therefore, ‘a’ has the dimensions of PV2.
Note that pressure P is force per unit ares and hence the dimensional formula for P is [ML–1T–2]. The dimensional formula for V2 is [L6]. Therefore, the dimensional formula for ‘a’ is [ML5T–2].

The dimensional formula for Stefan’s constant is
(a) [MT–3K–4]
(b) [ML2T–2K–4]
(c) [ML2T–2]
(d) [MT–2L0]
(e) [MT4L0]
This question appeared in Kerala Engineering Entrance (KEAM) 2009 question paper. You will find similar questions in other entrance test papers also.
We have E = σT4 where E is the energy radiated per second from unit area, σ is Stefan’s constant and T is the temperature of the black body.
Therefore, σ = E/T4
Note that energy has dimensional formula [ML2T–2]. Sine E is the energy radiated per second from unit area, the dimensional formula for E is [MT–3].
Therefore, the dimensional formula for Stefan’s constant σ is [MT–3K–4].

Pick out the correct dimensions in length of the quantity μ0ε0 from the following:
(a) 2
(b) 1
(c) 1
(d) 2
(e) zero

You know that the speed ‘c of electromagnetic waves in free space is given by
c = 1/√(μ0ε0)
Therefore μ0ε0 = 1/c2 and the dimensional formula for μ0ε0 is [L–2T2]. The quantity μ0ε0 therefore has – 2 dimensions in length [Option (d)].

 Which of the following is dimensionless?
(a) Stress
(b) Gas constant
(c) Frequency of sound
(d) Efficiency of heat engine
(e) Thermal conductivity

The correct option is (d) since the efficiency is the ratio of the output power to the input power.

AIEEE 2006 question paper:

Starting from the origin a body oscillates simple harmonically with a period of 2 s. After what time will the kinetic energy be 75% of the total energy?
(1) 1/12 s
(2) 1/6 s
(3) 1/4 s
(4) 1/3 s

This simple harmonic motion can be represented by the equation,
y = A sin ωt where y is the displacement at the instant t, A is the amplitude and ω is the angular frequency.
The instantaneous velocity v of the particle is given by
v = dy/dt = Aω cosωt
The maximum velocity vmax of the particle is evidently Aω and the maximum kinetic energy which is equal to the total energy is ½ mvmax2 where m is the mass of the particle. We have
½ mv2 = (¾)(½)mvmax2
Therefore, ½ m (Aω cosωt)2 = (¾)(½)m(Aω)2 from which cosωt = (√3)/2
Therefore, ωt = π/6 so that t = π/6ω = π/(6×2π/T ) = 1/6 s since the period T is 2 s.
 
The maximum velocity of a particle executing simple harmonic motion with amplitude 7 mm is 4.4 ms–1. The period of oscillation is
(1) 100 s
(b) 0.01 s
(c) 10 s
(d) 0.1 s

Since the maximum velocity vmax = Aω and the period T = 2π/ω we have
T = A/vmax = 2π×7×10–3/4.4 = 0.01 s

Multiple Choice Questions on Simple Harmonic Motion (SHM):


The essential formulae you have to remember in simple harmonic motion are the following:

(1) Equation of simple harmonic motion: y = Asinωt if initial phase and displacement are zero. Here ‘y’ is the displacement, ‘ω’ is the angular frequency and A is the amplitude.
y = Acosωt also represents simple harmonic motion but it has a phase lead of π/2 compared to the above one.
If there is an initial phase of Φ the equation is y = Asin(ωt + Φ).
y = Asinωt + Bcosωt represents the general simple harmonic motion of amplitude √(A2 + B2) and initial phase tan-1(B/A).

(2)
The differential equation of simple harmonic motion is d2y/dt2 = -ω2y
Note that ω =√(k/m) where ‘k’ is the force constant (force per unit displacement) and ‘m’ is the mass of the particle executing the SHM.

(3) Velocity of the particle in SHM, v = ω√(A2y2)
Maximum velocity, vmax = ωA
 
(4) Acceleration of the particle in SHM, a = - ω2y
Maximum acceleration, amax = ω2A
 
(5) Kinetic Energy of the particle in SHM, K.E. = ½ m ω2( A2 –y2)
Maximum Kinetic energy =
½ m ω2A2
Potential Energy of the particle in SHM, P.E. = ½ m ω2y2
Maximum Potential Energy = ½ m ω2A2
Total Energy in any position = ½ m ω2A2
Note that the kinetic energy and potential energy are maximum respectively in the mean position and the extreme position. The sum of the kinetic and potential energies which is the total energy is a constant in all positions. Remember this:
Maximum K.E. = Maximum P.E. = Total Energy = ½ m ω2A2
 
(6) Period
of SHM = 2π√(Inertia factor/ Spring factor)In cases of linear motion as in the case of a spring-mass system or a simple pendulum, period, T = 2π √(m/k) where ‘m’ is the mass and ‘k’ is the force per unit displacement.
In the case of angular motion, as in the case of a torsion pendulum,
T = 2π √(I/c) where I is the moment of inertia and ‘c’ is the torque (couple) per unit angular displacement.
You may encounter questions requiring calculation of the period of seemingly difficult simple harmonic oscillators. Understand that the question will become simple once you are able to find out the force constant in linear motion and torque constant in angular motion. Angular cases will be rare in Medical and Engineering Entrance test papers.
Let us now discuss some typical questions.

AIIMS 1998 test paper:

If a simple pendulum oscillates with an amplitude 50 mm and time period 2s, then its maximum velocity is
(a) 0.1 m/s (b) 0.15 m/s (c) 0.8 m/s (d) 0.16 m/s

Maximum velocity vmax = ωA where ‘ω’ is the angular frequency and ‘A’ is the amplitude. Therefore vmax = (2π/T)A = (2π/2)×50×10-3 = 0.157 m/s [Option (b)].

Kerala Engineering Entrance 2005 test paper:

A particle executes linear simple harmonic motion with an amplitude of 2 cm. When the particle is at 1 cm from the mean position, the magnitude of its velocity is equal to that of its acceleration. Then its time period in seconds is
(a) 1/ 2π√3 (b) 2π√3 (c) 2π/√3 (d) √3/2π (e) √3/π

The magnitudes of the velocity and acceleration of the particle when its displacement is ‘y’ are ω√(A2 –y2) and ω2y respectively. Equating them, ω√(A2 –y2) = ω2y, from which ω = [√(A2 –y2)]/y = √(4 –1) = √3. Period T = 2π/ω = 2π/√3.


Suppose you place a sphere of mass ‘m’ and radius ‘r’ inside a smooth, heavy hemispherical bowl of radius of 37r placed on a horizontal table. If the sphere is given a small displacement, what is its period of oscillation?
(a) 2π√(m/37rg) (b) 2π√(m/rg) (c) 12π√(r/g) (d) 2π√(r/g) (e) 2π√(37r/g)

The arrangement depicted in this question is similar to that of a simple pendulum. Instead of the usual string, you have a concave surface to confine the bob (sphere) to its path along the arc of a circle. The usual expression for the period, T = 2π√(L/g) holds here also, where the length L = 36r since the length of the pendulum is measured from the centre of gravity of the bob. The point of ‘suspension’ is evidently at the centre of the hemispherical bowl. The correct option is 12π√(r/g) given in (c).

What will be the period of oscillation of a simple pendulum of length 100 cm in a spaceship in a geostationary orbit?
Well, in any satellite orbiting the earth (in any orbit), the condition of weightlessness exists (effective g = 0), the pendulum does not oscillate and the period therefore is infinite.

Consider the following question:

A simple pendulum is arranged using a small metallic bob of mass ‘m’and a light rubber cord of length ‘L’ (on suspending the bob), area of cross section ‘A’ and Young’s modulus ‘Y’. [One should use inextensible cord only for simple pendulum!]. When this unconventional pendulum is at rest in its mean position, the bob is pulled slightly down and is released. Then, the period of the vertical oscillation of the bob is (assuming that the size of the bob is negligible compared to the length of the cord)
(a) 2π√2L/g (b) 2π√(mL/YA) (c) 2π√ (m/YAL) (d) 2π√ (L/g) (e) 2π√ (mY/AL)

The period as usual is given by T = 2π√(m/k). Here ‘m’ is the same as the mass of the bob. The force constant can be found by writing the expression for Young’s modulus (since it arises from the elastic force in the cord): Y = FL/A(δL) where δL is the increase in the length of the cord on pulling the bob down with a force F. Therefore, the force constant, F/(δL) = YA/L. On substituting this value, the period is 2π√(mL/YA).


The following MCQ on simple harmonic motion may generate a little confusion in some of you:


A sphere of mass M is arranged on a smooth inclined plane of angle θ, in between two springs of spring constants K1 and K2 . The springs are joined to rigid supports on the inclined plane and to the sphere (Fig). When the sphere is displaced slightly, it executes simple harmonic motion. What is the period of this motion?

(a) 2π[Mgsinθ/(K1-K2)]½ (b) 2π[M/{K1K2/(K1+K2)}]½ (c) 2π[Mgsinθ/(K1+K2)]½ (d) 2π[M/(K1+K2)]½ (e) 2π[(K1+K2)/M]½



You should note that gravity has no effect on the period of oscillation of a spring-mass system since the restoring force is supplied by the elastic force in the spring. (It can oscillate with the same period in gravity free regions also). So, whether you place the system on an inclined plane or a horizontal plane, the period is the same and is determined by the effective spring constant and the attached mass only. The effective spring constant is K1 + K2 since both the springs try to enhance the opposition to the displacement of the mass. The period of oscillation, as usual is given by, T = 2π√(Inertia factor/Spring factor) = 2π√[M/(K1 + K2)], given in option (d).


The following two questions (MCQ) appeared in Kerala Engineering Entrance 2006 test paper:


(1)The instantaneous displacement of a simple harmonic oscillator is given by y = A cos(ωt + π/4). Its speed will be maximum at the time
(a) 2π/ω (b) ω/2π (c) ω/π (d) π/4ω (e) π/ω

This question was omitted by a fairly bright student who got selected with a good rank. The question setter used the term speed (and not velocity) to make things very specific and to avoid the possible confusion regarding the sign. So what he meant is the maximum magnitude of velocity. The velocity is the time derivative of displacement: v = dy/dt = -Aω(sin ωt + π/4). Its maximum magnitude equal to Aω is obtained when ωt = π/4, from which t = π/4ω.
(2) A particle of mass 5 g is executing simple harmonic motion with an amplitude 0.3 m and time period π/5 s. The maximum value of the force acting on the particle is
(a) 5 N (b) 4 N (c) 0.5 N (d) 0.3 N (e) 0.15 N

If you remember the basic expression for period in the form, T = 2π√(m/k) where ‘k’ is the force constant, the solution becomes quite easy. From this, k = 4π2m/T2 = 4π2 ×5×10-3/(π/5)2 = 0.5. Since ‘k’ is the force for unit displacement, the maximum force is k times the maximum displacement (amplitude). Therefore maximum force = kA = 0.5×0.3 = 0.15N.

[If you remember that ω = √(k/m) you can arrive at the answer since T = 2π/ω]. 

MCQ on Bohr Model of Hydrogen Atom

(1) Suppose the energy required to remove all the three electrons from a lithium atom in the ground state is ‘E’ electron volt. What will be the energy required (in electron volt) to remove two electrons from the lithium atom in the ground state?
(a) 2E/3 (b) E – 13.6 (c) E – 27.2 (d) E – 40.8 (e) E – 122.4

The energy of the electron in a hydrogen like atom in the ground state is – 13.6Z2 electron volt. Therefore, after removing two electrons from the lithium (Z=3) atom, the third electron has energy equal to – 13.6×32 eV = 122.4 eV. The energy needed to remove two electrons from the lithium atom in the ground state is therefore equal to (E –122.4) eV.
 
(2) How many revolutions does the electron in the hydrogen atom in the ground state make per second? (h = 6.63×10-34 Js, mass of electron = 9.11×10-31 kg, Bohr radius = 0.53 A.U.)
(a) 6.55×10-15 (b) 3.28×10-15 (c) 3.28×10-16 (d) 1.64×10-15 (e) 9.11×10-15

The angular momentum (Iω) of the electron is an integral multiple of h/2π. Therefore, Iω = nh/2π, from which, for the first orbit (n=1), ω = h/2πI = h/2πme r2 . The orbital frequency of the electron is given by f = ω/2π = h/4π2m r2 = 6.55×10-15 per second, on substituting for h, m and r.
 
(3) The ionisation energy of the hydrogen atom is 13.6 eV. If hydrogen atoms in the ground state absorb quanta of energy 12.75 eV, how many discrete spectral lines will be emitted as per Bohr’s theory?
(a) 1 (b) 2 (c) 4 (d) 6 (e) zero

On absorbing 12.75 eV, the energy of the electron in the hydrogen atom will become (–13.6 + 12.75) eV which is – 0.85 eV. This is an allowed state (with n=4) for the electron, since the energy in the 4th orbit is – 13.6/n2 = – 13.6/42 = 0.85 eV. From the 4th orbit, the electron can undrego three transitions to the lower orbits (4→3, 4→2, 4→1). From the third orbit, the electron can undergo two transitions (3→2 and 2→1). The electron in the second orbit can undergo one transition (2→1). So, altogether 6 transitions are possible, giving rise to 6 discrete spectral lines [Option (d)].
You can work it out as n(n–1)/2 = 4(4–1)/2 =6.
 
(4) The electron in a hydrogen atom makes a transition from an excited state to the ground state. Which of the following siatements is true?
(a) Its kinetic energy increases and its potential and total energies decrease.
(b)Its kinetic energy decreases, potential energy increases and its total energy remains the same.
(c) Its kinetic and total energies decrease and its potential energy increases.
(d) Its kinetic, potential and total energies decrease.

This question appeared in IIT 2000 entrance test paper. The correct option is (a). You should note that the kinetic energy is positive while the potential energy and total energy are negative. Further, the kinetic energy and total energy are numerically equal and the numerical value is equal to half the potential energy.
The total energy is –13.6 Z2/n2. In lower orbits (with smaller n), the potential energy is smaller since it has a larger negative value. The total energy also is therefore smaller. But, the kinetic energy is greater since it has a larger positive value.

 
(5) If the binding energy of the electron in a hgydrogen atom is 13.6 eV, the energy required to remove the electron from the first excited state of Li++ is
(a) 122.4 eV (b) 30.6 eV (c) 13.6 eV (d) 3.4 eV
Questions of this type often appear in entrance test papers.
Since Li++ is a hydrogen like system with a single electron revolving round a nucleus of proton number Z = 3, the energy of the electron in orbit of quantum number n is
E = –13.6 Z2/n2 eV.
The energy in the first excited state (second orbit) is – 13.6×9/4 eV = – 30.6 eV. The energy to be supplied to the electron to remove it from the first excited state is therefore + 30.6 eV [Option (b)].
 
(6) The wave lengths involved in the spectrum of deuterium (1D2) are slightly different from that of hydrogen spectrum, because
(a) the attraction between the electron and the nucleus is different in the two cases
(b) the size of the two nuclei are different
(c) the nuclear forces are different in the two cases
(d) the masses of the two nuclei are different.

This MCQ appeared in AIEEE 2003 questionn paper. The answer to this will not be easy if you stick on to the elementary theory of the Bohr model in which the energy (En) of the electron of quantum number n (nth orbit) is given by En = – me4/8ε0n2h2 where m is the mass of the electron, e is its charge, ε0 is the permittivity of free space and h is Planck’s constant.
In the elementary theory we take ‘m’ as the mass of the electron on the assumption that the nucleus has a very large mass compared to the mass of the electron and hence the electron is moving round with the nucleus at the centre. The real situation is that both the electron and the nucleus are moving along circular paths with the centre of mass as the common centre. Instead of the actual mss of the electron, the reduced mass of the electron and the nucleus is to be substituted in the expression for energy. The modified form of the expression is En = – μ e4/8ε0 n2h2 where μ is the reduced mass of electron and the nucleus, given by
μ = Mme/(M+me), M and me being the masses of the nucleus and the electron respectively. [Generally, for a hydrogen like system with proton number Z, the expression for energy is En = – μ Z2e4/8ε0n2h2 ].
The nucleus of deuterium contains a proton and a neutron and has very nearly twice the mass of the hydrogen nucleus (proton). So, the reduced mass and the energy levels of deuterium are slightly greater than those of hydrogen and this is the reason for the difference in wave length. [The wave lengths are slightly shorter].The correct option is (d).
An important point you should remember in the light of the above discussion is the drastic change in the energy levels and the spectrum of positronium compared to hydrogen. Positronium is a highly unstable neutral atom with an electron revolving round a positron. [ You can as well say, a positron revolving round an electron!]. The concept of reduced mass is absolutely necessary in this case since the positron has the same mass as that of the electron so that the reduced mass of positronium is mm/(m+m) = m/2 where ‘m’ is the mass of positron as well as the electron.
Now consider the following MCQ:
 
A positronium atom undergoes a transition from the state n = 4 to n = 2. The energy of the photon emitted in this process is
(a) 1.275 eV (b) 2.55 eV (c) 3.4 eV (d) 5.1 eV (e) 13.6 eV

The expression for energy of positronium is En = – μ e4/8ε0n2h2 where μ is the reduced mass of positron and electron, given by μ = mm/(m+m) = m/2. Therefore, the mass of the electron (m) used in the expression for the energy of a hydrogen atom (Bohr’s theory) is to be replaced by m/2. All energy levels are therefore reduced to half of the hydrogen levels. Since the energies for states n=4 and n=2 for hydrogen are –13.6/16 eV(=
–0.85 eV) and –13.6/4 eV (= –3.4 eV) respectively, the energy of the photon emitted in the case of hydrogen is [(–0.85) – (–3.4)] eV = 2.55 eV. In the case of positronium, the energy will be half of this. So, the answer is 2.55/2 eV = 1.275 eV.

Two Questions on Specific Heat of Gases

(1) An ideal diatomic gas in a container is heated so that half of the gas molecules dissociate into atoms. The molar specific heats (at constant volume) of the sample of the gas in the container before and after heating are C1 and C2. Then C1/C2 is
(a) 3/7 (b) 5/7 (c) 7/9 (d) 9/10 (e) 15/11
 
The important point to note here is that on dissociation, each particle (diatomic molecule) with 5 degrees of freedom produces two particles (individual atoms) with 3 degrees of freedom. Therefore you have to use the value Cv = (5/2)R for the undissociated molecule and the value Cv = (3/2)R for the atoms formed on dissociation (Cv is the molar specific heat at constant volume and R is the universal gas constant).
Assuming that there are ‘n’ moles of the diatomic gas initially, the number of moles after dissociation is (3/2)n, with (n/2)×2 = n moles of atoms and n/2 moles of molecules.
The molar specific heat (at constant volume) before dissociation,
C1 = (5/2)R, appropriate for a diatomic gas.
The molar specific heat (at constant volume) after dissociation,
C2 = (Heat supplied for increasing the temperature through 1 K) /Number of moles
= [n×(3/2)R + (n/2)×(5/2)R] / (3/2)n = (11/6)R
Therefore, C1/C2 = (5/2)R/(11/6)R = 15/11.

(2) An ideal diatomic gas is heated at constant pressure. What is the fraction of the heat energy supplied, which increases the internal energy of the gas?
(a) 2/3 (b) 3/5 (c) 5/7 (d) 7/9   (e) 1/2

This is a simple question but it has appeared in various entrance test papers.
When you heat a gas at constant pressure, the gas expands, thereby doing work. When one mole of a diatomic gas is considered, the increase in the internal energy on heating it through 1 K is equal to its molar specific heat (molar heat capacity) at constant volume, which is (5/2) R where R is the universal gas constant. The total heat energy supplied in this case for increasing the internal energy and for doing the external work is the molar specific heat at constant pressure, which is (7/2) R. The fraction required in the question is therefore [(5/2)R] /[(7/2)R] = 5/7

Multiple Choice Questions (MCQ) involving Inductance:

(1) A long straight solenoid of cross section area10 cm2 has 10 turns per cm. A short 50 turn coaxial coil of cross section area 2 cm2 is fixed inside the solenoid at the middle. The mutual inductance of the solenoid and the coil is (in micro henry)
(a) 400 (b) 40 (c) 4π (d) π (e) ) 0.4π 

Mutual inductance between an infinitely long straight solenoid and a short secondary coil placed coaxially inside at the middle is given by
M = μ0nNA where μ0 is the magnetic permeability of free space (or air), n is the number of turns per metre of the solenoid, N is the total number of turns in the secondary coil and A is the cross section area of the secondary coil. (If the secondary coil is outside the solenoid, the cross section area of the solenoid will appear in place of the area of the secondary coil).
[Note that the mutual inductance as given by the above expression is the magnetic flux (linked with the secondary coil) per unit current in the solenoid: M = NBA where B is the magnetic field (μ0n×1) produced by unit current in the solenoid]
Substituting for the known quantities in the expression for mutual inductance, we have
M = 4π×10–7×1000 ×50 ×(2×10–4) = 4π×10–6 H = μH.
 
(2) The secondary coil in the above question is moved with uniform velocity coaxially over a distance of 5 cm in 2 seconds when a current of 2 ampere flows through the solenoid. The emf induced in the coil during the motion is
(a) 200π μV (b) 100π μV (c) 20π μV (d) 10π μV (e) zero

No emf will be induced in the coil since the same flux is linked with the coil throughout its motion and there is no flux change.

(3) In question No.1, suppose the solenoid carries a current of 4 A. If this current is switched off in 100 ms, the emf induced in the secondary coil will be
(a) 0.2π μV (b) 0.4π μV (c) π mV (d) 0.08π mV (e) 0.16π mV
We have ε = –M(dI/dt) = = 4π×10–6 [(4 – 0)/(100×10–3)] = 160π×10–6 volt = 0.16π mV.

4. A battery of emf V volt is connected in series with a coil of inductance L and resistance R at the instant t = 0. The current in the circuit when t = 2τ where τ is the time constant of the circuit is (base of natural logarithm = e)
(a) (V/R)[1 (1/e2)]
(b) (V/R (1/e2)
(c) (V/R)[1 (1/e)]
(d) (V/R)(1/e)]
(e) 2e(V/R)
The expression for the growth of current (with time) in an LR circuit is
I = I0[1– e–Rt/L]
where I0 is the final steady (maximum possible) current which is equal to V/R.
The time constant of the LR circuit is L/R. After 2 time constants, the current will be
I = I0[1– e–R× 2(L/R)/L] = (V/R) [1– e– 2] = (V/R)[1– (1/e2)]

Two Questions on Force of Buoyancy:

How does a balloon filled with helium rise in air? The weight of the balloon with helium is less than the force of buoyancy on the balloon and hence the balloon rises. Here is a question high lighting this principle:
A balloon of mass 6 kg is to be filled with helium so as to lift an instrument weighing 24 kg. The minimum volume of helium to be filled in the balloon is nearly (Density of air = 1.29 kg m–3, density of helium = 0.18 kg m–3 approximately)
(a) ) 22 m3 (b) 27 m3 (c) 30 m3 (d) ) 133 m3 (e) 166 m3
 
The balloon will start rising when the force of buoyancy on the balloon just exceeds the weight of the balloon with helium and the instrument. The force of buoyancy is equal to the weight of displaced air. Therefore, the minimum volume V of helium to be filled is given by
(6 + 0.18 V + 24)g = 1.29 Vg.
This gives V = 27 m3 approximately.
[The weight of air displaced by the instrument and the empty balloon is negligible and therefore not considered here].

A block of cork floats on the surface of water (in a container) with 30 cm outside water. If the system is placed on the moon where there is no atmosphere and the value of acceleration due to gravity is approximately one sixths that on the earth, the portion outside water will be
(a) 5 cm
(b) slightly less than 5 cm
(c) slightly less than 30 cm
(d) slightly greater than 30 cm
(e) slightly greater than 5 cm
 
If the acceleration due to gravity has a non-zero value, the exposed portion will remain unchanged since the weight of the floating body and the weight of displaced fluid are directly proportional to the value of the acceleration due to gravity. So, the exposed portion (or, the immersed portion) will remain the same whether you take the system to the moon or mars or inside a coal mine, when you consider the effect of ‘g’ only. But, you have to consider the effect of the atmosphere also. Since there is no atmosphere on the moon, there is no force of buoyancy due to the atmosphere and the net force of buoyancy (which is due to water in the container and due to the air above) is slightly reduced. Therefore, the block of cork gets immersed a little more and the exposed portion is slightly less than 30 cm.
[Note that if ‘g’ is zero, there is no weight and therefore no question of floatation].

Two Questions on Oscillations:

(1) A simple pendulum of period 2 s has a small bob of mass 50 g. The amplitude of oscillation of the bob is 10 cm and it is at a height of 45 cm from the ground in its mean position. While oscillating, the string breaks just when the bob is in its mean position. The horizontal distance R `from the mean position where the bob will strike the ground is nearly
(a) 35.2 cm (b) 23 cm (c) 15.3 cm
(d)12.4 cm (e) 9.4 cm

The angular frequency (ω) of oscillation of the pendulum is given by
ω = 2π/T = 2π/2 = π rad/s.
The bob has maximum velocity (vmax) in the mean position and is given by
vmax = ωA where A is the amplitude.
Therefore, vmax = π × 0.1 = 0.1 π ms–1
On getting detached from the string, the bob moves like a projectile shot horizontally from a height of 0.45 m with a velocity of 0.1 π ms–1. Its time of flight (t) is obtained from the vertical displacement of 0.45 m:
0.45 = 0 + ½ gt2. (Note that the initial vertical velocity is zero and the vertical acceleration is g, which we may take as 10 ms–2). This gives t2 = 0.09 so that t = 0.3 s.
The horizontal distance covered by the bob during this time is 0.1 π × 0.3 = 0.094 m = 9.4 cm.
[Note that the mass of the bob does not come into the picture and it just serves as a distraction].

A large horizontal surface moves up and down simple harmonically with an amplitude of 1 cm. If a mass of 3 kg (which is placed on the surface) is to remain continually in contact with it, the maximum frequency of the SHM will be
(a) 5 Hz (b) 2 Hz (c) 8 Hz (d) 10Hz (e) 15 Hz

The mass will remain in contact with the surface if the maximum acceleration produced in the simple harmonic motion does not exceed the acceleration due to gravity (g).
Therefore, we have
ω2A = g where ω is the angular frequency of the SHM.
From this ω = √(g/A) = √(10/ 0.01) = 10×√10.
The maximum frequency (linear) of oscillations is therefore given by
n = ω/2π = (10×√10)/2π = 5 Hz.
[Note that the mass of the body placed on the surface does not come into the picture and it just serves as a distraction].


Two questions on photo diodes:

(1) In using a photo diode as a photo detector, it is invariably reverse biased. Why?
(a) The power consumption is much reduced compared to reverse biased condition
(b) Electron hole pairs can be produced by the incident photons only if the photo diode is reverse biased
(c) Light variations can be converted into current variations only if the photo diode is reverse biased
(d) When photons are incident on the diode, the fractional change in the reverse current is much greater than the fractional change in the forward current
(e) The photo diode will be spoilt if it is operated under forward biased condition

Whether the photo diode is reverse biased or forward biased, the number of electron hole pairs produced by the incident photons is the same. In other words, the change in the diode current is the same in both cases. But in the reverse biased condition the current drawn by the diode in the absence of the photons is extremely small, of the order of nanoamperes or microamperes where as in the forward biased condition this current is significant, of the order of tens of milliamperes. The fractional change in the current because of the incident photons is therefore large and easily measurable if the photo diode is reverse biased. The correct option is (d). 

(2) The maximum wave length of photons that can be detected by a photo diode made of a semiconductor of band gap 2 eV is about
(a) 620 nm
(b) 700 nm
(c) 740 nm
(d) 860 nm
(e)1240 nm

The wave length λ (in Angstrom unit) of a photon of energy E (in electron volt) is given by
λE = 12400, very nearly.
Therefore, λ = 12400/E
[The above expression can be easily obtained by remembering that a photon of energy 1 eV has wave length 12400 Ǻ and the energy is inversely proportional to the wave length].
Since E = 2 eV we have λ = 12400/2 = 6200 Ǻ = 620 nanometre.
Photons with wave length greater than 640 nm will have energy less than 2 eV so that they will be unable to produce electron hole pairs in the semiconductor of band gap 2 eV. So the correct option is (a).


Two All India Institute of Medical Sciences (AIIMS) Questions on Optics :
The following questions which appeared in All India Institute of Medical Sciences (AIIMS) 2005 entrance question paper for admitting students to the MBBS Degree course are simple as usual. They are meant for checking your knowledge and understanding of fundamentals.

(1) The apparent depth of water in a cylindrical water tank of diameter 2R cm is reducing at the rate of x cm/minute when water is being drained out at a constant rate. The amount of water drained in c.c. per minute is (n1= refractive index of air, n2 = refractive index of water)
(a) x π R2 n1/n2 (b) x π R2 n2/n1 (c) 2 π R n1/n2 (d) π R2x
 
Since the refractive index is the ratio of real depth to the apparent depth, we have
Real depth = Apparent depth × refractive index.
Therefore, the rate at which the real depth is decreasing = xn2/n1 cm per minute.
The amount of water drained in c.c. per minute is therefore equal to x π R2 n2/n1, given in option (b).

(2) A telescope has an objective lens of focal length 200 cm and an eye piece with focal length 2 cm. If the telescope is used to see a 50 m tall building at a distance of 2 km, what is the length of the image of the building formed by the objective lens?
(a) 5 cm (b) 10 cm (c) 1 cm (d) 2cm

At the first glance this question may seem to be one involving the magnification produced by a telescope; but, this is quite simple since you are asked to consider the objective only.
The objective will produce the image of the building at the focus (which is at 2 m from the lens) and hence from the expression for magnification (M) we have
M = Distance of image/ Distance of object = Height of image/ Height of object
so that 2/ 2000 = x/50 where ‘x’ is the height of image in metre.
Therefore, x = 2×50/2000 = 0.05 m = 5 cm.

AP Physics Exam Resources- Two Questions (MCQ) on Moment of Inertia:

(1) Three circular discs of radii R, R and 2R are cut from a metallic sheet of uniform thickness and the smaller discs are placed symmetrically on the larger disc as shown in the figure. If the mass of a smaller disc is M, the moment of inertia of the system about an axis at right angles to the plane of the discs and passing through the centre of the larger disc is

(a) 5MR2 (b) 7MR2 (c) 9MR2
(d) 11MR2 (e) 12MR2
 
The mass of the larger disc is 4M (since its radius is twice that of the smaller disc) and its moment of inertia is (4M)×(2R)2/2 = 8MR2.
The moment of inertia of each smaller disc about the axis passing through the centre of the larger disc (as given by the parallel axis theorem) is MR2/2 + MR2 = 3MR2/2.
Note that the moment of inertia is a scalar quantity. Therefore, the total moment of inertia of the system of three discs is 8MR2 + 2×3MR2/2 = 11MR2.

(2) A small body of regular shape made of iron rolls up with an initial velocity ‘v’ along an inclined plane. It reaches a maximum height of 7v2/10g where ‘g’ is the acceleration due to gravity. The body is a
(a) ring (b) disc (c) solid sphere
(d) hollow sphere (e) cylindrical rod

The initial kinetic energy of the body is ½ Mv2 + ½ I ω2 where M is its mass and I is its moment of inertia about its axis (of rolling). The first term is its translational kinetic energy and the second term is its rotational kinetic energy.
Since the entire kinetic energy is used in gaining gravitational potential energy, we have
½ Mv2 + ½ I ω2 = Mgh where ‘h’ is the maximum height reached.
Therefore, ½ Mv2 + ½ I v2/R2 = Mg×7v2/10g, from which
I = (2/5)MR2.
The body is therefore a solid sphere.

Multiple Choice Questions on Elasticity:

Here are three questions (MCQ) on elasticity which appeared in Kerala Engineering and Medical Entrance 2004 Examination question papers:

(1) Wires A and B are made from the same material. A has twice the diameter and three times the length of B. If the elastic limits are not reached when each is stretched by the same tension, the ratio of energy stored in A to that in B is
(a) 2:3 (b) 3:4 (c) 3:2 (d) 6:1 (e) 12:1

The work (W) done in increasing the length of a wire or rod by ‘l’ by applying a force ‘F’ is given by
W = ½ Fl
[ Here is the proof for the above: The work dW done for increasing the length by dl is
F×dl. The total work done for increasing the length by ‘l’ is ∫F×dl where the limits of
integration are zero and ‘l’. Since the Young’s modulus, Y = (F/A)(L/l) where A is the
area of cross section and L is the length of the wire, we have F = YAl/L. The total work
done is therefore 0l (Yal/L)dl = ½ (Yal2/L) = ½ (YAl/L)×l = ½ Fl ]
The energy stored (which is equal to the work done) in a wire is therefore directly proportional to the increase in length. The ratio of energy stored is therefore W1/W2 = l1/l2 where l1 and l2 are the increases in length of the wires. But l1 = FL1/A1Y and l2 = FL2/A2Y so that
W1/W2 = l1/l2 = (L1/L2)×(A2/A1) = (3/1)×(1/4) = 3/4 [Option (b)]. 
 
(2) A wire of cross section 4 mm2 is stretched by 0.1 mm. How far will a wire of the same material and length but of area 8 mm2 stretch under the action of the same force?
(a) 0.05 mm (b) 0.01 mm (c) 0.15 mm (d) 0.2 mm (e) 0.25 mm
 
This question as well as the previous one appeared in Kerala Medical Entrance 2004 question paper.
Since the increase in length is inversely proportional to the area of cross section of the wire, the correct option is 0.05 mm.

(3) Compressibility of water is 5×10–10 m2/N. The change in volume of 100 ml of water subjected to 15×10–6 Pa pressure will be
(a) no change (b) increase by 0.75 ml (c) increase by 1.5 ml
(d) decrease by 1.5 ml (e) decrease by 0.75 ml

This question appeared in Kerala Engineering Entrance 2004 question paper. You must definitely remember the expression for bulk modulus ‘B’ as
B = P/(dV/V) where P is the pressure which produces a change in volume dV in a volume V. The negative sign indicates that an increase in pressure will produce a decrease in volume.
Compressibility is the reciprocal of bulk modulus. Therefore we have
1/(5×10–10) = (15×10–10 × 100×10–6)/dV, from which
dV = 0.75×10–6 m3 = 0.75 ml.
Since an increase in pressure produces a decrease in volume, the correct option is (e).


Multiple Choice Questions on Thermoelectric Effect:

Questions involving neutral temperature and temperature of inversion often appear in entrance examination question papers. You have to remember that the neutral temperature is a constant for a given thermo couple where as the temperature of inversion is not a constant. The temperature of inversion is dependent on the temperature of the cold junction and is always as much above the neutral temperature as the cold junction is below it.

Now, consider the following MCQ:

The temperature of the cold junction of a thermocouple is 0º C and its neutral temperature is 275º C. If the temperature of the cold is changed to 20º C, the neutral temperature and the temperature of inversion will be respectively
(a) 265º C and 550º C (b) 265º C and 530º C (c) 275º C and 530º C
(d) 275º C and 550º C (e) 275º C and 510º C
The neutral temperature will be unchanged (275º C). Since the cold junction is 255º C below the neutral temperature, the temperature of inversion has to be 255º C above the neutral temperature and will be 530º C. So, the correct option is (c).
Consider now the following simple question:
The metal which does not exhibit Thomson effect is
(a) iron (b) nickel (c) copper
(d) lead (e) bismuth

The correct option is (d). In determining thermo electric quantities, lead is often used as one member of the couple because of the absence of Thomson effect in it.



What is the unit of thermo electric power?
The term ‘thermoelectric power’ is a misnomer. Thermoelectric power is dV/dT where V is the thermo emf and T is the temperature of the hot junction. There is no ‘power’ involved in it and the unit is volt per Kelvin.
The following question appeared in Kerala Engineering 
Entrance 2007 question paper: 

A simple pendulum is released from A as shown. If m and l represent the mass of the bob and the length of the pendulum, the gain in kinetic energy at B is
(a) mgl/2
(b) mgl/√2
(c) (mgl√3)/2
(d) 2mgl/√3
(e) mgl





From position A, the bob of the pendulum has fallen through a distance l cos 30º (fig.). Therefore, the loss of potential energy is mg l cos 30º = (mgl√3)/2.
Therefore, the gain in kinetic energy by the bob (by the law of conservation of energy) is (mgl√3)/2.

The bob of a simple pendulum has mass 0.2 kg. The bob is drawn aside so that the string is horizontal and is released. When it swings through 60º its kinetic energy is 0.5 J. The kinetic energy when the sting becomes vertical will be
(a) √3 J
(b) 1/√3 J
(c) √3/2 J
(d) 2/√3 J
(e) 1 J
 
You can use the figure used with the previous question for working out the present question as well since the angle turned is 60º. The loss of potential energy by the bob on swinging through 60º is mg l cos30º itself. [Or, you may take it as mg l sin60º].
Since the loss of potential energy is equal to the gain of kinetic energy, we have
mg l cos30º = 0.5 J
Or, (mgl√3)/2 = 0.5 J.
The kinetic energy when the string becomes vertical will be mgl since the bob falls through a distance l and loses its entire initial potential energy mgl.
From the above equation mgl = 1/√3 J [Option (b)].
[Note that the mass of the bob given in the question just serves as a distraction].

IIT-JEE 2009 Multiple Choice Question on Elastic Collision

The following question which appeared in IIT-JEE 2009 question paper is based on the principle that in an elastic collision between two particles of the same mass, the velocities get interchanged: 

Two small particles of equal masses start moving in opposite directions from a point A in a horizontal circular orbit. Their tangential velocities are v and 2v respectively as shown in the figure. Between collisions, the particles move with constant speeds. After making how many elastic collisions, other than that at A, these two particles will again reach the point A?
(a) 4
(b) 3
(c) 2
(d) 1
      
Particle P1, after traversing one third (AB) of the circular track in the anticlockwise direction, will collide with particle P2 at position B. This collision occurs when the particle P2  has traversed two thirds (ACB) of the circular path (since the speed of P2 is twice that of P1). 

         Since the collision is elastic and the particles are of equal masses, their velocities are interchanged. P1 now travels in the clockwise direction with speed 2v and P2 travels in the anticlockwise direction with speed v. The second collision takes place at position C. (BAC is two thirds of the circular path while BC is one third of the path). The second collision at C results in the reversal of the velocities and P1 now travels in the anticlockwise direction with speed v where as P2 travels in the clockwise direction with speed 2v. Therefore, the third collision will occur at A.
Since the third collision at A is not to be counted, the answer is 2 [Option (c)].

Now suppose we modify the above question as follows:

Two small particles of equal masses start moving in opposite directions from a point A in a horizontal circular frictionless track. Their speeds are v and 3v respectively as shown in the figure. After making how many elastic collisions, other than that at A, these two particles will again reach the point A ?
(a) 4
(b) 3
(c) 2
(d) 1

[You may use the figure shown with the question above, replacing the velocity 2v with 3v]
You can easily arrive at the answer which is 3 [Option (b)].

Multiple Choice Questions on Diodes [Including EAMCET 2009 (Engineering) and AIEEE 2006 Questions]


(1) Currents flowing in each of the following circuits A and B respectively are
(1) 1A, 2 A
(2) 2 A, 1 A
(3) 4 A, 2 A
(4) 2 A, 4 A

This question appeared in EAMCET 2009 (Engineering) question paper.
In circuit A both diodes are forward biased and hence the circuit reduces to two 4 Ω resistors connected across the 8 V battery. Since the parallel combined value of the two resistors is 2 Ω, the current delivered by the battery is 8 V/2 Ω = 4 A.
In circuit B one diode is forward biased and the other diode is reverse biased and hence the circuit reduces to just one 4 Ω resistor connected across the 8 V battery. The current delivered by the battery is therefore 8 V/4 Ω = 2 A. The correct option is (3).

The following questions [No. (2) and (No. (3)] were included in AIEEE 2006 question paper:

(2) In the following, which one of the diodes is reverse biased?
You should note that all potentials are with respect to the ground. Therefore the diode in circuit (1) is reverse biased.
[In circuit (2) the anode of the diode is at a higher positive potential compared to its cathode and hence it is forward biased. In circuit (3) the cathode of the diode is at a higher negative potential and hence it is forward biased. In circuit (4) the cathode of the diode is at a negative potential and hence it is forward biased]. 
 
(3) The circuit has two oppositely connected ideal diodes in parallel. What is the current flowing in the circuit?
(1) 1.33 A
(2) 1.71 A
(3) 2.00 A
(4) 2.31 A 
 
Since the diode D1 is reverse biased, no current will flow through D1 and the 3 Ω resistor. The current delivered by the battery is limited by the 4 Ω and the 2 Ω resistors only and is equal to 12 V/(4+2)Ω = 2 A.
The following question is meant for checking your grasp of the behaviour of semiconductor diodes:
 
(4) In the circuit shown the diodes used are silicon rectifier diodes which require a forward bias of 0.7 volt for appreciable conduction. Their leakage current is negligible. The internal resistance of the battery is insignificant. The potential difference between the terminals A and B is very nearly equal to
(a) 6 V
(b) 3 V
(c) 5.3 V
(d) 0.7 V
(e) 0 V
 
The upper diode is reverse biased and can be ignored. The lower diode which is connected across the terminals A and B is forward biased and hence keeps the voltage across A and B at 0.7 V [Option (d)].

[You can use the voltage drop across a forward biased diode as a small reference voltage in electronic circuits just as you use the relatively larger breakdown voltages of reverse biased zener diodes]. 

Questions on Nuclear Physics


The following three questions from nuclear physics are simple but are useful in your preparation for entrance tests:

(1) Two radioactive samples S1 and S2 have half lives 3 hours and 7 hours respectively. If they have the same activity at a certain instant t, what is the ratio of the number of atoms of S1 to the number of atoms of S2 at the instant t?
(a) 9 : 49
(b) 49 : 9
(c) 3 : 7
(d) 7 : 3
(e) 1 : 1

If The number of atoms present at the instant t is N, we have
N = N0eλt where N0 is the initial number, e is the base of natural logarithms and λ is the decay constant.
Therefore, activity, dN/dt = λ N0eλt = λN
If N1 and N2 are the number of atoms of S1 and S2 respectively when the activities are the same, we have
λ1N1 = λ2N2 from which N1/N2 = λ2/λ1
But the decay constant λ is related to the half life T as T = 0.693/λ.
Therefore, N1/N2 = λ2/λ1 = T1/T2 = 3/7 [Option (c)].

(2) A nucleus ZXA has mass M kg. If Mp and Mn denote the mass (in kg) of proton and neutron respectively, the binding energy in joule is
(a) [ZMp + (A – Z)MnM]c2
(b) [ZMp + ZMnM]c2
(c) M – ZMp – (A – Z)Mn
(d) [M– ZMp – (A – Z)Mn]c2
(e) [AMn M]c2

Total mass of the Z protons is ZMp. Since the total number of nucleons is A, the total number of neutrons is (A – Z) and the total mass of the neutrons is (A – Z)Mn.
The mass defect M is the difference between the total mass of the nucleons (protons and neutrons together) and the mass of the nucleus: M =[ZMp + (A – Z)MnM].
Therefore, binding energy.= Mc2 where ‘c’ is the speed of light in free space.
Thus binding energy = [ZMp + (A – Z)MnM]c2

(3) If the aluminium nucleus 13Al27 has nuclear radius of about 3.6 fm, then the tellurium nucleus 52Te125 will have radius approximately equal to
(a) 3.6 fm
(b) 16.7 fm
(c) 8.9 fm
(d) 6.0 fm.
(e) 4.6

The nuclear radius R is given by
R = R0A1/3 where R0 is a constant (equal to 1.2×10–15 m, nearly) and A is the mass number of the nucleus.
If Rl and R2 are the radii of the given Al and Te nuclei respectively, we have
Rl = R0 (27)1/3 = 3R0 and
R2 = R0 (125)1/3 = 5R0
Dividing, Rl/R2 = 3/5
Therefore, R2 = 5R1/3 = (5×3.6)/3 fm = 6 fm.
By clicking on the label ‘nuclear physics’ below this post, you can access all posts related to nuclear physics on this site.

MCQs on Magnetism including EAMCET 2009 (Medical) Question:



(1) The period of oscillation of a magnetic needle in a magnetic field is T. If an identical bar magnetic needle is tied at right angles to it to form a cross (fig), the period of oscillation in the same magnetic field will be
(a) 21/4T
(b) 21/2T
(c) 2T
(d) T√3
(e) T/2

The period of oscillation (T) of the single magnetic needle is given by
T = 2π√(I/mB) where ‘I’ is the moment of inertia of the magnetic needle about the axis of rotation, ‘m’ is the magnetic dipole moment of the needle and ‘B’ is flux density of the magnetic field.
When two magnetic needles are tied together to form a cross, the moment of inertia becomes 2I and the magnitude of the magnetic dipole moment becomes √(m2 + m2) = m√2.

[Note that magnetic dipole moment is a vector quantity. Two identical vectors (each of magnitude m) at right angles will yield a resultant magnitude m√2].

The resultant magnetic moment will be directed along the bisector of the angle between the axes of the individual magnets since the magnets are identical. In the absence of a deflecting torque, the resultant dipole moment vector will align along the applied magnetic field B. On deflecting from this position, the system will oscillate with period T1 given by
T1 = 2π√(2I/mB√2) = 2π√(I√2/mB) = 21/4T

(2) Three identical magnetic needles each L metre long and of dipole moment m ampere metre are joined as shown without affecting their magnetisation. At points B and C unlike poles are in contact. The dipole moment of this system is
(a) m
(b) 2m
(c) 3m
(d) 3m/2
(e) 5m/2
 
The distance (AD) between the ends of the compound magnet is 2L. Since the pole strength is m/L, the dipole moment of the compound magnet is (m/L)2L = 2m
 
(3) A magnet of length L and moment M is cut into two halves (A and B) perpendicular to its axis. One piece A is bent into a semicircle of radiur R and is joined to the other piece at the poles as shown in the figure below:
Assuming that the magnet is in the form of a thin wire initially, the moment of the resulting magnet is given by
(1) M/2π
(2) M/π
(3) M(2 + π)/2π
(4) Mπ/(2 + π)
 
The above question appeared in EAMCET 2009 (Medicine) question paper.
The distance between the poles of the resulting magnet is (L/2) + 2R
Since the semicircular portion of radius R is made of the magnetised wire of length L/2, we have L/2 = πR so that R = L/2π and 2R = L/π
Therefore, length of the resulting magnet (L/2) + (L/π)
The pole strength (p) of the magnet is given by
p = M/L
Therefore, the dipole moment of the resulting magnet = Pole strength×Length = (M/L)[ (L/2) + (L/π)] = M(2 + π)/2π
  


Kerala Engineering Entrance Questions (MCQ) on Communication Systems


Questions on communication systems are generally simple and interesting and so you can easily score marks by attempting them. The following three multiple choice questions were included in Kerala Engineering Entrance 2005 question paper and will be able to answer them in less than three minutes:

(1) If a radio receiver is tuned to 855 kHz radio wave, the frequency of local oscillator in kHz is
(a) 1510
(b) 455
(c) 1310
(d) 1500
(e) 855

Since the tuned frequency is 855 kHz, the receiver is an AM (amplitude modulation) receiver. Modern AM receivers are super heterodyne receivers employing a higher frequency local oscillator. On mixing the local oscillator output with the incoming amplitude modulated carrier, an amplitude modulated wave at intermediate frequency (IF) of 455 kHz (by convention) is produced. Since the intermediate frequency is 455 kHz, it follows that the frequency of local oscillator is 855 kHz + 455 kHz = 1310 kHz.

(2) If n1 and n2 are the refractive indices of the core and the cladding respectively of an optical fibre,
(a) n1 = n2
(b) n1 < n2
(c) n2 < n1
(d) n2 = 2n1
(e) n2 = √(2n1)

Since the optical fibre confines the light signal within the fibre by total internal reflection, the refractive index of the cladding should be less than that of the core. Therefore, n2 < n1 [Option (c)]

(3) A TV tower has a height of 100 m. What is the maximum distance up to which TV transmission can be received? (Radius of the earth, R = 8×106 m)
(a) 34.77 km
(b) 32.7 km
(c) 40 km
(d) 40.7 km
(e) 42.75 km

We have maximum distance, d ≈ √(2Rh) where h is the height of the antenna.
Substituting given values, d ≈ √(2×8×106 ×100) = 40×103 m = 40 km.

[The mean radius of the earth is nearly 6400 km. The value is given as 8000 km in the problem to make your calculation simple.
You should remember that the height of the transmitting antenna (or receiving antenna) is the height with respect to the ground level. If an antenna is mounted on a mast of height h1 and the mast is erected on a hill or building of height h2, the height of the antenna will be h = h1 + h2]

Multiple Choice Questions (MCQ) on Alternating Currents:


(1) A short circuited coil is placed in a time varying magnetic field. Electric power is dissipated due to the current induced in the coil. If the number of turns were quadrupled and the wire radius halved, the electric power dissipated would be
(a) halved
(b) same
(c) doubled
(d) quadrupled

The above question appeared in IIT 2002 screening test paper.

Electric power dissipation is given by P = V2/R. where V is the voltage induced in the coil and R is the resistance of the coil. When the number of turns is quadrupled, the induced voltage V is quadrupled so that V2 becomes 16 times. But the resistance of the coil also becomes 16 times since resistance R = ρL /A where ρ is the resistivity (specific resistance), L is the length and A is the area of cross section of the wire. The length becomes 4 times when the number of turns is quadrupled and the area of cross section becomes one-fourth when the radius is halved. The resistance therefore becomes 16 times.
The power dissipated is therefore unchanged [Option (b)].

(2) An electric heater consumes 1000 watts power when connected across a 100 volt D.C. supply. If this heater is to be used with 200V, 50 Hz A.C.supply, the value of the inductance to be connected in series with it is
(a) 5.5 H
(b) 0.55 H
(c) 0.055 H
(d)1.1 H
(e)11 H 

The current drawn by the heater is 1000 W/100V = 10 A . When the heater is used with A.C. supply, it will draw 10 A itself. (Note that the current and the voltage values are R.M.S. values when you deal with electric power). If ‘L’ is the inductance required, the expression for the current I is
I = V/√(R2 + L2ω2)] where V, R and ω are respectively the alternating voltage, the resistance of the heater and the angular frequency of the A.C.
Substituting, 10 = 200/√[102 + L2 (100π)2], since ω = 2πf where f’ is the frequency of the A.C.
Squaring and rearranging, L2 = 300/100π)2 from which L = √3/10π = 0.055 H. 

(3) An alternating emf V = 6 cos1000t is applied across a series LR circuit of 3 mH inductance and 4 Ω resistance. The amplitude of the current is
(a) 0.6 A
(b) 1.2 A
(c) 1.4 A
(d) 1.8 A

Amplitude (maximum value or peak value) of current (Imax) is given by
Imax = Vmax/√(R2+L2 ω2) = 6/√[42+(3×10-3)2×10002] = 1.2 A
[Note that the values of Vmax and ω are obtained from the expression for the emf V which is in the form, V = Vmax cos ωt].

Here is a very simple question which you should answer carefully:

(4) The voltage V applied across an A.C. circuit and the current I flowing in it are given by
V = 12 cos ωt volt and I = 20 sin ωt milliampere respectively.
The power dissipated in the circuit is
(a) 120 watt
(b) 120 milliwatt
(c) 240 watt
(d) 249 milliwatt
(e) zero

In alternating current circuits the power is given by P = Vrms Irms cosΦ where Φ is the phase difference between the applied voltage and the resulting current. Since the voltage is a cosine function and the current is a sine function, the phase difference Φ is π/2. [Note that the voltage can be written as V = 12 sin(ωt + π/2) volt]. Therefore cosΦ (which is called the power factor) is zero. The correct option is (e).

Electronics - Multiple Choice Questions on Transistor Amplifiers


(1) The adjoining figure shows a common emitter transistor amplifier which uses a silicon transistor. If the quiescent emitter current is 1 mA what is the base biasing voltage?
(a) 4.7 V
(b) 3.7 V
(c) 2.7 V
(d) 1.7 V
(e) 0 V

Because of the emitter current the voltage drop across the 1 KΩ resistor connected to the emitter is 1 V.
[1 mA×1 KΩ = (1/1000) A×1000 Ω = 1 V].
The voltage drop across the base-emitter junction of the silicon transistor is 0.7 V. Therefore, the base voltage under no signal (quiescent) condition is 1 V + 0.7 V = 1.7 V.

(2) In the amplifier circuit shown in Question No.1 what is the function of the capacitor C1 connected across the 1 KΩ emitter resistor?
(a) To produce positive feed back.
(b) To produce negative feed back.
(c) To pass the excess signal to the ground.
(d) To act as filter capacitor for the transistor supply voltage.
(e) To bypass the signal current so that it will not flow through the emitter resistor.

The capacitor C1 provides an easy path (bypass) for the signal component of the emitter current. If C1 is absent the signal component of the emitter current will produce signal voltage drop across the emitter resistor, thereby reducing the signal output at the collector.
The correct option is (e).

(3) If the common emitter current gain βdc of the transistor used in the amplifier circuit shown in Question No.1 is 200, the quiescent base current of the transistor is very nearly equal to
(a) 1 mA
(b) 1 μA
(c) 2 μA
(d) 4 μA
(e) 5 μA

In the common emitter mode, the current amplification factor (current gain) under no signal condition (βdc) is given by
βdc = IC/IB where IC is the collector current and IB is the base current (both under no signal conditions).
Since the collector current is almost equal to the emitter current IE (because of large value of βdc), we have
βdc ≈ IE/IB
Therefore IB ≈ IE/βdc = 1 mA/200 = 0.005 mA = 5 μA.

(4) If the common emitter current gain βdc of the transistor used in the amplifier circuit shown in Question No.1 is 200, what is the voltage drop across the base biasing resistor R under quiescent conditions?
(a) 12 V
(b) 11 V
(c) 10.3 V
(d) 5.4 V
(e) 4.7 V

The quiescent base current is 5 μA as shown in answering question no.3 above. The base biasing voltage is 1.7 V as shown in answering question no.1. The power supply voltage is 12 V. Therefore, the voltage drop across the base biasing resistor R under quiescent conditions is 12 V – 1.7 V = 10.3 V.

(5) The base biasing resistor in the circuit shown in Question No.1 is
(a) 1 KΩ
(b) 4.7 KΩ
(c) 1.03 MΩ
(d) 1.87 MΩ
(e) 2.06 MΩ

The quiescent base current of the transistor is very nearly equal to 5 μA as shown in answering Question No.3. The voltage drop across the base biasing resistor R under quiescent conditions is 10.3 V as shown in answering Question No.4. Therefore, the base biasing resistor is given by
R = (10.3)V/(5 μA) = 2.06 MΩ


IIT JEE 2010 Question (MCQ) on Magnetic Force on Current Carrying Conductors


A thin flexible wire of length L is connected to two adjacent fixed points and carries a current I in the clockwise direction, as shown in the figure.

When the system is put in a uniform magnetic field of strength B going into the plane of the paper, the wire takes the shape of a circle. The tension in the wire is
(a) iBL
(b) iBL /π
(c) iBL /
(d) iBL /


Considering an elemental length dL of the wire, the magnetic force F acting on the element is given by
F = idLB = iRdθB, as is clear from the figure.
The tension T developed in the wire is resolved into rectangular components T sin(dθ/2) and T cos(dθ/2) as indicated in the figure.
The magnetic force F is related to the tension T as
F = iRdθB = 2T sin(dθ/2)
Since is a very small angle, sin(dθ/2) ≈ dθ/2
Therefore, the above equation gives T = iRB
But R = L /
Therefore, T = iBL /

Now, consider the following MCQ which is quite simple contrary to the impression of some of you:


Two straight, infinitely long parallel wires P and Q carry equal currents (I) in opposite directions (fig.). A square loop of side a carrying current i is arranged symmetrically in between the straight wires so that the plane of the loop is in the plane containing the wires P and Q. What is the torque acting on the current loop?
(a) Zero
(b) μ0Iia2/2πd
(c) (μ0Iia2/2π)[1/(d+a)]
(d) (μ0Iia2/2π)[1/d + 1/(d+a)]
(e) (μ0Iia2/2π)[1/d 1/(d+a)]

The resultant magnetic forces acting on the four sides of the current loop act at their centres. The forces on the opposite sides of the loop are oppositely directed as you can easily verify using Fleming’s left hand rule. Since the lines of action of the resultant forces on opposite sides of the loop coincide, there is no lever arm to produce a torque and hence the torque is zero.


Multiple Choice Questions on Elastic and Inelastic Collisions:


(1) A particle A of mass m moving along the positive x-direction with kinetic energy K suffers an elastic head-on collision with a stationary particle B of mass M. After the collision the particle A moves along the negative x-direction with kinetic energy K/9. What is the mass of particle B?
(a) 9 m
(b) 6 m
(c) 3 m
(d) 2 m
(e) m/3

Since the kinetic energy of A after collision is one-nineth of its initial kinetic energy, the momentum of A after collision is one-third of its initial momentum.
Since the momentum is to be conserved, we have
p = p p/3 where p is initial momentum of A and p’ is the momentum of B after the collision.
[The final momentum of particle A is negative since its direction is reversed].
Therefore, p’ = 4p/3
The kinetic energy gained by particle B due to the collision is p2/2M where M is the mass of particle B.
The kinetic energy lost by particle A due to the collision is (8/9)×p2/2m.
[Note that the initial kinetic energy of particle A is p2/2m and its final kinetic energy is (1/9) p2/2m].
Since the kinetic energy too is conserved in elastic collisions, the kinetic energy gained by particle B is equal to the kinetic energy lost by particle A. Therefore, we have
p2/2M = (8/9) p2/2m
Substituting for p’ = 4p/3, we have
(16/9) (p2/2M) = (8/9) p2/2m from which M = 2 m.

(2) A particle of mass m moving westwards with speed v collides with an identical particle moving northwards with speed v. If they stick together after the collision, what is their common speed after the collision?
(a) 4v
(b) 2√2 v
(c) 2v
(d) √2 v
(e) v/√2


The total momentum before collision is the vector sum of the initial momenta mv and mv at right angles (since one is westwards and the other is northwards) and is equal to √2 mv acting along the north-west direction (Fig.).
The combined mass is 2m. If V is the common speed, the total momentum after collision is 2m V, acting along the north-west direction.
Since the momentum is conserved, we have
2m V = √2 mv so that V = v/√2

IIT-JEE 2008 Reasoning Type Question on Rotation:


The following question appeared in IIT-JEE 2008 question paper under Reasoning Type Questions.

STATEMENT -1

Two cylinders, one hollow (metal) and the other solid (wood) with the same mass and identical dimensions are simultaneously allowed to roll without slipping down an inclined plane from the same height. The hollow cylinder will reach the bottom of the inclined plane first.

and

STATEMENT -2

By the principle of conservation of energy, the total kinetic energies of both the cylinders are identical when they reach the bottom of the incline.
(A) STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT -2 is a correct explanation for Statement-1
(B) STATEMENT -1 is True, STATEMENT -2 is True; STATEMENT -2 is NOT a correct explanation for STATEMENT -1
(C) STATEMENT -1 is True, STATEMENT -2 is False
(D) STATEMENT -1 is False, STATEMENT -2 is True
 
The acceleration (a) of a body of mass M rolling down an inclined plane is given by
a = g sinθ/[1 + (I/MR2)]
where θ is the angle of the plane (with respect to the horizontal), I is the moment of inertia (about the axis of rolling) and R is the radius of the body.
Since the moments of inertia of solid cylinder and hollow cylinder are respectively MR2/2 and MR2 the acceleration a is greater for the solid cylinder. Therefore, the solid cylinder will reach the bottom of the inclined plane first.
Since the cylinders have the same mass and are at the same height they have the same initial gravitational potential energy Mgh. This potential energy gets converted into translational and rotational kinetic energies (obeying the law of conservation of energy) when the cylinders roll down the incline and the total kinetic energies of both the cylinders are identical when they reach the bottom of the incline.
The correct option is (D).
 
Let us consider an ordinary type of multiple choice question now: 


A, B and C are three equal point masses (m each) rigidly connected by massless rods of length L forming an equilateral triangle as shown in the adjoining figure. The system is first rotated with constant angular velocity ω about an axis perpendicular to the plane of the triangle and passing through A. Next it is rotated with the same constant angular velocity ω about the side AB of the triangle. If K1 and K2 are the kinetic energies of the system in the first and the second cases respectively, the ratio K1/K2 is
(a) 2/3
(b) 4/3
(c) 8/3
(d) 10/3
(e) 16/3 

The rotational kinetic energy (K) is given by
K = ½ 2
Since the angular velocity is the same in the two cases, the ratio of kinetic energies must be equal to the ratio of moments of inertia.
Thereore, K1/K2 = I1/I2 = 2mL2/m(Lsin60º)2
[Note that in the second case the moment of inertia of the system is due to the single mass at C which is at distance Lsin60º from the axis AB].
Thus K1/K2 = 2/(√3/2)2 = 8/3.


AIPMT 2008 Question on Feed back Amplifier
 

The voltage gain of an amplifier with 9% negative feed back is 10. The voltage gain without feed back will be
(1) 10
(2) 1.25
(3) 100
(4) 90
 
If the voltage gain without feed back is Av and the feed back factor (fraction of output voltage fed back to the input) is β, the voltage gain (Afb) with feed back is given by
Afb = Av/(1 βAv)
In the case of negative feed back the sign of the feed back factor β is negative so that the voltage gain with feed back is given by
Afb = Av/(1+ βAv)
Since Afb = 10 and the magnitude of β is 9% = 0.09, we have on substituting,
10 = Av/(1+ 0.09Av)
This gives Av = 100.

Here is another question on feed back amplifiers:

Pick out the wrong statement:
When negative feed back is applied in a transistor amplifier
(1) its voltage gain is decreased
(2) its band width is decreased
(3) distortion produced by the amplifier is decreased
(4) the transistor current gain is unchanged

The second option alone is incorrect. The band width of a negative feed back amplifier will be greater than that of the same amplifier without the feed back. Since there is a reduction in the voltage gain consequent on the negative feed back, the 3 dB down frequency on the lower side will be shifted towards lower frequency and the 3 dB down frequency on the upper side will be shifted towards higher frequency.

Electromagnetic Induction- Two Multiple Choice Questions

 

(1) A long straight conductor carries a current that increases at a constant rate. The current induced in the circular conducting loop placed near the straight conductor (fig.) is
(a) zero
(b) constant and in the clockwise direction
(c) constant and in the anticlockwise direction
(d) increasing and in the clockwise direction
(e) increasing and in the anticlockwise direction
 
A current flows in the circular loop because of the increasing magnetic flux linked with it. This increase in the magnetic flux is to be opposed in accordance with Lenz’s law. The magnetic field lines produced by the straight conductor are directed normally into the plane of the circular coil (away from the reader). Therefore, the direction of the magnetic field lines produced by the induced current in the circular coil must be directed normal to the plane of the circular coil and towards the reader. The induced current should therefore flow in the anticlockwise sense.
The magnitude of the induced current is constant since the rate of increase of flux is constant. The correct option is (c).

(2) In the circuit shown a coil of inductance 0.5 H and negligible resistance is connected to a 12 V battery of negligible internal resistance using a resistive network. If I1 is the current delivered by the battery immediately after closing the switch and I2 is the current (delivered by the battery) long time after closing the switch,

(a) I1 = 0.5 A and I2 = 0.6 A
(b) I1 = 0.6 A and I2 = 0.6 A
(c) I1 = 0.5 A and I2 = 0.5 A
(d) I1 = 0 A and I2 = 0.5 A
(e) I1 = 0.6 A and I2 = 0.5 A
 
The current through the inductance cannot rise abruptly on closing the switch (because of the time constant L/R of the LR circuit). At the moment of switching the circuit on, the circuit behaves as though the inductance branch is absent. The initial current is therefore given by
I1 = 12 V/ (16+8) Ω = 0.5 A
The current through the inductance rises exponentially with time t from zero to the final steady value I [in accordance with the equation I = I0(1– eRt/L) with usual notations]. After a long time, the current through the inductance becomes steady and there is no voltage drop across the inductance. The inductance is in effect short circuited and the final steady current is given by
I2 = 12 v/(16+4) Ω = 0.6 A
[Note that the parallel combined value of 8 Ω and 8 Ω is 4 Ω]
The correct option is (a).

AIEEE 2008 Question Involving Kirchoff’s Laws:


A 5 V battery with internal resistance 2 Ω and a 2 V battery with internal resistance 1 Ω are connected to a 10 Ω resistor as shown in the figure.

The current in the 10 Ω resistor is
(a) 0.27 A, P1 to P2
(b) 0.27 A, P2 to P1
(c) 0.03 A, P1 to P2
(d) 0.03 A, P2 to P1


The circuit is redrawn, indicating the currents in the three branches. We have marked the directions of the currents I1 and I2 in the directions we normally expect the 5 V and the 2 V batteries to drive their currents. [Note that there can be situations in which the direction we mark is wrong.
There is nothing to be worried about such situations since you will obtain a negative current as the answer and you will understand that the real direction is opposite to what you have marked in the circuit].
Applying Kirchoff’s voltage law (loop law) to the loops ABP2P1 and P1P2CD we have respectively,
5 = 2×I1 + 10×(I1 I2) and
2 = 1×I2 10×(I1 I2)
The above equations can be rewritten as
12 I1 – 10 I2 = 5 and
10 I1 + 11 I2 = 2
These equations can be easily solved to give I1 = 2.34 A (nearly) and I2 = 2.31 A (nearly) so that the current (I1 I2) = 0.03 A.
Since the currents I1 and I2 are obtained as positive, the directions we marked are correct and the current flowing in the 10 Ω resistor is 0.03 A, flowing from P2 to P1 [Option (4)]. 
 
[Suppose we had marked the current I2 as flowing in the opposite direction. The current flowing in the 10 Ω resistor will then be (I1 + I2). We will then obtain I1 = 2.34 A and I2 = – 2.31 A, the negative sign indicating that the real direction of I2 is opposite to what we marked. The current flowing through the 10 Ω resistor will again be obtained correctly as (I1 + I2) = 2.34 A– 2.31 A = 0.03 A].

Geometrical Optics- Questions (MCQ) on Refraction at Prisms:


(1) The face AC of a glass prism of angle 30º is silvered. A ray of light incident at an angle of 60º on face AB retraces its path on getting reflected from the silvered face AC. If the face AC is not silvered, the deviation that can be produced by the prism will be

(a) 0º
(b) 30º
(c) 45º
(d) 60º
(e) 90º 
 
The deviation (d) produced by a prism is given by
d = i1 + i2 A where i1 and i2 are respectively the angle of incidence and the angle of emergence and A is the angle of the prism. Here we have i1 = 60º, i2 = 0º (since the ray falling normally will proceed undeviated from the face AC if it is not silvered) and A = 60º.
Therefore, d = 30º.

(2) In the above question, what is the refractive index of the material of the prism?
(a) 1.732
(b) 1.652
(c) 1.667
(d) 1.5
(e) 1.414 

In the triangle ADN angle AND is 60º since angle DAN = 30º and angle DNA = 90º. Therefore, the angle of refraction at D is 30º. The refractive index of the material of the lens (n) is given by
n = sin i1/ sinr1 = sin 60º/ sin 30º = √3 = 1.732

(3) Two thin (small angled) prisms are combined to produce dispersion without deviation. One prism has angle 5º and refractive index 1.56. If the other prism has refractive index 1.7, what is its angle?
(a) 3º
(b) 4º
(c) 5º
(d) 6º
(e)

Since the deviation (d) produced by a small angled prism of angle A and refractive index n is given by
d = (n – 1)A, the condition for dispersion without deviation on combining two prisms of angles A1 and A2 with refractive indices n1 and n2 respectively is
(n1 – 1)A1 = (n2 – 1)A2
Therefore, 0.56×5 = 0.7×A2 so that A2 = 4º

Questions (MCQ) on Newton’s Laws:


(1) Two identical frictionless pulleys carry the same mass 2m at the left ends of the light inextensible strings passing over them. The right end of the string carries a mass 3m in the case of arrangement (i) where as a force of 3mg is applied in the case of arrangement (ii) as shown in the adjoining figure. The ratio of the acceleration of mass 2m in case (i) to the acceleration of mass 2m in case (ii) is

(a) 2:3
(b) 1:1
(c) 5:3
(d) 3:5
(e) 2:5 

In both cases the net driving force is 3mg – 2mg = mg. But in case (i) the total mass moved is 5m where as in case (ii) the total mass moved is 2m. The acceleration in case (i) is mg/5m = g/5 where as the acceleration in case (ii) is mg/2m = g/2.
The ratio of accelerations = (g/5)/ (g/2) = 2/5 [Option (e)].

(2) An object of mass 4 kg moving along a horizontal surface with an initial velocity of 2 ms–1 comes to rest after 4 seconds. If you want to keep it moving with the velocity of 2 ms–1, the force required is
(a) zero
(b) 1 N
(c) 2 N
(d) 4 N
(e) 8 N 
 
The acceleration ‘a’ of the body is given by
0 = 2 + a×4, on using the equation, vt = v0 + at
Therefore, a = – 0.5 ms–2
The body is retarded (as indicated by the negative sign) because of forces opposing the motion. The opposing force has magnitude ma = 4×0.5 = 2 N. Therefore, a force of 2 N has to be applied opposite to the opposing forces to keep the body moving with the velocity of 2 ms–1.

IIT-JEE 2008 Question (MCQ) on Young’s Double Slit


In a Young’s double slit experiment, the separation between the two slits is d and the wave length of the light is λ. The intensity of light falling on slit 1 is four times the intensity of light falling on slit 2. Choose the correct choice(s).
(A) If d = λ, the screen will contain only one maximum.
(B) If λ < d < 2λ, at least one more maximum (besides the central maximum) will be observed on the screen.
(C) If the intensity of light falling on slit 1 is reduced so that it becomes equal to that of slit 2, the intensities of the observed dark and bright fringes will increase.
(D) If the intensity of light falling on slit 2 is reduced so that it becomes equal to that of slit 1, the intensities of the observed dark and bright fringes will increase


The path difference [S2N = (S2P S1P) in fig.] between the interfering beams is d sinθ. Therefore for obtaining a maximum at the point P on the screen the condition to be satisfied is
dsinθ =
When d = λ, sinθ = n
Since sinθ ≤ 1, n ≤ 1
The possible values of n are 0 and 1. But the maximum corresponding to n = 1 cannot be observed since θ cannot be 90º. Therefore, the central maximum of order zero corresponding to zero path difference (n = 0) alone can be observed. Option A is therefore correct
If λ < d < 2λ, the limiting values of d are λ and 2λ. In addition to the usual central maximum, fringes of order up to n satisfying the equation, dsinθ = will be obtained, where d < 2λ.
If d = 2λ, we have 2λ = nλ/sinθ so that n = 2 sinθ.
Since sinθ should be less than 1 for observing the fringe, the maximum value of n must be 1. Therefore, the central maximum and the maximum of order n = 1 will be observed. Option B too is therefore correct
Options C and D are obviously incorrect since the intensity of the dark fringes will become zero when the light beams passing through the slits are of equal intensity.
 
Now consider another question on Young’s double slit:

Suppose the wave length λ and the double slit separation d in a Young’s double slit experiment are such that the 6th dark fringe is obtained at point P shown in the above figure. The path difference (S2P – S1P) will be
(a) 5 λ
(b) 5 λ/2
(c) 6 λ
(d) 3 λ
(e) 11 λ/2
 
The dark fringe of order 1 (1st dark fringe) is formed when the path difference is λ/2. The dark fringe of order 2 is formed when the path difference is 3λ/2 (and not 2λ/2) and the dark fringe of order 3 is formed when the path difference is 5λ/2 (and not 3λ/2). Generally, the dark fringe of order n is formed when the path difference is (2n 1)λ/2. The 6th dark fringe is therefore formed when the path difference is 11λ/2.