Two Questions (MCQ) on Angular Momentum

The following two questions are similar in that both require the calculation of angular momentum in central field motion under inverse square law forces.

(1) An artificial satellite of mass ‘m’ is orbiting the earth of mass ‘M’ in a circular orbit of radius ‘r’. If ‘G’ is the gravitational constant, the orbital angular momentum of the satellite is

(a) [GMm²r]^1/2 (b) [GMmr]^1/2 (c) [GMm/r]^1/2 (d) [GMm²/r²]^1/2 (e) [GMm²r³]^1/2

The orbital angular momentum of a satellite is mvr where ‘v’ is the orbital speed. [Angular momentum = Iω = mr²ω = mr²v/r = mvr where ‘I’ is the moment of inertia and ‘ω’ is the angular velocity of the satellite].

The centripetal force required for the circular motion of the satellite is supplied by the gravitational pull so that we have

mv²/r = GMm/r²

From this, m²v²r² = GMm²r so that angular momentum mvr = [GMm²r]^1/2

(2) In a hydrogen atom in its ground state, the electron of mass ‘m’ is moving round the proton in a circular orbit of radius ‘r’. The orbital angular momentum of the electron is (with usual meaning for symbols)

(a) [m²e²r/4πε₀]^1/2 (b) [m²er/4πε₀]^1/2 (c) [me²r²/4πε₀]^1/2

(d) [me²r/4πε₀]^1/2 (e) [me²r³/4πε₀]^1/2

The steps for finding the orbital angular momentum of the electron are similar to those in question No.1, with the difference that the centripetal force is supplied in this case by the electrostatic attractive force between the proton and the electron.

We have mv²/r = (1/4πε₀)e²/r² from which m²v²r² = (1/4πε₀) ×me²r, so that orbital angular momentum, mvr = [me²r/4πε₀]^1/2

Two Multiple Choice Questions on Digital Circuits

Most of you will like the section on digital circuits, especially because at the class 12 level you have to study very simple circuits. Consider the following MCQ:

For the circuit shown, logic level 1 is +5 volts and logic level 0 is 0 volt. This circuit is

(a) an EXOR gate (b) an AND gate (c) an INVERTER

(d) an OR gate (e) a NOR gate

If both inputs A and B are zero, the diodes will not conduct and the output point will be at ground potential so that the output Y = 0.

If at least one input is at logic 1 level (+5 volts), the diode connected to that input will conduct. The diode connected to the output also will conduct making the output high (+5V). Thus the output Y=1. The same thing happens if both inputs are high.

So, the circuit is an OR gate.

Now, consider the following question:

The digital circuit shown in the figure implements

(a) OR operation (b) EXOR operation

(c) NAND operation (d) NOR operation (e) AND operation

The first gate is a NAND gate. The second gate also is a NAND gate whose inputs are shorted. But when the inputs of a NAND gate are shorted, it becomes an inverter (NOT gate). So, the circuit is a NAND followed by an inverter which is altogether an AND gate [Option (e)].

An IIT-JEE 2007 Question on Centre of Mass

Here is an assertion-reason type MCQ (involving centre of mass motion) which appeared in IIT-JEE 2007 question paper:

STATEMENT-1

If there is no external torque on a body about its centre of mass, then the velocity of the centre of mass remains constant

because

STATEMENT-2

The linear momentum of an isolated system remains constant.

(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

This is a very simple question meant for checking your understanding of basic principles. But if you are not careful, you are liable to pick out a wrong answer!

Generally, if there are external torques or forces or both, the velocity of the centre of mass will change. So, the condition of no external torque alone is not sufficient to ensure the constancy of the velocity of the centre of mass. Statement-1 is therefore false.

[You should also note that an external torque about the centre of mass will not change the velocity of the centre of mass].

Statement-2 is evidently true.

The correct option therefore is (d).

IIT-JEE 2007 Questions on Rotational Motion

The following straight objective type multiple choice question appeared in IIT-JEE 2007 question paper:

A small object of uniform density rolls up a curved surface with an initial velocity v. It reaches up to a maximum height of 3v²/4g with respect to the initial position. The object is

(a) ring (b) solid sphere (c) hollow sphere (d) disc

The body has translational and rotational kinetic energies and these are completely converted in to gravitational potential energy at the maximum height so that we can write

½ Mv² + ½ Iω² = Mgh Where M is the mass, I is the moment of inertia and ω is the angular velocity of the body and h is the maximum height reached. Since ω = v/R where R is the radius of the rolling body, the above equation can be rewritten as

½ Mv² + ½ I(v²/R²) = Mg×(3v²/4g)

From this, I = MR²/2, which is the value for a disc [Option (d)].

The following three questions which appeared in IIT-JEE 2007 question paper are Linked Comprehension Type multiple choice questions:

Two discs A and B are mounted coaxially on a vertical axle. The discs have moments of inertia I and 2I respectively about the common axis. Disc A is imparted an initial angular velocity 2ω using the entire potential enargy of a spring compressed by a distance x₁. Disc B is imparted an angular velocity ω by a spring having the same spring constant and compressed by a distance x₂. Both the discs rotate in the clockwise direction.

Question(1):

The ratio x₁/x₂ is

(a) 2 (b) ½ (c) √2 (d) 1/√2

Equating the potential energies of the springs to the kinetic energies of the discs, we have

½ k x₁² = ½ I×4ω² and

½ k x₂² = ½ ×2I×ω² for the two cases. Here ‘k’ is the spring constant.

From these equations, x₁/x₂ = √2

Question 2:

When disc B is brought in contact with disc A, they acquire a common angular velocity in time t. The average frictional torque on one disc by the other during this period is

(a) 2Iω/3t (b) 9Iω/2t (c) 9Iω/4t (d) 3Iω/2t

Since the angular momentum is conserved, we have

I×2ω + 2I×ω = (I+2I)×ω’ where ω’ is the common angular velocity of the discs. [We have added the angular momenta since they are in the same direction].

From this, ω’ = (4/3)ω

Disc A will have an angular retardation of magnitude ‘α₁’ during the time ‘t’ where as disc B will have an angular acceleration of different magnitude ‘α₂’ during the time ‘t’.

Considering disc A, we have ω’ = 2ω – α₁t from which α₁ = (2ω – ω’)/t = 2ω/3t since ω’ = (4/3)ω.

The average frictional torque exerted on A by B = Iα₁ = 2Iω/3t

[An equal and opposite torque will be exerted on B by A. Check by finding α₂ and hence 2Iα₂].

Question 3:

The loss of kinetic energy during the above process is

(a) Iω²/2 (b) Iω²/3 (c) Iω²/4 (d) Iω²/6

Loss of kinetic energy = Initial kinetic energy – Final kinetic energy

= ½ I×(2ω)² + ½ ×2I×ω² – ½ ×3I×(4ω/3)² = (Iω²)/3

Sound – Two Questions on Beats

Questions similar to the following one have appeared in various entrance tests:

A set of 31 tuning forks are so arranged that each gives 5 beats per second with the previous one. If the frequency of the last tuning fork is double that of the first, the frequency of the third tuning fork is

(a) 250 Hz (b) 220 Hz (c) 180 Hz (d) 160 Hz (e) 150 Hz

Don’t be scared by the relatively large number of forks. This is a very simple question.

If n₁, n₃ and n₃₁ are the frequencies of the 1^st, 3^rd and 31^st forks, we have

n₃₁ = 2n₁ since the frequency of the last fork is double that of the first.

Further, n₃₁ = n₁ + 30×5 since there are 30 increments in frequency (each of 5 Hz) from the first fork to the 31^st fork.

Thus we have 2n₁ = n₁ + 150 from which n₁ = 150 Hz.

The frequency of the 3^rd fork, n₃ = n₁ + 2×5 = 150 + 10 = 160 Hz.

Now, consider the following MCQ:

When a tuning fork is excited, molecules of air vibrate in accordance with the equation x = A cos(512πt). When this tuning fork and another identical tuning fork loaded with a little wax are excited together, 4 beats are heard. The frequency of the second fork loaded with wax is

(a) 516 Hz (b) 508 Hz (c) 384 Hz (d) 260 Hz (e) 252 Hz

When the first tuning fork is excited, the vibrations of the air molecules are simple harmonic with angular frequency ω = 512π as is evident from the form of the equation, x = A cos(512πt).

The linear frequency of vibration of the first fork is n = ω/2π = 512π/2π = 256 Hz.

The frequency of the 2^nd tuning fork before loading with wax was therefore 256 Hz. After loading with wax, its frequency is lowered. Since the beat frequency is 4 Hz, its frequency (after loading) is 256 – 4 = 252 Hz.

Bohr Model of Hydrogen Atom – An IIT-JEE 2007 question

The following MCQ appeared in IIT-JEE 2007 question paper:

The largest wave length in the ultraviolet region of the hydrogen spectrum is 122 nm. The smallest wave length in the infra red region of the hydrogen spectrum (to the nearest integer) is

(a) 802 nm (b) 823 nm (c) 1882 nm (d) 1648 nm

The wave number 1/λ, which is the number of waves per meter length in the case of hydrogen spectrum is given by Rydberg’s relation,

1/λ = R(1/n₁² – 1/n₂²) where R is Rydberg’s constant and n₁ and n₂ are integers.

Ultraviolet radiations are obtained in the Lyman series of hydrogen spectrum when electron transitions take place from higher orbits (of quantum number n>1)to the innermost orbit (of quantum number n=1). So, for the Lyman series, n₁=1 and n₂ = 2,3,4,…etc. The largest wave length in the Lyman series is obtained when the transition is from 2^nd orbit (n₂=2) to the first orbit (n₁=1).

The smallest wave length in the infra red region is obtained when electron transition occurs from the outermost orbit (n₂ = ∞) to the third orbit (n₁ = 3) and this spectral line is the shortest wave length line in the Paschen series ( for which n₁ = 3 and n₂ = 4,5,6….etc.).

For the largest wave length ultraviolet line we have (expressing the wave length in nanometer),

1/122 = R(1/1² – 1/2²) = 3R/4

For the smallest wave length(λ') infrared line we have

1/λ' = R(1/3² – 1/∞) = R/9

Dividing the first equation by the second, λ'/122 = 3×9/4, from which λ' = 823 nm.

Let us consider another similar question which appeared in Kerala Medical Entrance 2001 question paper:

Given that the longest wave length in Lyman series is 1240 Ǻ, the highest frequency emitted in Balmer series is

(a) 8×10¹⁴ Hz (b) 8×10¹² Hz (c) 8×10¹⁰ Hz (d) 8×10³ Hz (e) 8×10² Hz

In the Rydberg’s relation, 1/λ = R(1/n₁² – 1/n₂²), n₁=1 and n₂ = 2,3,4,…etc., for the Lyman series. For the Balmer series (which is in the visible region), n₁=2 and n₂ = 3,4,5….etc.

The longest wave length in Lyman series is obtained for n₂ = 2 and the highest frequency (shortest wavelength) in Balmer series is obtained for n₂ = ∞.

Rydberg’s relation for the above two cases are (expressing wave lengths in Angstrom),

1/1240 = R(1/1² – 1/2²) = 3R/4 and

1/ λ' = R(1/2² – 1/∞) = R/4

Dividing the first equation by the second,

λ'/1240 = 3, from which λ' = 3720 Ǻ.

The frequency (ν) of this line is given by

ν = c/λ' = (3×10⁸) /(3720×10^–10) = 8×10¹⁴ Hz.

IIT-JEE 2007- Three Questions from Optics

The following three questions appeared in IIT-JEE (Part I) 2007 question paper:

(1) In an experiment to determine the focal length (f) of a concave mirror by the u-v method,a student places the object pin A on the principal axis at a distance x from the pole P. The student looks at the pin and its inverted image from a distance, keeping his/her eye in line with PA. When the student shifts his/her eye towards left, the image appears to the right of the object pin. Then,

(a) x less than f (b) f less than x less than 2f (c) x = 2f (d) x greater than 2f

This question is a simple one and is intended to check your understanding of techniques used in experimental physics in addition to your theoretical knowledge. Since the image appears to the right of the object when the student shifts his eye towards the left, the image is nearer to the student and hence the image distance ‘v’ is greater than the object distance ‘u’. This is possible only if the object is placed between f and 2f. So the correct option is (b).

[The image is real since it is inverted as mentioned in the question. The problem can be worked out even if this fact is not mentioned in this problem].

(2) A ray of light traveling in water is incident on its surface open to air. The angle of incidence is θ, which is less than the critical angle. Then there will be

(a) only a reflected ray

(b) only a refracted ray and no reflected ray

(c) a reflected ray and a refracted ray and the angle between them would be less

than 180° – 2θ

(d) A reflected ray and a refracted ray and the angle between them would be

greater than 180° –2θ

If the angle of incidence is not equal to or greater than the critical angle, there will be partial transmission and partial reflection. As is evident from the figure, the angle between the reflected ray and the refracted ray (angle SQR) is less than 180° – 2θ. The correct option is (c).

(3) STATEMENT-1

The formula connecting u,v and f for a spherical mirror is valid only for mirrors
whose sizes are very small compared to their radii of curvature

because

STATEMENT-2

Laws of reflection are strictly valid for plane surfaces, but not for large spherical surfaces.

(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

This is an assertion-reason type MCQ. You might have noted that the formula connecting u, v and f was derived considering rays close to the principal axis so that only a small portion of the mirror surrounding the pole is involved. Statement-1 is therefore true. Statement-2 is false since the laws of reflection are applied to a ray which is incident at a given point. The size of the mirror and its curvature are not involved here.

You can find all the posts in Optics on this site by clicking on the label 'optics' below this post.

MCQ involving Magnetic Force on Current carrying Conductors

Questions on magnetic force on current carrying conductors are interesting and often simple. Occasionally you may find questions which may baffle you. See the following question:

A copper wire is bent in the form of a sine wave of wave length ‘λ’ andpeak to peak value ‘a’ as shown in figure. A magnetic field of flux density ‘B’ tesla acts perpendicular to the plane of the figure in the entire region. If the wire carries a steady current ‘I’ ampere, the magnetic force on the wire is

(a) I√(a²+λ²) B (b) IaB (c) I(a+λ)B (d) I(λ– a)B (e) IλB

This is a very simple question and the correct option is (e).

This is a very simple question and the correct option is (e).

Imagine the wire to be made of a large number of horizontal and vertical elements as shown. There are as many vertical elements carrying current upwards as there are those carrying currents downwards. The magnetic forces on them will be leftwards and rightwards and they will get canceled. But the magnetic forces on the horizontal elements will be in the same direction ( either upwards or downwards, depending in the directions of the magnetic field and the current) and they will get added to produce a net force IλB.

The following MCQ appeared in IIT–JEE Screening 2003 question paper:

A conducting loop carrying a current ‘I’ is placed in a uniform magnetic field pointing in to the plane of the paper as shown. The loop will have a tendency to

(a) contract

(b) expand

(c) move towards positive X-axis

(d) move towards negative X-axis

The magnetic force everywhere on the loop is radially outwards as given by Fleming’s left hand rule. So, the loop has a tendency to expand [Option (b)].

In the above question, suppose that in place of the option ‘expand’, you had the option, ‘move towards the positive Z-direction’. In that case also, the correct option would be (b) because the loop will act as a magnetic dipole whose south pole is the nearer face. The loop will therefore move towards the reader.

MCQ on Viscosity – Poiseuille’s Flow

Poiseuille’s formula for the volume ‘V’ of a liquid of density ‘ρ’ flowing in time ‘t’ through a capillary tube of length ‘L’ and radius ‘r’ under a pressure difference ‘P’ between the ends of the tube is

V = (πPr⁴t)/ (8Lη) where ‘η’ is the coefficient of viscosity of the liquid. [You should note that this formula holds good only if the flow is slow and steady (stream-lined)].

The rate of flow (volume flowing per second) ‘Q’ is given by

Q = (πPr⁴)/ (8Lη)

Here P = hρg where ‘h’ is the height of the liquid column which produces the pressure difference so that

Q = (π hρg r⁴)/ (8Lη)

You will often find questions based on Poiseuille’s formula in entrance tests for admitting students to various courses. Consider the following question:

A capillary tube of length ‘L’ and radius ‘r’ is joined to another capillary tube of length L/4 and radius r/2 A liquid flows through this series combination. If the pressure difference between the ends of the first tube is P, that between the ends of the second tube is

(a) P (b) 4P (c) 8P (d) 16P (e) P/4

Since the tubes are in series, the rates of flow through the tubes are equal so that from Poiseuille’s equation

πPr⁴/8Lη = πP'(r/2)⁴/[8(L/2)η], where P’ is the pressure difference between the ends of the second tube.

From this P' = 16P/2 = 8P.

Now, consider the following MCQ:

Under a constant pressure head, the volume of a liquid flowing per second through a capillary tube of radius 1 mm and length 16 cm is 4 cm³. If another tube of radius 0.5 mm and length 8 cm is connected in series with it and the same pressure head is applied across the combination, the volume of liquid flowing per second will be ( in cm³)

(a) 5/3 (b) 5/6 (c) 4/3 (d) 4/5 (e) 4/9

When tubes are connected in series, the net rate of flow (Q_net) under a given pressure head is given by the reciprocal relation,

1/Q_net = 1/Q₁ + 1/Q₂ + 1/Q₃ +…..etc. where Q₁, Q₂, Q₃.....etc. are the individual rates of flow when the tubes are connected separately to the same pressure heaed.

In the present case, since there are two tubes only, 1/Q_net = 1/Q₁ + 1/Q₂ so that

Q_net = Q₁Q₂/(Q₁+Q₂). Here Q₁ = 4 cm³.

Since the rate of flow through a tube is given by Q = (πPr⁴)/ (8Lη), we have Q α r⁴/L.

Since the radius of the second tube is half that of the first, the rate of flow is reduced to (½)⁴ = 1/16 of that through the first tube on account of this. Since the length of the second tube is half that of the first, the rate of flow is increased to twice that through the first tube on this account. Therefore, rate of flow through the second tube is given by

Q₂ = Q₁×(2/16) = Q₁/8 = 4/8 cm³ = 0.5 cm³.

The net rate of flow is therefore given by Q_net = (4×0.5)/(4+0.5) = 4/9 cm³ [Option (e)].

Surface Tension–Questions involving excess of pressure inside a bubble

Here is a question which is popular among question setters:

Two spherical soap bubbles of radii r₁ and r₂ in vacuum combine under isothermal conditions. The resulting bubble has a radius equal to

(a) (r₁+r₂)/2 (b) √(r₁+r₂) (c) r₁r₂/(r₁+r₂) (d) (r₁+r₂)/√2 (e) √(r₁²+r₂²)

As the temperature is constant, we have P₁V₁+ P₂V₂ =.PV where P₁ and P₂ are the pressures inside the separate bubbles, V₁ and V₂ are their volumes, P is the pressure inside the combined bubble and V is its volume.

Since the bubbles are located in vacuum, the pressure inside the bubble is equal to the excess of pressure 4T/r, where T is the surface tension and ‘r’ is the radius so that we have (4T/r₁)×[(4/3)πr₁³]+ (4T/r₂)×[(4/3)πr₂³] = (4T/R)×[(4/3)πR³] where R is the radius of the combined bubble.

This yields R = √(r₁²+r₂²).

[You can work out this problem by equating the surface energies: 4πr₁²T+4πr₂²T= 4πR²T, from which R = √(r₁²+r₂²)]

Now, consider the following MCQ:

Excess of pressure inside one soap bubble is four times that inside another. Then the ratio of the volume of the first bubble to that of the second is

(a) 1:16 (b) 1:32 (c) 1:64 (d) 1:2 (e) 1:4

If r₁ and r₂ are the radii of the bubbles and T is the surface tension of soap solution, we have (4T/r₁)/(4T/r₂) = 4.

Therefore, r₁/r₂ = ¼. Since the volume is directly proportional to the cube of the radius, the ratio of volumes V₁/V₂ = (r₁/r₂)³ = (¼)³ = 1/64 [Option (d)].

The following MCQ appeared in AIEEE 2004 question paper. This question is popular among question setters and has appeared in other entrance test papers as well:

If two soap bubbles of different radii are connected by a tube,

(a) air flows from the bigger bubble to the smaller bubble till the sizes become equal

(b) air flows from the bigger bubble to the smaller bubble till the sizes are interchanged

(c) air flows from the smaller bubble to the bigger

(d) there is no flow of air

The excess of pressure inside the smaller bubble is greater than that inside the bigger bubble. (Remember, P = 4T/r and hence excess of pressure ‘P’ is inversely proportional to the radius ‘r’). Therefore, air will flow from the smaller bubble to the bigger bubble [Option (c)].

MCQ on Vibration of Strings & Sonometer

In continuation of the post dated 18th March 2007, let us discuss two more questions involving transverse waves in stretched strings. The following MCQ appeared in Kerala Engineering Entrance Examination 2006 question paper:

Two stretched strings of same material are vibrating under same tension in fundamental mode. The ratio of their frequencies is 1:2 and ratio of the lengths of the vibrating segments is 1:4. Then the ratio of the radii of the strings is
(a) 2:1 (b) 4:1 (c) 3:2 (d) 8:1 (e) 4:5
You know that the frequency (n) of vibration of a stretched string in the fundamental mode is given by
n = (1/2L)√(T/m) = (1/2L)√(T/πr²ρ) where L is the length of the vibrating segment of the wire, T is the tension and ‘m’ is the linear density (mass per unit length) which is πr²ρ where ‘r’ is the radius and ρ is the density of the material of the wire.
Therefore, if n₁ and n₂ are the frequencies of the two wires having lengths L₁ and L₂ and radii r₁ and r₂ respectively, we have
n₁/n₂ = [(1/2L₁)√(T/πr₁² ρ)] / [(1/2L₂)√(T/πr₂²ρ) = L₂r₂/L₁r₁. [Note that the strings have the same density since they are of the same material].
Since n₁/n₂ = ½ and L₁/L₂ = ¼ as given in the question,
½ = 4r₂/r₁ from which r₁/r₂ = 8 [Option (d)].

Now, consider the following question:
A sonometer wire is kept stretched by suspending a 60 kg mass from the free end of the wire. The suspended mass has a volume of 0.01 m³. The fundamental frequency of the wire in this condition is 300 Hz. If the suspended mass is completely immersed in water, the fundamental frequency of the wire will be approximately
(a) 285 Hz (b) 273 Hz (c) 300 Hz (d) 330 Hz (e) 355 Hz
The frequency will decrease since the tension is decreased (because of the decreased weight of the mass in water).
When the mass is in air, the tension,T₁ = mg = 60×g where ‘g’ is the acceleration due to gravity.
When the mass is in water, tension, T₂ = mg – up thrust = 60×g – 0.01×1000×g = 50×g. [Note that the up thrust is the weight of 0.01 m³ of water displaced by the suspended mass].
Since the frequency is given by n = (1/2L)√(T/m) with usual notations, n α √T.
If n₁ and n₂ are the frequencies in the two cases, n₁/n₂ = √(T₁/T₂) or,
300/n₂ = √(60/50) = √1.2 = 1.1 (nearly) so that n₂ = 273 (nearly).

Multiple Choice Questions on Radioactivity

The following Question appeared in Karnataka CET 2003 question paper:
Half life of a radioactive substance is 20 min. The time between 20% and 80% decay will be
(a) 25 min. (b) 30 min. (c) 40 min. (d) 20 min
If the initial activity is A0, the activity ‘A’ after ‘n’ half lives is given by
A = A₀/2ⁿ.
Let us take the initial activity as 100 units. After 20% decay, the activity becomes 80 units and after 80% decay, the activity becomes 20 units. These two cases can be stated as
80 = 100/2ⁿ and
20 = 100/2^m
where ‘n’ and ‘m’ are the numbers of half lives required for 20% decay and 80% decay respectively.
Dividing, 80/20 = 2^m/2ⁿ = 2^(m–n). Or, 2^(m-n) = 4, from which (m–n) = 2 half lives = 2×20 min. = 40 min.

Now consider the following MCQ which appeared in Karnataka CET 2004:
A count rate meter shows a count of 240 per minute from a given radioactive source. One hour later the meter shows a count rate of 30 per minute. The half life of the source is
(a) 80 min. (b) 120 min. (c) 20 min. (d) 30 min.
From the equation, A = A₀/2ⁿ, we have 30 = 240/2ⁿ so that n = 3. Therefore one hour is equal to 3 half lives which means the half life of the substance is 20 min.

MCQ on Oscillation of Magnets

The period (T) of angular oscillations of a magnet suspended freely in a magnetic field of flux density ‘B’ is given by
T = 2π √(I/ mB) where ‘I’ is the moment of inertia of the magnet about the axis of the angular motion and ‘m’ is the magnetic dipole moment. You will find questions based on this relation in your entrance examination question papers. Here is a question which is meant also for checking your grasp of the moment of inertia and the vector property of the magnetic dipole moment:
Two identical magnets are placed one above the other and tied together so that their like poles are in contact. The period of oscillation of this combination (on suspending horizontally using torsionless suspension fibre) in a horizontal magnetic field is 2 s. What will be the period of oscillation if the magnets are placed one above the other such that they are mutually perpendicular and bisect each other?
(a) 2 s (b) √2 s (c) 2√2 s (d) 2^¼ s (e) 2×2^¼ s

The period of oscillation is given by T = 2π √(I/ mB) with usual notations.
In the both cases, the moment of inertia is twice that of one magnet. In the first case, the net dipole moment is twice that of one magnet since the like poles are pointing in the same direction. In the second case, the net dipole moment is √2 times the moment of one magnet since the moment vectors at right angles get added.
The equations for the period in the two cases are therefore
T1 = 2π √(2I/ 2mB) = 2π √(I/ mB) = 2 s (as given in the question) and
T2 = 2π √(2I/ √2 mB) = 2π √(√2 I/ mB) = T1 √(√2) = 2×2^¼ s.
The following MCQ appeared in EAMCET (Med) A.P.2003 question paper:
The period of oscillation of a magnet at a place is 4 seconds. When it is remagnetised so that the pole strength becomes 4 times the initial value, the period of oscillation in seconds is
(a) ½ (b) 1 (c) 2 (d) 4

The period of oscillation is given by T = 2π √(I/ mB) with usual notations. The period is therefore inversely proportional to the square root of the dipole molent ‘m’. When the pole strength is made 4 times the initial value, the dipole moment is made 4 times the initial value so that the period becomes half the initial value. The correct option is 2 seconds.

Two Questions Involving Gravitation

Most of you may be knowing that the moon does not possess an atmosphere because the thermal velocity acquired by gas molecules on the moon when heated by the solar radiations is significant compared to the escape velocity on the moon’s surface (2.4 km/s). The escape velocity on the earth’s surface is 11.2 km/s which is much greater than the velocity acquired by oxygen and nitrogen gas molecules on getting heated by solar radiations. (In the case of hydrogen molecules, this is not the case). It is enough that the most probable velocity

[√(2RT/M)] of a gas molecule is in excess of about 20% of the escape velocity, for the molecule to escape to outer space.
Now, consider the following question:

The radius of the earth is 6400 km and the acceleration due to gravity on the earth’s surface is 9.8 ms^–2. The universal gas constant is 8.4 J mol^–1 K^–1. The temperature at which the r.m.s. velocity of oxygen gas molecules becomes equal to the velocity of escape from the surface of the earth is
(a) 1.59×10⁶ K (b) 1.59×10⁵ K (c) 1.59×10⁴ K (d) 1.59×10³ K (e) 1.59×10² K
The escape velocity is given by v_e = √(2gR_E) where ‘g’ is the acceleration due to gravity and ‘R_E’ is the radius of the earth.
On substituting for ‘g’ and ‘R_E’, the escape velocity, v_e = 11.2×10³ m/s
The molecular velocity (r.m.s.) is given by v = √(3RT/M) where ‘R’ is universal gas constant, ‘T’ is the temperature (in Kelvin) and ‘M’ is the molar mass of the gas (oxygen in the present case).
Therefore, √(3RT/M) = 11.2×10³. Substituting for R = 8.4 and M = 0.032 kg, the temperature works out to be 1.59×10⁵ K.
Now, consider the following question which is based on Kepler’s law:
A planet moves around the sun. When it is farthest away from the sun at distance r₁, its speed is v₁. When it is closest to the sun at distance r₂ its speed will be
(a) r₁v₁/r₂ (b) (r₁/r₂)² v1 (c) √(r₁/r₂) ×v₁ (d) r₂v₁/r₁ (e) √(r₂/r₁)× v₁
According to Kepler’s law, the line joining the planet to the sun sweeps out equal areas in equal intervals of time. If we consider a very small time interval δt, the areas swept when the planet is at apogee (farthest away) and at perigee (closest to the sun) will be triangles whose areas are directly proportional to v₁r₁ and v₂r₂ respectively. [The bases of the triangular areas swept in the time δt are v₁δt and v₂δt and the altitudes are r₁ and r₂ respectively].
Therefore, from Kepler’s law, r₁ v ₁ = r₂ v₂ so that v₂ = r₁v₁ /r₂

Vibration of Strings

The fundamental frequency of vibration (n) of a string (or wire) is given by
n = (1/2L)√(T/m) where L is the length of the wire, T is the tension and ‘m’ is the linear density (mass per unit length) of the string.
You may get questions based on this relation. See the following MCQ:
A sonometer wire and a tuning fork are excited together. Four beats are heard when the length of the wire is 60 cm as well as 62 cm. The frequency of the tuning fork is ( in Hz)
(a) 512 (b) 488 (c) 384 (d) 256 (e) 244
Since the frequency of vibration of the wire is inversely proportional to its length, we can write, in the two cases,
(n + 4) α 1/60 and
(n – 4) α 1/62 where ‘n’ is the frequency of the tuning fork.
[Note that the frequency of the wire is greater than that of the fork when its length is smaller].
Dividing, (n + 4)/ (n – 4) = 62/60 from which n = 244Hz.
Now, consider the following questionwhich appeared in EAMCET (Med) 2003 question paper:
Two uniform wires are vibrating simultaneously in their fundamental modes. The tensions, lengths, diameters and the densities of the two wires are in thr ratio 8:1, 36:35, 4:1 and 1:2 respectively. If the note of the higher pitch has a frequency 360 Hz, the number of beats produced per second is
(a) 5 (b) 10 (c) 15 (d) 20
Since the frequency of vibration of a stretched wire is given by
n = (1/2L)√(T/m) = (1/2L)√(T/πr²ρ) where L is the length of the wire, T is the tension and ‘m’ is the linear density (mass per unit length) which is πr²ρ where ‘r’ is the radius and ρ is the density of the material of the wire, we have,
n₁/n₂ = (L₂/L₁) √[(T₁/T₂)( r₂ /r₁)² (ρ₂/ρ₁)]
= (35/36) √[(8/1)( 1 /16) (2/1)]
= (35/36)×1
This means that n₂ is the higher frequency. Since the higher frquency is given as 360 Hz, we have n₁/360 = 35/36 from which n₁ = 350 Hz.
Therefore, beat frequency = 360 – 350 = 10 Hz.

Questions (MCQ) on Direct Current Circuits

Here is a question which you can work out using Ohm’s law only:
Two constantan wires P and Q have their lengths in the ratio 1:2 and radii in the ratio 2:1. They are connected in series and potentials 2V and 20V are applied at the free ends of P and Q respectively. The potential at the junction of the wires is
(a) 2V (b) 4V (c) 9V (d) 12V (e) 16V

The resistances of P and Q are in the ratio 1:8 since the length of Q is twice that of P and the area of cross section of Q is a quarter if that of P. [The resistance is given by R = ρL/A where ρ is the resistivity, L is the length and A is the area of cross section].
The potential drops across P and Q (when current flows in them) are in the ratio 1:8. The potential difference applied across the series combination of P and Q is (20–2) = 18V. Therefore, the potential drop across Q = 18×8/(1+8) = 16V.
The potential at the junction of P and Q is (20V– 16V) = 4V.

Now, consider the following MCQ:

In the circuit shown, the power dissipated in the 2Ω resistor is 9W. What is the power dissipated in the 4Ω resistor?

(a) 18W (b) 12W (c) 9W (d) 3W (b) 2W

If the current through the 2Ω resistor is ‘I’, the current through the 4Ω resistor is I/3 since the total resistance (9Ω) in the 4Ω resistor branch is 3 times the total resistance (3Ω) in the 2Ω resistor branch. The expressions for power dissipation in 2Ω and 4Ω are respectvely
P = I² ×2 and
P' = (I/3)² ×4 so that P'/P = 2/9 from which P'= P×(2/9) = 9×2/9 = 2W.

Vibration of Air Columns – Resonance Column & Organ Pipe

In Acoustics a closed pipe or organ pipe means a tube closed at one end. An open pipe or organ pipe means a tube open at both ends. When a standing wave (stationary wave) is formed in an organ pipe, the closed end will be a node and the open end will be an antinode. This is why the length of the pipe in the fundamental mode is equal to λ/4 (which is the distance between neighbouring node and antinode) in a closed pipe. In an open pipe, in the fundamental mode, the length of the pipe is equal to λ/2 since the consecutive antinodes are located at the ends of the tube.
You should remember that a closed pipe can support odd harmonics only where as an open pipe can support all harmonics. In other words, the frequencies of vibration of the air column in a closed pipe are in the ratio 1: 3 : 5 : 7 : etc., while those in an open pipe are in the ratio 1 : 2 : 3 : 4 : 5 : etc.
Now, consider the following MCQ:
An open organ pipe and a closed organ pipe have the same length. The ratio of their fundamental frequencies is
(a) 1 (b) 2 (c) 3 (d) 4 (e) 3/4
If ‘L’ is the length of the pipe, we have, L = λ/2 for the open pipe and L = λ'/4 for the closed pipe where λ and λ' are the wave lengths of sound in the two cases (in the fundamental mode).
The corresponding fundamental frequencies are n = v/ λ = v/2L and n' = v/λ' = v/4L, from which n/n' = 2 [Option (b)].
Let us consider another question:
Almost the entire length of an aluminium pipe of length 1.1m is dipped vertically in water contained in a tall jar. An excited tuning fork of frequency 500 Hz is held at the upper end of the pipe and the pipe is gradually raised. How many discrete resonance conditions are possible? (Speed of sound in air = 330 m/s)
(a) 1 (b) 2 (c) 3 (d) 4 (e) 5
The arrangement mentioned in this problem makes a simple resonance column apparatus. The wave length of sound emitted by the fork, λ = v/n = 330/500 = 0.6666 m.
The first resonance (fundamental mode) is obtained when the exposed length of the pipe is λ/4. The second resonance is obtained when the exposed length is 3 λ/4. These two are definitely possible since the length of the pipe is 1.1 m and λ = 0.6666 m.
The third resonance will be obtained when the exposed length is 5λ/4 = 5×0.6666/4 = 0.83 m. This too is possible.
The fourth resonance will be obtained when the exposed length of the pipe is 7λ/4 = 7×0.6666/4 = 1.16 m. This is not possible since the length of the entire pipe is 1.1 m only.
So, the correct option is (c).

MCQ on Communication Systems

Questions on communication systems at the higher secondary/plus two level are simple. Here is a question on optical communication systems:
An optical communication system operates at a wave length of 750 nm. The available channel band width for optical communications is only 1% of the optical source frequency. How many TV signals can the system accommodate if each signal requires a band width of 5 MHz?
(a) 8×10⁵ (b) 7.5×10⁵ (c) 6×10⁵ (d) 5×10⁵ (a) 4×10⁵
The optical source frequency, f = c/λ = 3×10⁸/(750×10^–9)= 4×10¹⁴ Hz.
Total band width available in the system = 1% of 4×10¹⁴ Hz = 4×10¹² Hz.
Therefore, no. of TV signals that can be accommodated = (4×10¹² Hz)/ (5×10⁶ Hz) = 8×10⁵.
Now, consider the following MCQ:
A 9 MHz signal is transmitted from a ground transmitter at a height of 300m. The maximum electron density of the ionosphere is 1.44×10¹². A receiver at a distance of 60 km can receive the signal by
(a) space wave only (b) sky wave only (c) space wave and sky wave (d) satellite transponder only (e) sky wave and satellite transponder
The maximum line of sight distance possible is given by d = √(2Rh) where ‘R’ is the radius of the earth (6400 km) and ‘h’ is the transmitter height. Therefore, d = √(2×6400×10³×300) = 62×10³ m = 62 km.
Reception by space wave is possible since the receiver is at 60 km.
The upper frequency limit for ionospheric reflection (critical frequency) is given by
f_c = 9√N_max = 9√(1.414×10¹²) = 10.8×10⁶ Hz =10.8 MHz.
The transmitter frequency is 9 MHz only so that the waves can be reflected by the ionosphere.
So, reception by sky wave also is possible and the correct option is (c).
[Note that satellite transponder doesn’t come into the picture since the waves cannot penetrate through the ionosphere].
Here is another MCQ:
A photo detector is made using a semiconductor having a band gap of 1.55 eV. The maximum wave length it can detect is nearly
(a) 500 nm (b) 600 nm (c) 750 nm (d) 800 nm (e) 850 nm
If the wave length is too large, the photon energy will be too small and the incident light will not be able to produce charge carriers in the semiconductor. With a given semiconductor therefore, there is an upper limit for the detectable wave length. Note that the product of the wave length in Angstrom and the energy in electron volt of any photon is 12400 (very nearly). So, the 1.55 eV photon has wave length equal to (12400/1.55) Ǻ = 8000 Ǻ = 800 nm. This is the maximum wave length this semiconductor can detect [Option (d)].

JAWAHARLAL INSTITUTE OF POSTGRADUATE MEDICAL EDUCATION AND RESEARCH (JIPMER)- Admission to M.B.B.S. Course- 2007

Jawaharlal Institute of Post-graduate Medical Education and Rrsearch (JIPMER) has invited applications for admission to the first year MBBS course (Session 2007-2008). Request for supply of application form and prospectus by post should reach the REGISTRAR (ACADEMIC) JIPMER, PUDUCHERRY - 605 006 on or before 14th March, 2007, along with a Crossed Demand Draft, drawn in favour of ‘ACCOUNTS OFFICER, JIPMER’, PAYABLE AT PUDUCHERRY (PONDICHERRY-605 006) and self addressed stamped envelope for Rs.50/- & of size 26 cm × 32 cm (to send the prospectus with application to the candidates). The Bank draft should be Rs. 350/- for General candidates and Rs.250/- for Scheduled Caste / Scheduled Tribe candidates.
Filled in Application Form should be sent to the REGISTRAR (ACADEMIC), JIPMER, PUDUCHERRY-605006 so as to reach him on or before 28th March 2007, 4.30 P.M.

PROSPECTUS AND APPLICATION FORM CAN ALSO BE OBTAINED IN PERSON AT THE BANK OF BARODA, EXTENSION COUNTER, JIPMER, PUDUCHERRY-6 ON PAYMENT OF RS.350/- (IN RESPECT OF GENERAL CANDIDATES) AND RS.250/- (IN RESPECT OF SC/ST CANDIDATES) IN CASH DURING OFFICE (BANK) HOURS TILL 28th MARCH, 2007 (WEDNESDAY) ON ALL WORKING DAYS BETWEEN 9.30 AM TO 3.00 PM ( On Saturday between 9.30 am to 12.00 noon).

Candidates can also download the Prospectus and application from the web site www.jipmer.edu and submit the application ‘ONLINE’. However, they should take a print out in A4 Size Paper and affix the Photograph and sign the application and send the same along with the Demand Draft (Rs.350/- for General Candidates and Rs.250/- for SC/ST Candidates drawn in favour of the Accounts Officer, JIPMER, Puducherry-6 (Pondicherry-605 006). The D.D. should be drawn on any Nationalized Bank) payable at Puducherry (Pondicherry - 605 006) and attested copy of Community Certificate in case of SC/ST candidates and Medical Certificate in case of Physically Handicapped candidates (if applicable), so as to reach the REGISTRAR (ACADEMIC), JIPMER, PUDUCHERRY-605006 on or before 28th March 2007 (4.30 PM) (Wednesday).
The Entrance Examination will be conducted on Sunday the 27th May, 2007 from 10.00 a.m. to 12.30 p.m. at the following centres:
(1) Puducherry (Pondicherry) (2) Chennai, (3) Hyderabad, (4) Delhi, (5) Kolkata and (6) Thiruvananthapuram.
Further details can be obtained from the prospectus as well as from the website www.jipmer.edu. Make it a point to visit the site for information updates.

Charged Particles in Magnetic and Electric Fields

Here is a multiple choice question which appeared in AIEEE 2002 question paper:
If an electron and a proton having same momenta enter perpendicular to a magnetic field, then
(a) curved path of electron and proton will be same (ignoring the sense of revolution)
(b) they will move undeflected
(c) curved path of electron is more curved than that of the proton
(d) path of proton is more curved
The radius of the circular path of the electron is obtained by equating the magnetic force to the centripetal force: qvB = mv²/r. The radius ‘r’ is therefore given by r = mv/qB. The radius is therefore directly proportional to the momentum (mv) and inversely proportional to the charge (q) of the particle.

The momenta are given as equal in the problem. Since the proton and the electron have the same charge magnitudes and are moving in the same magnetic field B, they will follow paths of the same radius[Option (a)].
You might have noted that the path of a charged particle in an electric field is generally parabolic. This is because of the fact that in a uniform electric field, the electric force on the particle has the same direction everywhere. The motion is similar to the projectile motion in a gravitational field. Now, consider the following question:
A particle of charge ‘+q’ and mass ‘m’ is projected with a velocity ‘v’at an angle ‘θ’ with respect to the horizontal, in an electric field ‘E’ which is directed vertically downwards. If there are no gravitational or magnetic fields, the horizontal range of the particle is
(a) (v²sin2θ)/E (b) (v²sin2θ)/qE (c) (mv²sin2θ)/E (d) (mv²sin2θ)/qE (e) (qmv²sin2θ)/E

In the case of the motion of a projectile in a gravitational field, the expression for horizontal range is R = (v² sin2θ)/g. In the present case, the gravitaional acceleration ‘g’ is replaced by the acceleration produced by the electric field.
Acceleration produced by the electric field = Force/ Mass = qE/m. The correct option therefore is (d).
The following MCQ appeared in AIIMS 2004 question paper:
The cyclotron frequency of an electron gyrating in a magnetic field of 1 T is approximately
(a) 28 MHz (b) 280 MHz (c) 2.8 GHz (d) 28 GHz
The cyclotron frequency is the frequency with which a charged particle describes circular path in a magnetic field and is given by f =qB/2πm with usual notations. [You can get it this way: qvB = mrω² where ω is the angular frequency. Substituting v = ωr in this, we get ω = qB/m. Frquency f = ω/2π =qB/2πm].

Substituting for the mass and charge of the electron, we have

f = (1.6×10^–19 ×1)/ (2π×9.1×10^–31 ) = 28×10⁹ Hz = 28 GHz.

Questions on Polarisation -Brewster’s law

Most of you might have noted that the transverse nature of light wave was proved by the phenomenon of polarisation. You should remember that sound wave cannot be polarised since it is longitudinal.
You will often find questions based on Brewster’s law in the section on polarisation. When unpolarised light proceeding through a rarer medium is incident at an angle ‘i’ on a denser medium, the reflected beam is plane polarised if tan i = n where ‘n’ is the refractive index of the denser medium with respect to the rarer medium.. This is Brewster’s law. [ The transmitted beam will be partially plane polarised].
Here is a simple question based on Brewster’s law:When unpolarised beam of light is incident on a glass slab, the rflected beam is found to be completely plane polarised. The angle between the reflected beam and the transmitted beam is
(a) 30° (b) 45° (c) 60° (d) 90° (e) dependent on the refractive indexThe correct option is (d). You can easily prove this as follows:
Since n = sin i/sin r, we have tan i = sin i/ sin r so that sin i/ cos i = sin i/sin r. Therefore, cos i = sin r. Therefore, r = 90° – i so that i + r = 90°. With reference to the figure, the angle BOC ( which is the angle between the reflected and transmitted beams) is therefore equal to 90°.
Now consider the following question:
When unpolarised light proceeding through air is incident at an angle of 60° on a transparent slab, the reflected rays are found to be completely plane polarised. The refractive index of the slab and the angle of refraction into the slab are respectively
(a) 1.7, 30° (b) 1.414, 40° (c) 1.5, 30° (d) 1.3, 45° (e) 1.732, 30°If you are in too much hurry, you may pick out option (a). But the correct option is (e). Refractive index, n = tan i = tan 60° = √3 = 1.732. Angle of refraction r = 90° – i = 90° – 60° = 30°.