IIT-JEE 2011 Questions (MCQ) on Simple Harmonic Motion

“Do not worry about your problems with mathematics, I assure you mine are far greater.”

– Albert Einstein

The following two questions on simple harmonic motion were included in the IIT-JEE 2011 question paper. Both questions are simple for those who have mastered the fundamental principles:

(1) A wooden block performs SHM on a frictionless surface with frequency, ν₀. The block carries a charge +Q on its surface. If now a uniform electric field E is switched on as shown, then the SHM of the block will be

(A) of the same frequency and with shifted mean position

(B) of the same frequency and with the same mean position

(C) of changed frequency and with shifted mean position

(D) of changed frequency and with the same mean position

When the uniform electric field is switched on, the positively charged block experiences a constant force in the direction of the electric field and so the mean position of the block is shifted. But the frequency (ν) of oscillation of the block is unchanged since it depends only on the mass m of the block and the force constant k of the spring, in accordance with the relation

ν = (1/2π) [√(k/m)]

The correct option is (A).

(2) A point mass is subjected to two simultaneous sinusoidal displacements in x-direction, x₁(t) = A sin ωt and x₂(t) = A sin(ωt + 2π/3). Adding a third sinusoidal displacement x₃(t) = B sin(ωt + φ) brings the mass to a complete rest. The value of B and φ are

(A) √2 A, 3π/4

(B) A, 4π/3

(C) √3 A, 5π/6

(D) A, π/3

The particle in the above question is subjected to two simultaneous simple harmonic motions (SHM) of the same frequency and amplitude in the same direction. the resultant motion of the particle is simple harmonic as is evident by adding x₁(t) and x₂(t):

x₁(t) + x₂(t) = A sin ωt + A sin(ωt + 2π/3) = A sin(ωt +π/3)

The amplitude of the resultant simple harmonic motion is A itself but the initial phase of the motion is now π/3.

As the particle remains at rest on adding the third simple harmonic motion, x₃(t) = B sin(ωt + φ) the amplitude (B) of the third SHM must be A itself, but it must be 180º (or, π radian) out of phase. In other words, the initial phase (φ) of the third simple harmonic motion must be π/3 + π = 4π/3

The correct option is (B).

You can find a useful post on simple harmonic motion here.

IIT-JEE 2011 – Paragraph Type Multiple Choice (Single Answer) Questions Involving Phase Space Diagrams

“The earth provides enough to satisfy every man’s needs, but not every man’s greed.”

– Mahatma Gandhi

The practice of asking two or three multiple choice questions (often single answer type) based on a given paragraph, is resorted to in many entrance examinations. The following three questions[(1), (2) and (3)] which appeared in IIT-JEE 2011 question paper are relatively simple:

Paragraph for questions (1), (2) and (3)

Phase space diagrams are useful tools in analyzing all kinds of dynamical problems. They are especially useful in studying the changes in motion as initial position and momentum are changed. Here we consider some simple dynamical systems in one–dimension. For such system, phase space is a plane is which position is plotted along horizontal axis and momentum is plotted along vertical axis. The phase space diagram is x (t) vs p(t) curve in this plane. The arrow on the curve indicates the time flow. For example, the phase space diagram for a particle moving with constant velocity is a straight line as shown in the figure. We use the sign convention in which position or momentum upwards (or to right) is positive and downwards (or to left) is negative.

(1) The phase space diagram for a ball thrown vertically up from the ground is

The arrow on the diagram shows the flow of time. Initially the ball is on the ground and its velocity and momentum have maximum positive values (since it moves upwards). At the highest point of its trajectory the displacement of the ball has maximum positive value but its momentum is zero. Finally when the ball just hits the ground, its displacement is zero but its momentum has maximum negative value(since it is moving downwards).

The correct option is obviously (D).

(2) The phase space diagram for simple harmonic motion is a circle centered at the origin. In the figure, the two circles represent the same oscillator but for different initial conditions, and E₁ and E₂ are the total mechanical energies respectively. Then

(A) E₁ = √2 E₂

(B) E₁ = 2 E₂

(C) E₁ = 4 E₂

(D) E₁ = 16 E₂

When the total energy of the simple harmonic motion (SHM) is E₁, the amplitude is 2a and when the total energy is E₂, the amplitude is a, as is evident from the phase space diagrams. The total energy E of a particle of mass m in simple harmonic motion of amplitude A and angular frequency ω is given by

E = ½ mω²A²

So when the amplitude is doubled, the total energy is quadrupled.

Thus E₁ = 4 E₂ as given in option (C).

[The phase space diagram given in the above question represents an undamped simple harmonic motion since respective amplitudes appropriate to the initial conditions remain constant. If the simple harmonic motion is a damped one, the curve will be a spiral, proceeding inwards].

(3) Consider the spring-mass system, with the mass submerged in water, as shown in the figure. The phase space diagram for one cycle of this system is

The correct option is (B) on account of the following reasons:

(i) The amplitude of oscillations must decrease with time [which is not the case in diagrams (C) and (D)] as the viscous forces in water damps the motion of the system.

(ii) When the displacement is maximum positive value (corresponding to maximum positive position co-ordinate), the momentum is zero and is going to increase in magnitude in the negative direction [unlike in the case of diagram (A)].

All India Pre-Medical/Pre-Dental Entrance Examination (AIPMT) 2011 Questions on Magnetic Force

“I have no special talents. I am only passionately curious.”

– Albert Einstein

Two questions involving magnetic force on current carrying conductors and moving charges were included in the AIPMT 2011 (Preliminary) question paper. Here are those questions with solution:

(1) A current carrying closed loop in the form of a right angled isosceles triangle ABC is placed in a uniform magnetic field acting along AB. If the magnetic force on the arm BC is F, the force on the arm AC is

(1) √2 F

(2) –√2 F

(3) – F

(4) F

Since the current loop is placed in a uniform magnetic field, the net force acting on the loop is zero.

[Note that a current loop in a uniform magnetic field will experience a torque but no net force. If the magnetic field is non-uniform there will be a torque as well as a net force].

Since the arm AB of the loop is parallel to the magnetic field, the magnetic force on AB is zero. Since the magnetic force on the arm BC is F (as given in the question), the magnetic force on the arm Ac must be – F (so as to make the net force on the loop zero).

(2) A uniform electric field and a uniform magnetic field are acting along the same direction in a certain region. If an electron is projected in the region such that its velocity is pointed along the direction of the fields, then the electron

(1) will turn towards left of the direction of motion

(2) will turn towards right of the direction of motion

(3) speed will decrease

(4) speed will increase

The magnetic force will be zero since the electron is projected parallel to the field. The electric field will exert a force opposite to the direction of motion of the electron and hence its speed will decrease [Option (3)].

All India Pre-Medical/Pre-Dental Entrance Examination (Preliminary) 2010 (AIPMT 2010) Multiple Choice Questions on Centre of Mass

The following questions (MCQ) on centre of mass were included in the AIPMT 2010 question paper:

(1) A man of 50 kg mass is standing in a gravity free space at a height of 10 m above the floor. He throws a stone of 0.5 kg mass downwards with a speed 2 ms^–1. When the stone reaches the floor, the distance of the man above the floor will be

(1) 20 m

(2) 9.9 m

(3) 10.1 m

(4) 10 m

You can work out this question either by using the concept of centre of mass or by applying the law of conservation of momentum.

The position of the centre of mass of the man-stone system will be unchanged as there are no external forces. If x is the distance of the man from the centre of mass when the stone reaches the floor, we have

50×x = 0.5×10 from which x = 0.1 m

As the centre of mass of the system is initially at a distance of 10 m above the floor, the distance of the man when the stone reaches the floor will be 10 + x = 10.1 m.

[You can apply the law of conservation of momentum and obtain the ‘recoil speed’ v of the man from the equation, 50×v = 0.5×2, from which v = 0.02 ms^–1.

The time taken by the stone to reach the floor is 10/2 = 5 s. During this time the man will move up through a distance 0.02×5 = 0.1 m so that when the stone reaches the floor, the man will be at a distance 10 + 0.1 = 10.1 m above the floor].

(2) Two particles which are initially at rest move towards each other under the action of their internal attraction. If their speeds are v and 2v at any instant, then the speed of centre of mass of the system will be

(1) v

(2) 2 v

(3) zero

(4) 1.5 v

Since the particles are initially at rest, the speed of the centre of mass of the system is zero. Since there are no external forces on the system, the centre of mass will continue to be at rest even though the particles are moving under the action of their internal attraction. Therefore the correct option is (3).

By trying a search for ‘centre of mass’ using the search box provided on this page, you can access all posts on centre of mass on this site.

You will find some additional posts on centre of mass at AP Physics Resources.

Kerala Engineering Entrance [KEAM (Engineering)] 2007 Question on Percentage Error

In continuation of the previous post (dated 23^rd March 2011) related to errors in measurements, I give you today the following question which appeared in KEAM (Engineering) 2007 question paper:

The value of two resistors are R₁ = (6 ± 0.3) kΩ and R₂ =(10 ± 0.2) kΩ. The percentage error in the equivalent resistance when they are connected in parallel is

(a) 5.125 % (b) 2 % (c) 3.125 %

(d) 7 % (e) 10.125 %

When resistances R₁ and R₂ are connected in parallel the equivalent resistance R is given by

R = R₁R₂/(R₁ + R₂)

The fractional error in R is ∆R/R, which you can find by taking logarithm and differentiating:

∆R/R = (∆R₁/R₁) + (∆R₂/R₂) + ∆(R₁ + R₂)/(R₁ + R₂)

[Here ∆R₁ is the error in R₁, ∆R₂ is the error in R₂ and ∆(R₁ + R₂) is the error in (R₁ + R₂). On differentiation the third term is negative but we have taken the sign of the third term as positive since we have to consider the worst case in which the maximum possible error is obtained].

Therefore, ∆R/R = (0.3/6) + (0.2/10) + (0.5/16)

Percentage error in R = percentage error in R₁ + percentage error in R₂ + percentage error in (R₁ + R₂) = (0.3/6)×100 + (0.2/10)×100 + (0.5/16)×100 = 5 % + 2 % + 3.125 % = 10.125 %.

As an extension of the above problem consider the following question:

In the above question what will be the percentage error in the heat produced per second when a current of (4 ± 0.1) ampere is passed through the parallel combination of R₁ and R₂?

(a) 3.5 %

(b) 7 %

(c) 10.125 %

(d) 15.125 %

(e) 20.25 %

The heat H produced is given by H = I²R, where I is the current.

The fractional error in the heat produced per second is

∆H/H = 2×(∆I/I) + ∆R/R

Percentage error in H is 2×(∆I/I) ×100 + (∆R/R) ×100

Therefore, percentage error in H = 2×(0.1/4)×100 + 10.125 = 15.125%

Percentage Change and Percentage Error in a Physical Quantity

“Everything that is really great and inspiring is created by the individual who can labour in freedom.”

– Albert Einstein

The fractional change in a physical quantity Q is (ΔQ)/Q where ΔQ is the actual change in the quantity. The percentage change in the quantity = Fractional change × 100.

Let us consider a few multiple choice questions to make things clear:

(1) A copper wire having resistance 10 Ω is stretched so that its radius decreases by 3%. What is the resistance of this wire after stretching?

(a) 10 Ω (b) 10.3Ω (c) 10.6Ω (d) 11.2Ω (e) 11.8Ω

The resistance R of a wire is given by

R = SL/A where S is the resistivity (specific resistance), L is the length and A is the area of cross section. Since S is constant for a given material, R depends only on L and A. When the radius of the wire is decreased by 3 %, its cross section area A is decreased by 6 % (since A = πr²). The resistance increases by 6% on account of this. When A decreases by 6 %, the length L increases by 6 % since the volume of the wire is unchanged. On account of the 6% increase in length there is an additional increase in resistance by 6%. So the total increase in resistance is 12%. The resistance after stretching = 10 + 1.2 =11.2 Ω [Option (d)]

(2) A nichrome wire has resistance 20 Ω. What will be resistance of another nichrome wire whose length is greater by 5 % and radius is greater by 1%?

(a) 20.2 Ω (b) 20.4 Ω (c) 20.6 Ω (d) 20.8 Ω (e) 21 Ω

There is a 5 % increase in resistance due to the increase in length and a 2 % decrease in resistance due to the increase in radius.

[Note that 1% increase in radius produces 2 % increase in the area of cross section].

The net increase in resistance = 5 % – 2 % = 3 % = 20 × 3/100 = 0.6 Ω. The correct option therefore is (c).

In the above questions, we have considered the changes in certain quantities (length and radius in the above cases). The changes may be positive (increments) or negative (decrements). The final result may increase or decrease because of the individual changes.

(3) A wire has mass m = 0.2 ± 0.002 g, radius r = 0.25 ± 0.005 mm and length L = 6 ± 0.06 cm. The maximum percentage error in the measurement of the density ρ of the material of this wire is

(a) 1 (b) 2 (c) 4 (d) 6 (e) 8

Here the percentage error in mass = (0.002/0.2)×100 = 1 %. Percentage error in radius = (0.005/0.25) × 100 = 2 %. Percentage error in length = (0.06/6)×100 = 1%. Since the density ρ = mass/volume = m/πr²L, the maximum percentage error in density = percentage error in mass + 2×percentage error in radius + percentage error in length = 1+ (2×2) + 1 = 6%.

[The percentage error in r is multiplied by 2 since the power of r is 2. This follows thus:

We have ρ = m/πr²L. On taking logarithms and differentiating, we have

dρ/ρ = dm/m – 0 – (2 dr/r) – dL/L

dρ/ρ, dm/m, dr/r and dL/L are respectively the fractional errors in density, mass, radius and length. Therefore we have

Percentage error in density = Percentage error in mass + 2×percentage error in radius + percentage error in length.

Note that we have taken the signs of all errors as positive since we want to make the errors add up to obtain the maximum percentage error].

Multiple Choice Practice Questions on Electrostatics

“Live as if you were to die tomorrow. Learn as if you were to live for ever.”

– Mahatma Gandhi

If you grasp the fundamental principles well, you will be able to answer multiple choice questions within the stipulated time. If you are confused about the fundamental principles, you will be tempted to waste your time while dealing with even simple questions just because of some simple distractions. See the following questions:

(1) A battery of emf 12 V is connected in series with two initially uncharged capacitors, each of value 100 μF (Fig.). The capacitors are fully charged. If the total energy spent by the battery for charging the capacitors is E joule, the energy of each capacitor is

(a) E

(b) E/2

(c) 2 E

(d) E/4

(e) E/8

If the total charge supplied by the battery is Q coulomb, the total energy E supplied by the capacitor is given by

E = VQ joule where V is the emf of the battery (which is 12 volt).

The two identical 100 μF capacitors in series is equivalent to a single capacitor of capacitance C = 50 μF. The energy of this charged capacitor combination is ½ CV² joule = ½ VQ joule since Q = CV.

Therefore, the energy of the charged capacitor combination is ½ E.

Since there are two identical capacitors in the combination, the energy of each capacitor is E/4.

[So the capacitors have gained a total energy of E/2, even though the battery has delivered a total energy E. What happens to the other half of the energy? Well, it is irrecoverably lost as heat generated in the resistance of the circuit].

(2) Charges of equal magnitude are fixed at the diagonally opposite corners A and C of a non conducting square frame ABCD (Fig.). In case (i) the charges at A and C are positive and a third free negative charge q is moved from B to D. In case (ii) the charges at A and C are negative and a third free negative charge q is moved from B to D. In case (iii) the charges at A and C are of opposite sign and a third free negative charge q is moved from B to D. Pick out the correct statement regarding the work done for moving the charge q:

(a) In case (i) the work done is positive

(b) In case (ii) the work done is positive

(c) In case (i) the work done is negative

(d) In case (iii) the work done is positive

(e) In all cases the work done is zero

In case (i) the potentials at B and D are positive, but of equal value. In case (ii) the potentials at B and D are negative, but of equal value. In case (iii) the potentials at B and D are zero. Therefore, in all cases the potential difference between points B and D is zero so that the work done in moving the charge q from B to D is zero.

You will find some useful questions (MCQ) with solution here.

Two Questions on Zener Diode Voltage Regulators

“It is unwise to be too sure of one’s own wisdom. It is healthy to be reminded that the strongest might weaken and the wisest might err”

– Mahatma Gandhi

Zener diodes, as you know, are widely used as reference voltage elements in a variety of voltage regulator circuits. Today I give you a couple of multiple choice questions involving the use of zener diodes in simple voltage regulators:

(1) In the circuit shown, D₁ is a silicon diode which has a voltage drop of 0.7 V while in full conduction under forward bias. The zener diode D₂ has a breakdown voltage of 6.8 V. What is the current through the 450 Ω resistor?

(a) 1 mA

(b) 10 mA

(c) 20 mA

(d) 100 mA

(e) 0 mA

You can use the voltage drop across a forward biased ordinary silicon diode as a reference voltage in voltage regulator circuits. In the circuit shown, a reverse biased zener diode (of breakdown voltage 6.8 V) and a forward biased silicon diode (of voltage drop 0.7 V) in series make a reference voltage of 7.5 V. The output voltage of the circuit is thus 7.5 V.

Since the input voltage to the regulator circuit is 12 V, the voltage drop across the 450 Ω resistor is 12 V – 7.5 V = 4.5 V.

The current through the 450 Ω resistor is (4.5 V)/(450 Ω) = 0.01 A = 10 mA.

(2) The circuit shown in the adjoining figure is the simplest shunt voltage regulator using a zener diode of breakdown voltage 6 V. What is the power dissipated in the zener diode?

(a) 100 mW

(b) 120 mW

(c) 240 mW

(d) 360 mW

(e) 480 mW

Since the input voltage to the regulator circuit is 10 V and the regulated output voltage is 6 V, the voltage drop across the 40 Ω resistor is 10 V – 6 V = 4 V.

Therefore, the current through the 40 Ω resistor (current limiting resistor) is (4 V)/ 40 Ω = 0.1 A = 100 mA. This is the total current flowing into the parallel combination of the zener diode and the100 Ω load resistor.

The current through the load resistor of 100 Ω is (6 V)/(100 Ω) = 0.06 A = 60 mA.

Therefore, the current flowing through the zener diode is 100 mA – 60 mA = 40 mA.

Power dissipated in the zener diode is 6 V×40 mA = 240 mW.

Kerala Medical Entrance 2008 (KEAM 2008) Questions on Communication Systems

The following multiple choice questions on communication systems appeared in the Kerala Medical Entrance 2008 (KEAM-2008) examination question paper:

(1) A 1000 kHz carrier wave is modulated by an audio signal of frequency range 100-5000 Hz. Then the width of the channel in kHz is

(a)10

(b) 20

(c) 30

(d) 40

(e) 50

Let us assume that the system uses amplitude modulation of the usual double sideband type. (It should have been mentioned in the question)

The channel width is twice the highest modulating signal frequency and is therefore equal to 2×5000 Hz = 10000 Hz = 10 kHz.

[Remember that in the standard AM sound broadcast systems the channel band width allotted to a station is 10 kHz].

(2) If the critical frequency for sky wave propagation is 12 MHz, then the maximum electron density in the ionosphere is

(a) 1.78×10¹²/m³

(b) 0.178×10¹⁰/m³

(c) 1.12×10¹²/m³

(d) 0.56×10¹²/m³

(e) 0.148×10¹²/m³

The critical frequency f_c for reflection by the ionosphere is given by

f_c = 9 N^1/2 where N is the maximum electron number density.

Therefore, N = f_c²/81 = (12×10⁶)² /81 = 1.78×10¹²/m³