Sunday, October 29, 2006

M.C.Q. on A. C. Circuits

Problems in alternating current circuits are generally interesting and if you have a thorough idea about the vector method (phasor method), most problems become quite simple. To illustrate the method, let us consider the following MCQ:
In a series LCR circuit, the voltages across the inductor and capacitor are 18V and 12V respectively. If the output of the alternating voltage source connected to this series LCR circuit is 10V, what is the voltage drop across the resistor?
(a) 6V (b) 8V (c)10V (d) 12V (e) 4V
In the vector method, alternating currents and voltages are treated as vectors. The currents and voltages used here can be peak values or RMS values. Let us take RMS values since all currents and voltages mentioned in alternating current problems are RMS values (by convention) unless specified otherwise. You should note that the current and voltage are in phase in the case of a resistor. But the voltage across an inductor leads the current by π/2 where as the voltage across a capacitor lags behind the current by π/2. This means that the voltage drops (vectors) across an inductor and a capacitor are oppositely directed. The resultant voltage drop across the series LC combination in the above question is therefore 18V-12V= 6V and this voltage vector leads the voltage (vector) across the resistor by π/2. The supply voltage vector (10V) is the vector sum of the voltage drop across the resistor and the net voltage drop across the LC combination. [All these vectors are shown in the figure in which the current in the circuit is considered as a vector I in the positive X-direction. Note that this current vector is common for all the components since the circuit is a series circuit.. The voltage drop (vector VR) across the resistor also is taken along the positive X-direction since the current and voltage are in phase in the case of a resistor. The voltage across the inductor is indicated by the vector VL which leads the current vector I by π/2. It is therefore shown in the positive Y-direction. The voltage across the capacitor is indicated by the vector VC which lags behind the current vector I by π/2. It is therefore shown in the negative Y-direction.]
Therefore we have,
10 = √(VR2 + 62) where VR is the voltage drop across the resistor. From this we obtain VR= 8V.
In general, the supply voltage (V) is given by
V = √[VR2 +(VL-VC)2]
Here is another simple question:
An inductor is connected in series with a resistor. When a 15V, 50 Hz alternating voltage source is connected across this series combination, the voltage drop across the inductor is 5V. The voltage drop across the resistor will be
(a) 15V (b) 20V (c) 10V (d) 2√10 V (e) 10√2 V
The voltage drop vector VR in the case of the resistor is in phase with the current vector. But the voltage drop vector VL in the case of the inductor leads the current vector by π/2. Hence we have two vectors of magnitudes VR volts and 5 volts with an angle π/2 between them. Their vector sum gives the supply voltage vector of magnitude 15 volts. Hence we have,
15 = √ [VR2 + 52], from which VR = √200 = 10√2 volts.
What do you grasp from the above discussion? Certainly this: You cannot add voltage drops in AC circuits as scalar quantities. You have to perform vector addition taking into account the phase relation between the current and the voltage in the components. Here you have to forget your habit of scalar addition you perform in the case of direct current circuits!
The following MCQ appeared in the Karnataka CET (Medical) 2002 question paper:
When 100V D.C. is applied across a coil, a current of 1A flows through it. When 100V, 50Hz A.C. is applied to the same coil, only 0.5 A flows. The inductance of the coil is
(a) 5.5 mH (b) 0.55 mH (c) 55 mH (d) 0.55 H
Instead of giving you the resistance of the coil, you are given the direct voltage and the resulting direct current to enable you to calculate the resistance: R = V/I =100/1 =100 Ω. When the alternating voltage is applied, we have I = V/Z = V/√(R2 + L2 ω2 ) so that 0.5 = 100/√[1002 + L2 ×(100π)2]. Squaring, 0.25 = 104/ (104 + 104π2L2). Since π2 = 10 (nearly), this reduces to 0.25 = 1/(1 + 10L2), from which L = 0.55henry = 55 mH.
Now consider the following multiple choice question:
A series LCR circuit is connected to a 10V, 1 kHz A.C. supply. If L= 40mH, R= 50Ω and the current in the circuit is 0.2A, the voltage drop across the capacitor (in volts) is approximately
(a) 10V (b) 15V (c) 20V (d) 25V (e) 50V

The value of the capacitor is not given and you are asked to find the voltage drop across it. This is a problem involving resonance, indicated by the current of 0.2A (=10/50) which is limited by the resistor only. At resonance, the voltage drop across the capacitor is equal (in magnitude) to the voltage drop across the inductor, which is ILω = 0.2X0.04 × 2π×1000 = 16π = 50V, nearly. Therefore, the correct option is (e).
Let us now discuss the following question which appeared in the AIIMS 2003 question paper:
A capacitor of capacitance 2μF is connected in the tank circuit of an oscillator oscillating with a frequency of 1 kHz. If the current flowing in the circuit is 2 mA, the voltage across the capacitor will be
(a) 0.16V (b) 0.32V (c) 79.5V (d) 159V

This is an example of how simple a question can be sometimes. The voltage drop across the capacitor = I/Cω = (2×10-3 )/(2 ×10-6×2π×1000) =0.16V.
The following MCQ also is a simple one, but some of you may be tempted to give the wrong answer.
In an oscillating LC circuit the maximum charge on the capacitor is Q. The charge on the capacitor when the energy is shared equally between the electric and magnetic fields is
(a) Q (b) Q/2 (c) Q/√3 (d) Q/√2
Energy in the electric field means the energy of the capacitor and the energy in the magnetic field means the energy of the inductor. The total energy output of the oscillator is the maximum energy stored in the capacitor (or inductor), which is equal to Q2/2C. When the energy is shared equally between the electric and magnetic fields, the energy of the capacitor = ½ × Q2/2C. If ‘q’ is the charge on the capacitor in this condition, we have,
q2/2C = ½ × Q2/2C, so that q = Q/√2.
To conclude this post, let us discuss one more question:
In a series LCR circuit the voltage across each of the components L,C, and R is 80V. The supply voltage and the voltage across the LC combination are respectively
(a) 80V, 160V (b) 240V, 0V (c) 160V, 0V (d) 80V, 160V (e) 80V, 0V
Since the voltage across the capacitor and the inductor are equal in magnitude, the circuit is at resonance. Since the voltages (vectors!) across these components are in opposition, the net voltage across them (LC combination) is zero. The entire supply voltage will appear across the resistor. This is equal to 80V, as given in the question. So, the correct option is (e).
Very simple questions may tempt you to be careless while answering. Don’t be careless while answering this simple question which appeared in H.P.P.M.T. 2005 question paper:
An ideal transformer has Np turns in the primary and Ns turns in the secondary. If the voltage per turn is Vp in the primary and Vs in the secondary, Vs/Vp is equal to
(a) 1 (b) Ns/Np (c) Np/Ns (d) (Np/Ns)2
You should note that the magnetic flux linked per turn is the same for the primary and the secondary, since the coupling coefficient between them is almost unity. The rate of change of flux per turn also is the same, which means the induced voltage per turn is the same for the primary and the secondary. Therefore, the correct option is (a).
The following M.C.Q. appeared in Kerala Medical Entrance 2006 test paper:
A 16 μF capacitor is charged to 20 volt potential. The battery is then disconnected and pure 40 mH coil is connected across the capacitor so that LC oscillations are set up. The maximum current in the coil is
(a) 0.2 A (b) 40 mA (c) 2 A (d) 0.4 A (e) 0.8 A
‘Pure’ 40 mH coil means the resistance of the coil is zero. So, there is no energy loss due to Joule heating and the entire initial energy of the capacitor is handed over to the inductor when the current in the circuit is maximum. Therefore we have, ½CV2 = ½ L I2 with usual notations so that I = √[CV2/L] = √[(16×10-6×400)/ (40×10-3)] = 0.4 A

Monday, October 23, 2006

Magnetic Force on Moving Charges

"You may never know what results come of your actions, but if you do nothing, there will be no results."
– Mahatma Gandhi

Most of you might be remembering the expression for the magnetic force ‘F’ on a charge ‘q’ moving with velocity ‘v’ in a magnetic field of flux density ‘B’:
F = qvB sinθ where θ is the angle between v and B. Retain the order qvB. It helps you to remember this expression in its vector form:
F = qv×B
The vector form gives you the magnitude (qvB sinθ) of the force and its direction which is along the direction of the cross product vector v×B. You should remember that the direction of v is the direction of motion of a positive charge. If you have a negatively charged particle such as an electron, you should reverse the direction of v to get the direction of the force.
If there is an electric field (E) also in the region, the net force on the charge is given by
F = q(v×B + E).
This is the Lorentz force equation.
Let us consider the following MCQ which appeared in the IIT-JEE Screening 2003 question paper:
For a positively charged particle moving in XY plane initially along the X-axis, there is a sudden change in its path due to the presence of electric and/or magnetic fields beyond P. The curved path is shown in the XY plane and is found to be non-circular.Which one of the following
combinations is possible? (i,j,k are unit
vectors along X,Y,Z directions)
(a) E = 0; B =bi + c
(b) E = ai; B =ck + ai
(c) E = 0; B =cj + bk(d) E = ai; B =ck + bj
As the curved path is confined to the XY plane,
the component of magnetic field effective in
deflecting the particle is the Z-component only.
The other component should be the X-component since the X-component cannot deflect the charged particle proceeding along the X-direction. There has to be an electric field since the path of the particle is non-circular. All these conditions are satisfied by option (b).
Now consider the following multiple choice questions:
(1)An electron is projected horizontally from south to north in a uniform horizontal magnetic field acting from west to east. The direction along which it will be deflected by this magnetic field is
(a) vertically upwards 

(b) vertically downwards 
(c) northwards 
(d) southwards 
(e) eastwards
Since the electron is negatively charged, the direction of its velocity is to be reversed for finding the direction of the cross product vector v
×B. The correct option therefore is (a). You can use Fleming’s left hand rule (motor rule) also for finding the direction of the magnetic force. But, when you apply the rule, remember that the direction of the conventional current is that of positive charge.
(2) An electron enters a uniform magnetic field of flux density B=2i+6j-√8 k with a velocity V=i+3j-√2 k where i,j and k are unit vectors along the X, Y and Z directions respectively. Then
(a) both speed and path will change 

(b) speed alone will change 
(c) the path will become circular 
(d) the path will become helical (e) neither speed nor path will change
The vectors B and V are parallel as the components of B are twice the components of V. So the magnetic force on the electron is zero and the correct option is (e).
Generally, in problems of the above type, you will have to find the angle between the vectors V and B using cosθ = V.B/ VB. In the present case we have, cosθ = (2+18+4)/ √(12×48) = 1 so that θ= zero. If in a similar problem, the value of θ works out to be 90˚(if cos
θ works out to be zero), the correct option would be (c). If the value of θ were neither zero nor 90˚, the path would be helical and the correct option would be (d).
(3) Doubly ionized atoms X and Y of two different elements are accelerated through the same potential difference. On entering a uniform magnetic field, they describe circular paths of radii R1 and R2. The masses of X and Y are in the ratio
(a) R1/R2 (b) √(R1/R2) (c) √(R2/R1) (d) (R2/R1)2 (e) (R1/R2)2

Since both atoms are doubly ionized, they have the same charge ‘q’ and since they are accelerated by the same potential difference ‘V’, they have the same kinetic energy, qV. If m1 and m2 are the masses and v1 and v2 are the velocities of X and Y respectively, we have, ½ m1v12 = ½ m2v22 from which m1/m2 = v22/v12 = [qBR2/m2]2/ [qBR1/m1]2, on substituting for the velocities from the centripetal force equation, mv2/R = qvB.
Therefore, m1/m2 = (R1/R2)2, given by option (e).

Thursday, October 19, 2006

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).
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.
The following simple question appeared in the 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)].
The following question appeared in 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].

Sunday, October 15, 2006

All India Pre-Medical/Pre-Dental Entrance Examination (AIPMT) 2007

Central Board of Secondary Education (CBSE), Delhi has announced the dates pertaining to the All India Pre-Medical/Pre-Dental Entrance Examination (AIPMT) 2007 for admission to 15% of the total seats for Medical/Dental Courses in all Medical/Dental colleges run by the Union of India, State Governments and Municipal or other local authorities in India except in the States of ANDHRA PRADESH AND JAMMU & KASHMIR. The dates of the Examination are:
(1) Preliminary Examination : 1st April 2007 (Sunday)
(2) Final Examination : 13th May 2007 (Sunday)
The Information Bulletin and Application Form costing Rs.400/- for General Category candidates and Rs.200/- for SC/ST candidates inclusive of examination fee can be obtained against Cash Payment from designated branches of Canara Bank and Regional Offices of CBSE up to 18-11-2006. Designated branches of Canara Bank in Kerala are:
KOTTAYAM : P.B. No.122, K.K.Road, Kottayam-686 001
QUILANDY : Fasila Buidling, Main Road, Quilandy-673 305
TRIVANDRUM : Ist Floor, Ibrahim Co. Bldg., Challai, Trivandrum-695 023
TRIVANDRUM : Plot no.2, PTP Nagar Trivandrum-695 038
TRIVANDRUM : TC No.25/1647, Devaswom Board Bldg. M.G. Road, Trivandrum-695 001
ERNAKULAM : Shenoy’s Chamber, Shanmugam Road, Ernakulam, Cochin-682 031
CALICUT : 9/367-A, Cherooty Road, Calicut-673 001
TRICHUR : Trichur Main Ramaray Building, Round South, Trichur-680001
QUILON : Maheshwari Mansion, Tamarakulam, Quilon-691 001
PALGHAT : Market Road, Big Bazar, 20/68, Ist floor,Palghat-678 014.

The Information Bulletin and Application Form can also be obtained by Speed Post/Registered Post by sending a written request with a Bank Draft/Demand Draft for Rs.450/- for General Category and Rs.250/- for SC/ST Category payable to the Secretary, Central Board of Secondary Education, Delhi along with a Self Addressed Envelope of size 12” x 10”. The request must reach the Deputy Secretary (AIPMT), CBSE, 2, Community Centre, Preet Vihar, Delhi-110 092 on or before 10-11-2006. The request should be super scribed as “Request for Information Bulletin and Application Form for AIPMT, 2007”.
Completed Application Form is to be dispatched by Registered Post/Speed Post only. Last Date for receipt of completed Application Forms in CBSE is 20-11-2006. You can obtain complete information at

Saturday, October 14, 2006

M.C.Q. on Circular Motion, Moment of Inertia & Rigid Body Rotation

The section on rotational motion may appear to be uninteresting to some of you, but your attitude is to be corrected. Once you understand the basic principles, this section becomes interesting and easy to score high marks. Let us begin with a M.C.Q. meant for checking your understanding of basic principles:
A small pendulum bob of mass ‘m’ is tied to one end of a light inextensible string and is revolved in a vertical circle of radius ‘r’. If T is the tension in the string and ‘v’ is the speed of the stone at the highest point, the net force on the stone at the highest point is
(a) mg+T (b) T-mg (c) mg (d) mg + T + mv2/r (e) mg – T +mv2/r

Even though this is a simple question, you are likely to pick out the wrong answer if you have not understood the basic things. The net force is the centripetal force acting on the bob and is equal to mg + T. You should note that at the highest point both the tension and the gravitational pull are towards the centre of the circle and they add up to produce the net force.
If you were asked to find out the net force at the lowest point of the circle, your answer would be (T-mg) because T acts towards the centre (upwards) and mg as usual acts downwards.
Suppose the speed of the bob at the highest point were just sufficient to keep the bob moving along the circle. Then T would be zero. The speed v1 of the bob will be given by mv1 2 /r = mg since the centripetal force is then supplied by the gravitational pull alone. This critical speed at the top of the circle is therefore v1 = √(rg). What will be the speed of the bob at the bottom of the vertical circle if the bob were just to move along the circular path? The kinetic energy of the bob at the bottom is higher by mg×2r because it loses gravitational potential energy when it falls from top to bottom. The total kinetic energy at he bottom is ½ mv12 + mg×2r = ½ mgr + 2mgr = (5/2) mgr. If v2 is the speed of the bob at the bottom, we have ½ mv22 = (5/2) mgr. The critical speed at the bottom of the circle is
v2 = √(5rg). You should remember this: If a body is to move in a vertical circle it should have a minimum horizontal speed of √(5rg) at the bottom of the circle. So if you arrange a simple pendulum and push the bob to have a horizontal speed of √(5rg), the bob will just move along a vertical circle.
When the bob is moving along a vertical circle, the difference between the tensions in the sting when the bob is at the top and bottom of the circle is 6mg. You may prove this by way of a simple exercise for you.
Now consider the following mcq which will be quite easy for you now:
A small steel sphere is suspended using light inextensible string to form a simple pendulum. This bob is initially at rest. What is the minimum speed with which another identical steel sphere should hit the pendulum bob so that the bob just moves along a vertical circle? Treat the collision as elastic.
(a) √(rg) (b) √(2rg) (c) √(2.5rg) (d) √[(5/4)rg] (e) √(5rg)
The correct option is the last one: √(5rg). Since the spheres are identical, the sphere which is moving will come to rest at the instant of hitting the pendulum bob, transferring entire kinetic energy to the bob and the bob will move forward with the same speed. Since the critical speed (at the bottom) for the motion along a vertical circle is √(5rg), this is the minimum speed of the moving sphere.
Consider another multiple choice question:
A stone of mass ‘m’ is projected with a velocity ‘u’ making an angle of 45˚with the horizontal. The magnitude of the angular momentum of the projectile about an axis perpendicular to the plane of projection and passing through the point of projection is
(a) mu2/√2g (b) mu3/4√2 g (c) mu3/√2 g (d) mu2/2g (e) zero
The maximum height ‘H’ reached by the projectile is (u2sin2θ)/2g with usual notations. The velocity of the projectile at the maximum height is ucosθ, which is the horizontal component of its velocity that remains unaltered through out its motion. The magnitude of the angular momentum of the projectile at the highest point is = magnitude of linear momentum×lever arm = mucosθ×H = mucosθ × u2sin2θ/2g = mu3cos45×(sin245)/2g = mu3/4√2 g [Option (b)].
Let us now disuss the following questions which appeared in Kerala Engineering Entrance 2001 test paper:
(1) If ‘I’ is the moment of inertia and ‘E’ is the kinetic energy of rotation of a body, then its angular momentum will be
(a) √(EI) (b) 2EI (c) E/I (d) √(2EI) (e) EI
Since the rotational kinetic energy E = ½ I ω2 we can write E = ½ I2 ω2/I from which the angular momentum Iω = √(2EI).
It will be convenient If you remember the expression for rotational kinetic energy as E = L2/2I where L is the angular momentum. This equation is similar to that for translational kinetic energy in the form E = p2/2m where ‘p’ is the linear momentum.
(2) A circular thin disc of mass 2 kg has a diameter 0.2 m. Calculate the moment of inertia about an axis passing through the edge and perpendicular to the plane of the disc (in kgm
(a) 0.01 (b) 0.03 (c) 0.02 (d) 3 (e) 2
The moment of inertia of the disc about its central axis, perpendicular to its plane is MR2/2. Therefore the moment of inertia about a parallel axis passing through its edge, on applying the parallel axis theorem is MR2/2 + MR2 = (3/2)MR2 = (3/2) ×2×(0.1)2 = 0.03 kgm2.
(3) A torque of 50 Nm acting on a wheel at rest rotates it through 200 radians in 5 seconds. Calculate the angular acceleration produced (in rad/sec2)
(a) 8 (b) 4 (c) 16 (d) 12 (e) 10
The torque given in the question is not required for working it out. It just serves the purpose of a distraction. Just as you remember the equation, s = ut + ½ at2 in linear motion, you should remember its angular counter part as θ = ω0t + ½ αt2 where θ is the angular displacement, ω0 is the initial angular velocity (which is zero in the problem), and α is the angular acceleration. Substituting the values given, we have, 200 = 0 + ½×α×25 from which α = 16 rad/sec2.
The following M.C.Q. appeared in the All India Pre-Medical/Dental Entrance Exam.(C.B.S.E.)-2004 question paper:
A round disc of moment of inertia I2 about its axis perpendicular to its plane and passing through its centre is placed over another disc of moment of inertia I1 rotating with an angular velocity ‘ω’ about the same axis. The final angular velocity of the combination of discs is
(a) I2ω/(I1 + I2) (b) ω (c) I1ω/(I1+ I2) (d) (I1 + I2)ω/I1
This is a simple question, the answer of which is based on the law of conservation of angular momentum. The initial angular momentum of the system is I1 ω. The final angular momentum is (I1 + I2)ω' where ω' is the final angular velocity of the combination of the discs. On equating the initial and final angular momenta, we obtain, ω' = I1ω/(I1 + I2).
The following question has appeared in different forms in Medical and Engineering Entrance tests:
If the earth suddenly shrinks to one-eighths its present volume without changing its mass and spherical shape, the duration of the day will
(a) decrease by 6 hours (b) increase by 6 hours (c) decrease by 12 hours (d) decrease by 18 hours (e) remain unchanged

When the volume of a sphere becomes one-eighths, its radius becomes half. The angular momentum of the earth is conserved in spite of the shrinkage so that we have I1ω1= I2ω2 where I1and I2 are the moments of inertia and ω1 and ω2 are the angular velocities of the earth before and after the shrinkage respectively. Substituting for I1 [= (2/5)MR2] and I2 [= (2/5)M (R2/4)] we obtain ω2 = 4ω1. Since the angular velocity changes to 4 times the initial value, the spin period of the earth (T= 2π/ω) changes to one-fourth of the initial value. So, the duration of the day will become 24/4 = 6 hours. The duration therefore decreases by 18 hours.
Generally, if the radius of the earth becomes ‘n’ times the present value, the duration of the day becomes 24n2 hours.
Let us now consider the following interesting question which appeared in the IIT Screening 2000 question paper:
A long horizontal rod has a bead which can slide along its length, and initially placed at a distance ‘L’ from one end A of the rod. The rod is set in angular motion about A with constant angular acceleration ‘α’. If the coefficient of friction between the rod and the bead is ‘μ’, and gravity is neglected, then the time after which the bead starts slipping is
(a) √(μ/α) (b) (μ/α) (c) 1/√(μα) (d) infinitesimal
The bead gets pressed against the rod because of the tangential acceleration of the rod at the location of the bead. The force with which the bead presses against the rod is F = ma = mLα. where ‘m’ is the mass of the bead and ‘a’ is the tangential acceleration which is equal to Lα. The frictional force therefore is μmLα. It is the centrifugal force which tries to push the bead outwards. Just when the slipping begins, we have μmLα = mv2/L. Initially ‘v’ is zero and as time elapses, its value increases in accordance with the equation, v = 0 + at = Lαt where ‘t’ is the time at which slipping begins. Substituting this value of ‘v’ in the above condition for slipping, we obtain μ = αt2 from which t = √(μ/α).

Wednesday, October 11, 2006

Two Kerala Entrance 2006 Questions on Electrostatics


The following mcq appeared in the Kerala Engineering Entrance
Test 2006 question paper:
A network of six capacitors, each of value C, is made as shown in Fig 1. The quivalencapacitance between the points A and B is
(a) C/4 (b) 3C/4 (d) 3C/2 (d) 3C (e) 4C/3
The network shown in Fig 1 is redrawn in the second figure shown side by side with Fig1. You can easily understand that the network has two identical branches, each containing a parallel combination of two capacitors in series with a single capacitor.The parallel combined value 2C in series with C makes a net value 2C.C/(2C+C) = 2C/3. The parallel combined value of the two branches is the equivalent capacitance between A and B and is equal to 4C/3 [Option (e)].
The following mcq appeared in the Kerala Medical Entrance Test 2006 question paper:
The total electrical flux leaving a spherical surface of radius ‘r’ metre enclosing an electric dipole of charge ‘q’ is
(a) zero (b) q/ε0 (c) 8π
r2q/ε0 (d) 2q/ε0 (e) 4π r2q/ε0

This is a very simple question which you will occasionally get. Since the closed surface contains a dipole, the electric flux leaving the surface is zero. Note that the outward flux due to the positive charge of the dipole is balanced by the inward flux due to the negative charge of the dipole.

Monday, October 09, 2006

Multiple Choice Questions on Electrostatics:

The following M.C.Q. appearered in the question paper of A.I.E.E.E.2004 (It may appear to be time consuming on the first reading, but it is simple):Two spherical conductors B and C having equal radii and carrying equal charges on them repel each other with a force F when kept apart at some distance. A third spherical conductor A having the same radius as that of B but uncharged is brought in contact with B, then brought in contact with C and finally removed away from both. The new force of repulsion between B and C is
(a) F/4 (b) 3F/4 (c) F/8 (d) 3F/8
If Q is the initial charge on B and C and ‘r’ is the distance between their centres, the initial repulsive force between B and C is F= (1/4πε0)(Q2/ r2). When the uncharged sphere A is brought in contact with B, they share the charge Q and each will have a charge Q/2 since they are identical spheres. When the sphere A is then brought in contact with C they will share the total charge Q + Q/2 = 3Q/2 equally and each will have a charge 3Q/4. On removing the sphere A we have sphere B with charge Q/2 and the sphere C with charge 3Q/4. The new repulsive force between B and C is evidently (1/4πε0)(Q/2)(3Q/4)/r2 = (1/4πε0)3Q2/8r2 = 3F/8 [Option (d)].
Take a special note of the following M.C.Q. which often finds a place in entrance test papers:
A charge ‘q’ is placed at the centre of the line joining two equal point charges, each equal to +Q. This system of three charges will be in equilibrium if 'q' is equal to
(a) +Q (b) +Q/2 (c) –Q/2 (d) +Q/4 (e) –Q/4

+Q____ q_____+Q

The force on the charge ‘q’ placed at the centre will be zero irrespective of the sign of ‘q’. For the system to be in equilibrium, the net force on the charge +Q should be zero. Therefore, (1/4πε0)[Q2 /d2 + Qq/(d/2)2] = 0 where ‘d’ is the separation between the charges Q&Q. From this we get q = -Q/4 [Option (e)].The following M.C.Q is quite simple because of the symmetry in the arrangement of charges. Similar questions can be seen in entrance test papers:Equal charges of value -Q each are arranged at the eight vertices of a non-conducting skeleton cube of side ‘a’. If a point charge +Q is placed at the centre of the cube, the electrostatic force exerted by the eight negative charges on the positive charge at the centre is
(a) Q2/3πε0a2 (b) 4Q2/3πε0a2 (c) 8Q2/3πε0a2 (d) 16Q2/3πε0a2 (e) zero

The net electrostatic force on the charge Q at the centre is zero since the force on Q due to each negative charge is balanced by the force due to the negative charge at the diagonally opposite corner of the cube.
The following M.C.Q. which appeared in the A.I.E.E.E.2004 question paper is worth noting:

Four charges equal to –Q are placed at the four corners of a square and a charge ‘q’ is at the centre.If the system is in equilibrium, the value of ‘q’ is
(a) –Q(1+2√2)/4 (b) Q(1+2√2)/4 (c) –Q(1+2√2)/2 (d) Q(1+2√2)/2
The system will be in equilibrium if the attractive force on the charge –Q at a corner due to the central charge ‘q’ is balanced by the net repulsive force on the same charge –Q due to the remaining three charges (-Q each) at the remaining three corners. Hence we have (if ‘a’ is the side of the square),
(1/4πε0) [Qq/(√2 a/2)2] = (1/4πε0) [Q2/(√2 a)2] + √2 × [(1/4πε0) ×(Q2/a2)].
The term on LHS is the attractive force between –Q and q which are separated by the distance half of √2 a. The first term on RHS is the repulsive force between –Q and the diagonally opposite charge –Q (which are separated by √2 a). The second term on RHS is the net repulsive force between –Q and the remaining two charges (-Q each).
Hence we obtain 4q = Q + Q(1+2√2) from which q = Q(1+2√2)/4 [Option (b)].

The following question was asked at the Kerala Medical Entrance test of 2006:Three identical charges each of 2μC are placed at the vertices of a triangle ABC. If AB+AC=12cm and AB.AC=32cm2, the potential energy of the charge at A is
(a) 1.53J (b) 5.31J (c) 3.15J (d) 1.35J (e) 3.51J

The electrostatic potential energy of the charge at A ia (1/4πε0)(Q.Q/AB + Q.Q/AC) since the charges are of the same value, Q=2μC. Since AB.AC=32, AB = 32/AC. Substituting this in the equation, AB + AC = 12, we obtain 32/AC + AC = 12. Rearranging, (AC)2 – 12AC + 32 = 0, which yields AC = 8 or 4. Since AB.AC = 32 cm2, if AC = 8 cm, AB = 4 cm and vice versa.
Therefore potential energy = (1/4πε0)[(4×10-12)/(4×10-2) +(4×10-12)/(8×10-2)] joule.
Since 1/4πε0 =9×109, the potential energy works out to 1.35 J, given in option (d).
Now consider the following M.C.Q.:
Two isolated copper spheres of radii 3cm and 6cm carry equal positive charges of 30 units each. If they are connected by a thin copper wire and then the wire is removed, what will be the charge on the smaller sphere?
(a) 60 units (b) 40 units (c) 30 units (d) 20 units (e) 10 units
The total charge (Q1+Q2) in the system is 60 units. Since the connecting wire is thin, its capacitance can be neglected. Even though the potentials of the spheres are different initially, their potentials will become the same when they are connected by the copper wire.
The capacitance of a spherical conductor being 4πε0 r, the charges on the spheres will be directly proportional to their radii, since Q = CV and V is the same. Therefore we have Q1/Q2 = 3/6 = 1/2 and Q1+Q2 = 60. So, Q1 = 60×1/3 = 20 units. The potential of the spheres will change after removing the connecting wire; but the charges on them will not change. So, the correct option is (d).
Let us discuss another M.C.Q.:
An infinite number of charges each equal to –q coulomb are placed on a straight line at x = 1m, 2m, 4m, 8m,16m,…….. What will be the potential at x = 0 due to these charges?
(a) infinite (b) 2q/4πε0 (c) –q/4πε0 (d) –2q/4πε0 (e) zero
The potential is certainly negative since a negative charge will produce negative potential only. You should supply the sign of the charge in the expression for the net potential. The potential at x = 0 is given by, V = (-q/4πε0)(1+ ½ + ¼ + 1/8 + 1/16 + ………). Since the infinite series yields a value equal to 2, the answer is –2q/4πε0.
Very simple questions are often likely to mislead even very bright students. Be careful. Here is a very simple question:Nine negative charges, each of magnitude Q are arranged symmetrically along the circumference of a circle of radius R. The electric field at the centre of the circle is
(a) (1/4πε0) Q/R2 (b) (1/4πε0) (9Q/R2) (c) -(1/4πε0) (9Q/R2)
(d) -(1/4πε0) (9Q/R2) cos(2π/9) (e) zero
When you place a positive test charge at the centre of the circle, it will experience zero net force since it is pulled equally by the nine charges arranged symmetrically all around. Therefore, the electric field at the centre is zero.
You will find all the posts on Electrostatics in this site by clicking on the label ‘electrostatics’ below this post.

Thursday, October 05, 2006

Multiple Choice Questions (MCQ) on electromagnetic induction

Questions based on electromagnetic induction are simple and interesting. Usually you will get questions from this section in any Medical and Engineering Entrance Test. Consider the following two questions:
(1) Lenz’s law is a consequence of the law of conservation of
(a) linear momentum (b) charge (c) angular momentum (d) energy (e) none of the above

According to Lenz’s law, the induced current will always oppose the change which causes it. Indeed, the induced current has to oppose the change because, otherwise the change which causes the current will persist and the current will continue to flow once it is started. You will then have a supply of energy (in the form of electric current) without any external agency doing any work. This will then violate the law of conservation of energy, which is impossible. So, Lenses law holds good in accordance with the law of conservation of energy. Option (d) is the correct answer.
(2) The electrical entity inductance can be compared to the mechanical entity
(a) energy (b) impulse (c) momentum (d) torque (e) inertia

The correct option is (e). Inertia in Mechanics is the property by which a body (or, a mechanical system) tries to oppose any change in its state of rest or of uniform motion. Inductance is the property by which an electric circuit tries to oppose any change of current flowing in it. So, inductance and inertia are comparable.
The above two questions high light two basic points. Now consider the following:
A bar magnet is released into a copper ring which is directly below it. What about the acceleration of the magnet? Greater than ‘g’ or equal to ‘g’ or less than ‘g’?
The acceleration is less than ‘g’ since the falling magnet will generate an induced current in the copper ring and the induced current will oppose the motion of the magnet.
Now, consider the following M.C.Q.:
A jet plane is flying horizontally at a speed of 1800 km/hour. What is the potential difference developed between the tips of its wings if the wing span is 25m? Earth’s magnetic field at the location is 0.4 gauss and the angle of dip is 30˚.
(a) 25mV (b) 250mV (c) 500mV (d) 2.5V (e) 5V
The motional emf developed between the tips of the wings is given by V = BvLv where Bv is the vertical component of the earth’s magnetic flux density, L is the distance between the tips of the wings (wing span) and ‘v’ is the velocity. [Note that this emf is produced because of the cutting of the vertical field lines and this is why we use the vertical component of the field]. We have Bv = Bsin30 = 0.4×10-4×½ = 0.2×10-4 tesla. Also, L = 25m and v = 500m/s. The emf then works out to be 0.25V = 250mV. [Note that gauss is the cgs unit of magnetic flux densitywhich is often used. One tesla = 104 gauss].
Let us consider another question involving Faraday’s disc, the first electric generator:
A circular copper disc 10cm in diameter rotates 1800 times per minute about a central axis at right angles to the plane of the disc. A uniform magnetic field of 1 tesla is applied perpendicular to the plane of the disc. The voltage induced between the centre and the edge of the disc is
(a) 0.235V (b) 0. 47V (c) 2.35V (d)4.7V (e) zero

The motional emf induced when a conductor moves perpendicular to a magnetic field is the product of the area swept per second (by the conductor) and the magnetic field. Therefore, in the present case, induced voltage = area swept per secong by a radius × B = n πr2B = (1800/60) × π ×(0.05)2 ×1= 0.2356V. The correct option therefore is (a).
Let us modify this question as follows:
A copper rod 10cm long rotates 1800 times per minute about an axis passing through one end at right angles to the rod. A uniform magnetic field of 1 tesla is applied perpendicular to the plane of rotation. The voltage induced between the ends of the rod is
(a) 0.235V (b) 0. 942V (c) 2.35V (d)4.7V (e) zero
This mcq is similar to the previous one. The voltage induced will be four times the previous value since the area swept is four times. The correct option is (b).
Now let us modify the above ‘rod problem’ as follows:
An aluminium rod 10 cm long rotates 1800 times per minute about an axis passing through its centre at right angles to it. A uniform magnetic field of 1 tesla is applied perpendicular to the plane of rotation. The voltage induced between the ends of the rod is
(a) 0.235V (b) 0. 47V (c) 2.35V (d)4.7V (e) zero
This is a simple question but there is chance of committing a mistake! The correct option is neither (a) nor (b). The potential at the ends (with respect to the mid point of the rod) will be the same (0.235V) so that the potential difference between the ends will be zero [Option (e)].
You should note that the motional emf is generated because of the shifting of mobile charge carriers due to the Lorentz force.
The following two questions (mcq) appeared in the Kerala Engineering Entrance test paper of 2006:
(1) A copper disc of radius 0.1m is rotated about its centre with 20 revolutions per second in a uniform magnetic field of 0.1T with its plane perpendicular to the field. The emf induced across the radius of the disc is
(a) π /20 volt (b) π /10 volt (c) 20π milli volt (d) 10π milli volt (e) 2π milli volt
The emf induced = n πr2B = 20π(0.1)2 ×0.1 = 0.02 π volt = 20π milli volt [Option (c)].
(2) A varying magnetic flux linking a coil is given by Φ = Xt2. If at time t=3s, the emf induced is 9V, then the value of X is
(a) 0.66 Wb.s-2 (b) 1.5 Wb.s-2 (c) -0.66 Wb.s-2 (d) -1.5 Wb.s-2 (e) -0.33 Wb.s-2
The induced emf = -dΦ/dt. Therefore, -2Xt = 9 when t=3s so that -6X = 9, from which X = -1.5. The correct option is (d).