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

“I object to violence because when it appears to do good, the good is only temporary; the evil it does is permanent.”

– Mahatma Gandhi

The following question which appeared in IIT JEE 2010 question paper is worth noting:

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

When the system is put in a uniform magnetic field of strength B going into the plane of the paper, the wire takes the shape of a circle. The tension in the wire is

(a) iBL

(b) iBL /π

(c) iBL /2π

(d) iBL /4π

Considering an elemental length dL of the wire, the magnetic force F acting on the element is given by

F = idLB = iRdθB, as is clear from the figure.

The tension T developed in the wire is resolved into rectangular components T sin(dθ/2) and T cos(dθ/2) as indicated in the figure.

The magnetic force F is related to the tension T as

F = iRdθB = 2T sin(dθ/2)

Since dθ is a very small angle, sin(dθ/2) ≈ dθ/2

Therefore, the above equation gives T = iRB

But R = L /2π

Therefore, T = iBL /2π

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

Two straight, infinitely long parallel wires P and Q carry equal currents (I) in opposite directions (fig.). A square loop of side a carrying current i is arranged symmetrically in between the straight wires so that the plane of the loop is in the plane containing the wires P and Q. What is the torque acting on the current loop?

(a) Zero

(b) μ₀Iia²/2πd

(c) (μ₀Iia²/2π)[1/(d+a)]

(d) (μ₀Iia²/2π)[1/d + 1/(d+a)]

(e) (μ₀Iia²/2π)[1/d – 1/(d+a)]

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

Apply for Joint Entrance Examination 2011 (JEE 2011) for Admission to IITs

Notification regarding the Joint Entrance Examination 2011 (JEE 2011) for Admission to IITs has been issued. This Joint Entrance Examination is for admission to the under graduate programmes in the IITs at Bhubaneswar, Bombay, Delhi, Gandhinagar, Guwahati, Hyderabad, Indore, Kanpur, Kharagpur, Madras, Mandi, Patna, Rajasthan, Roorkee and Ropar as well as at IT-BHU Varanasi and ISM Dhanbad.

Important Dates for IIT JEE 2011:

November 01 to December 15, 2010: Online application process
November 12 to December 15, 2010: Sale of Off-line application forms.

Last date for receiving application forms (Off line or printed copy of Online) at IITs: 17:00 hrs, Monday, December 20, 2010

Date of Exam of IIT JEE 2011:

Sunday, April 10, 2011 Paper 1: 09:00 to 12:00 hrs;

Paper 2 : 14:00 to 17:00 hrs

All information regarding online application and offline application and other important details about IIT JEE 2011 can be obtained from the following websites:

IIT Bombay: http://www.jee.iitb.ac.in
IIT Guwahati: http://www.iitg.ernet.in/jee/
IIT Delhi: http://jee.iitd.ac.in
IIT Kharagpur: http://www.iitkgp.ernet.in/jee
IIT Kanpur: http://www.jee.iitk.ac.in
IIT Roorkee:http://jee.iitr.ernet.in/

IIT Madras: http://jee.iitm.ac.in

Multiple Choice Questions on Elastic and Inelastic Collisions

Today we will discuss two questions on collisions. The first question pertains to elastic collision in one dimension and the second question pertains to inelastic collision in two dimensions:

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

(a) 9 m

(b) 6 m

(c) 3 m

(d) 2 m

(e) m/3

Since the kinetic energy of A after collision is one-nineth of its initial kinetic energy, the momentum of A after collision is one-third of its initial momentum.

Since the momentum is to be conserved, we have

p = p’ – p/3 where p is initial momentum of A and p’ is the momentum of B after the collision.

[The final momentum of particle A is negative since its direction is reversed].

Therefore, p’ = 4p/3

The kinetic energy gained by particle B due to the collision is p’²/2M where M is the mass of particle B.

The kinetic energy lost by particle A due to the collision is (8/9)×p²/2m.

[Note that the initial kinetic energy of particle A is p²/2m and its final kinetic energy is (1/9) p²/2m].

Since the kinetic energy too is conserved in elastic collisions, the kinetic energy gained by particle B is equal to the kinetic energy lost by particle A. Therefore, we have

p’²/2M = (8/9) p²/2m

Substituting for p’ = 4p/3, we have

(16/9) (p²/2M) = (8/9) p²/2m from which M = 2 m.

(2) A particle of mass m moving westwards with speed v collides with an identical particle moving northwards with speed v. If they stick together after the collision, what is their common speed after the collision?

(a) 4v

(b) 2√2 v

(c) 2v

(d) √2 v

(e) v/√2

The total momentum before collision is the vector sum of the initial momenta mv and mv at right angles (since one is westwards and the other is northwards) and is equal to √2 mv acting along the north-west direction (Fig.).

The combined mass is 2m. If V is the common speed, the total momentum after collision is 2m V, acting along the north-west direction.

Since the momentum is conserved, we have

2m V = √2 mv so that V = v/√2

AIEEE 2010 Questions (MCQ) from Nuclear Physics

“I am a friend of Plato, I am a friend of Aristotle, but truth is my greater friend”

– Sir Isaac Newton

Multiple choice questions based on a given paragraph are seen in many entrance exam question papers. Here are two such questions on nuclear physics (which appeared in AIEEE 2010 question paper):

Directions : Questions (i) and (ii) are based on the following paragraph:

A nucleus of mass M + Δm is at rest and decays into two daughter nuclei of equal mass M/2 each. Speed of light is c.

(i) The speed of daughter nucleus is

(1) c√[Δm/(M + Δm)]

(2) c [Δm/(M + Δm)]

(3) c√(2Δm/M)

(4) c√(Δm/M)

The energy released in the process is Δmc² in accordance with Einstein’s mass-energy relation. This energy is carried by the daughter nuclei (as kinetic energy).

Therefore we have

½ (M/2)v² + ½ (M/2)v² = Δmc² where v is the speed of each daughter nucleus.

Or, Mv²/2 = Δmc² from which v = c√(2Δm/M), as given in option (3).

[Note that the momentum is to be conserved in the process. Therefore, the daughter nuclei have to travel in opposite directions with the same speed since they have the same mass. If you were asked to give the velocities of the daughters, the answer would be v and – v]

(ii) The binding energy per nucleon for the parent nucleus is E₁ and that for the daughter nuclei is E₂. Then

(1) E₁ = 2 E₂

(2) E₂ = 2 E₁

(3) E₁ > E₂

(4) E₂ > E₁

The binding energy per nucleon for the daughter nuclei is greater than that for the parent nucleus since energy is released in the process[Option (4)].

The following question (MCQ) also was included in the AIEEE 2010 question paper:

A radioactive nucleus (initial mass number A and atomic number Z) emits 3 α-particles and 2 positrons. The ratio of number of neutrons to that of protons in the final nucleus will be

(1) (A – Z – 4)/(Z – 2)

(2) (A – Z – 8)/(Z – 4)

(3) (A – Z – 4)/(Z – 8)

(4) (A – Z – 12)/(Z – 4)

When an α-particle is emitted, the nucleus loses two protons and two neutrons. When a positron is emitted by the nucleus, a proton in the nucleus gets converted into a neutron. Since the number of neutrons in the original neucleus is A – Z, the number of neutrons in the final nucleus will be (A – Z) – (3×2) + 2 = A – Z – 4.

The protons in the final nucleus will be Z – (3×2) – 2 = Z – 8

Therefore, the ratio of number of neutrons to that of protons in the final nucleus will be(A – Z – 4)/(Z – 8), as given in option (3).

EAMCET (Engineering Entrance) 2010 Questions on Capacitors

“Non-violence is the greatest force at the disposal of mankind. It is mightier than the mightiest weapon of destruction devised by the ingenuity of man”

– Mahatma Gandhi

Today we will discuss three questions which appeared in EAMCET (Engineering Entrance) 2010 question paper. Questions (1) and (2) are from electrostatics while question no.3 is from transient phenomena.

(1) Two capacitors of capacities 1 μF and C μF are connected in series and the combination is charged to a potential difference of 120 V. If the charge on the combination is 80 μC, the energy stored in the capacitor of capacity C in microjoule is

(1) 1800

(2) 1600

(3) 14400

(4) 7200

The effective capacitance of the series combination is 1×C/(1+C) μF = C/(1+C) μF.

[Remember that when capacitors C₁, C₂, C₃,….etc are connected in series, the effective capacitance C_eff is given by the reciprocal relation, 1/C_eff = 1/C₁ + 1/C₂ + 1/C₃ + …. etc.]

We retain the unit of capacitance as μF for convenience.

Since the charge carried by the combination is 80 μC, we have (from Q = CV with usual notations)

80 = [C/(1+C)]×120

[We have substituted the charge in microcoulomb itself since the capacitance is in microfarad].

This gives 80 C + 80 = 120 C from which C = 2 μF

When capacitors are in series, all capactors carry the same charge. Therefore, the charge on the 2 μF capacitor is 80 microcoulomb.

The energy U of the 2 μF capacitor is given by

U = Q²/2C = 80²/(2×2) = 1600 microjoule.

[Since Q is in microfarad and C is in microcoulomb, the energy is in microjoule].

(2) The potential difference between two parallel plates is 10⁴ volts. If the plates are separated by 0.5 cm, the force on an electron between the plates is

(1) 32×10^–13 N

(2) 0.32×10^–13 N

(3) 0.032×10^–13 N

(4) 3.2×10^–13 N

This is a very simple question. The electric field (E) between the plates is given by

E = V/d where V is the potential difference and d is the separation between the plates

The force F on the electron (of caharge e = 1.6×10^–19 coulomb) is given by

F = Ee = Ve/d = (10⁴×1.6×10^–19) /(0.5×10^–2) = 3.2×10^–13 N

(3) A capacitor of capacity 0.1 μF connected to a resistor of 10 MΩ is charged to a certain potential and then made to discharge through the resistor. The time in which the potential will take to fall to half its original value is (Given log₁₀ 2 = 0.3010)

(1) 2 sec

(2) 0.693 sec

(3) 0.5 sec

(4) 1 sec

If V₀ is the initial voltage across the capacitor of capacitance C, the voltage V at time t during the discharge through the resistance R is given by

V = V₀ e^–t/RC where e is the base of natural logarithms.

Therefore V/V₀ = e^–t/RC

Since V/V₀ = ½ in the question, we have

½ = e^–t/RC

Or, e^t/RC = 2

Taking logarithm, t/RC = ln 2

Therefore, t = RC ln 2 = 10×10⁶×0.1×10^–6×ln 2

If you remember that ln 2 = 0.693, you can write the answer immediately as 0.693 sec.

[You should definitely know that the relation between natural logarithm (with base e) and common logarithm (base 10) of quantity x is

ln_e x = 2.303 log₁₀ x

So ln 2 = 2.303×0.3010 = 0.693]

Karnataka CET 2010 Questions on Alternating Currents

Science without religion is lame; religion without science is blind.

– Albert Einstein

Questions on alternating currents are generally interesting and often not tough. By clicking on the label ‘alternating current’ below this post, you can access all posts related to this section on this site.

Here are two questions on alternating currents which appeared in Karnataka CET 2010 question paper:

(1) In a series R-L-C circuit, the voltage across R is 100 V and the value of R = 1000 Ω The capacitance of the capacitor is 2×10^–6 F; angular frequency of A.C. is 200 rad s^–1. Then the P.D. across the inductance coil is

(a) 100 V

(b) 40 V

(c) 250 V

(d) 400 V

Since the voltage drop across the 1000 Ω resistor is 100 V, the current (I) in the R-L-C circuit is 100 V/1000 Ω = 0.1 A.

At resonance the magnitude of the voltage across the inductance (V_L) is the same as that across the capacitor (V_C).

Or, V_L = V_C = I/Cω where ω is the angular frequency of A.C.

Therefore, V_C = 0.1/(2×10^–6×200) = 250 V.

(2) A capacitor and an inductance coil are connected in separate AC circuits with a bulb glowing in both the circuits. The bulb glows more brightly when

(a) an iron rod is introduced into the inductance coil

(b) the number of turns in the inductance coil is increased

(c) separation between the plates of the capacitor is increased

(d) a dielectric is introduced into the gap between the plates of the capacitor

The reactance of the inductor is Lω where L is the inductance and ω is the angular frequency of AC. In options (a) and (b) the value of L is increased and hence the inductive reactance is increased. The current through the bulb is therefore decreased, thereby decreasing the brightness.

The capactive reactance is 1/Cω where C is the capacitance. When the separation between the plates is increased as given in option (c), the capacitance is decreased. The capacitive reactance is therefore increased, thereby decreasing the current and the brightness.

On introducing a dielectric slab between the plates, the capacitance is increased, thereby decreasing the capacitive reactance. The brightness of the bulb is then increased because of the increased current. So the correct option is (d).

IIT-JEE 2010 Questions on Thermodynamics (Multiple Correct Answer Type and Integer Type)

The following questions on thermodynamics appeared in the IIT-JEE 2010 question paper:

[Question No.1 is Multiple Correct Answer Type. Question No.2 is Integer Type, the answer to which is a single digit integer ranging from 0 to 9].

(1) One mole of an ideal gas in initial state A undergoes a cyclic process ABCA, As shown in figure. Its pressure at A is P₀. Choose the correct option(s) from the following:

(A) Internal energies at A and B are the same

(B) Work done by the gas in process AB is P₀V₀ ln 4

(C) Pressure at C is P₀/4

(D) Temperature at C is T₀/4

AB is an isothermal process (A and B are at the same temperature). Therfore the internal energies at A and B are the same.

The work (W) done by the gas in the isothermal process AB is given by

W = nRT ln(V₂/V₁) where R is universal gas constant, ‘n’ is the number of moles in the sample of the gas, T is the temperature at which the process occurs, V₁ is the initial volume and V₂ is the final volume.

Therefore W = 1×RT₀ ln (V₂/V₁) = RT₀ ln(4V₀/V₀)

Since RT₀ = P₀V₀ we obtain

W = P₀V₀ ln 4

Thus options A and B are correct.

We have PV/T = constant

Assuming that the line BC passes through the origin, the temperature at C must be T₀/4. Considering the states at A and C we have

P₀V₀ /T₀ = P_CV₀/(T₀/4)

The pressure at C is therefore given by

P_C = P₀/4

So all options are correct.

(2) A diatomic ideal gas is compressed adiabatically to 1/32 of its initial volume. If the initial temperature of the gas is Ti (in Kelvin) and the final temperature is aTi, the value of a is:

In the case of an adiabatic process we have

TV^γ–1 = constant where γ is the ratio of specific heats of the gas.

Therefore, T_iV^γ–1 = aT_i(V/32)^γ–1

For an ideal diatomic gas γ = 7/5 and hence we have

T_iV^2/5 = aT_i(V/32)^2/5

This gives a = 32^2/5 = [2⁵]^2/5 = 2² = 4.

Two Karnataka CET 2010 Multiple Choice Questions on Direct Current Circuits

The following questions appeared in Karnataka CET 2010 question paper:

(1) A battery of e.m.f. E has internal resistance ‘r’. A variable resistance R is connected to the terminals of the battery. A current I is drawn from the battery. V is the terminal P.D. If R alone is graually reduced to zero, which of the following best describes I and V?

(a) I approaches zero, V approaches E

(b) I approaches E/r, V approaches zero

(c) I approaches E/r, V approaches E

(d) I approaches infinity, V approaches E

The current I is given by

I = E/(r+R)

When R is reuced to zero, I = E/r

Since the terminal P.D. is given by V = ER/(r+R), we have V = 0 when R = 0.

The correct option is (b).

(2) Three voltmeters A, B and C having resistances R, 1.5 R and 3 R respectively are used in a circuit as shown. When a P.D. is applied between X and Y, the readings of the voltmeters are V₁, V₂ and V₃ respectively. Then

(a) V₁ = V₂ = V₃

(b) V₁ <>2 = V₃

(c) V₁ > V₂ > V₃

(d) V₁ > V₂ = V₃

The effective resistance of the voltmeters B and C in parallel is 1.5R× 3R/(1.5R+3R) = R.

Therefore, the same P.D. exists across A and the parallel combination of B and C and the voltmeters show the same reading [Option (a)].