CBSE to conduct National Eligibility Cum Entrance Test (NEET) 2013 for admission to MBBS and BDS Courses

“The pursuit of truth and beauty is a sphere of activity in which we are permitted to remain children all our lives.”

– Albert Einstein

Most of the students aspiring for admission to MBBS and BDS courses in India will be happy about the recent notification that admission to MBBS and BDS Courses in the institutions approved by the Medical Council of India and Dental Council of India are subject to merit position of candidates in the National Eligibility Cum Entrance Test (Under-Graduate), 2013.

The Central Board of Secondary Education (C.B.S.E.) will conduct the National Eligibility cum Entrance Test (NEET) for admission to MBBS and BDS Courses on Sunday, the 5^th May 2013.

The syllabus for the National Eligibility cum Entrance Test (NEET) is available on the Medical Council of India website www.mciindia.org

Lorentz force – An IIT-JEE 2012 Multiple Correct Answer(s) Type Question

“Making the simple complicated is commonplace; making the complicated simple, awesomely simple, that’s creativity.”

– Charles Mingus    

IIT-JEE Multiple Correct Answer(s) Type Questions are designed to have four choices (A), (B), (C) and (D) out of which ONE or MORE are correct. The following question appeared in IIT-JEE 2012 question paper. The question is meant for checking your understanding of Lorentz forces.

Consider the motion of a positive point charge in a region where there are simultaneous uniform electric and magnetic fields E = E₀ ĵ and B = B₀ ĵ. At time t = 0, this charge has velocity v in the x-y plane, making an angle θ with the x-axis. Which of the following option(s) is (are) correct for t > 0?

(A) If θ = 0°, the charge moves in a circular path in the x-z plane.

(B) If θ = 0°, the charge undergoes helical motion with constant pitch along the y-axis.

(C) If θ = 10°, the charge undergoes helical motion with its pitch increasing with time, along the y-axis.

(D) If θ = 90°, the charge undergoes linear but accelerated motion along the y-axis.

Note that the electric and magnetic fields are along the positive y-direction.

If θ = 0, the charged particle is moving at right angles to the electric and magnetic fields (Fig.). The magnetic force (qv×B₀) will make the particle move along a circle and the electric force (qE₀) will push it along the positive y-direction. The path of the particle will therefore be a helix of increasing pitch.

If θ = 10°,the magnetic field will make the particle move along the positive y-direction, following a helical path of constant pitch; but since the electric force (qE₀) pushes it along the positive y-direction, the helical path has increasing pitch.

If θ = 90°, the charged particle is moving parallel to the electric and magnetic fields. The magnetic field has no action on the motion since there is no magnetic force. But the electric force (qE₀) will push the particle along the positive y-direction. The particle is thus in accelerated linear motion. 

Therefore options (C) and (D) are correct.

Multiple Choice Questions on Electrostatics

“Example isn't another way to teach, it is the only way to teach.”

– Albert Einstein

Today we will discuss a few questions from electrostatics. You will find many questions (with solution) in this section discussed earlier on this site. You can access all those questions by clicking on the label ‘electrostatics’ below this post.

(1) A 6 μF capacitor is connected in series with a 2 μF capacitor. The 6 μF capacitor can withstand a maximum voltage of 3 kV where as the 2 μF capacitor can withstand a maximum voltage of 6 kV. The maximum voltage that the parallel combination can withstand is

(a) 2 kV

(b) 3 kV

(c) 6 kV

(d) 8 kV

(e) 12 kV

We have capacitors C₁ and C₂ (let us say) having values 6 μF and 2 μF. If the maximum voltage that the parallel combination can withstand is V_max, the voltage V₁ across the 6 μF capacitor on applying this voltage across the series combination is given by

            V₁ = V_max C₂/(C₁ + C₂)

[The charge Q on each capacitor on connecting the voltage V_max across the series combination is given by Q = C₁C₂ V_max/(C₁ + C₂), remembering that the effective capacitance of the series combination is C₁C₂/(C₁ + C₂)].

Therefore, V₁ = V_max×2/(6+2) = V_max/4

Since C₁ can withstand a maximum voltage of 3 kV we have V_max/4 = 3 kV

This gives V_max = 12 kV.

[Do not jump to a conclusion at this stage, You have to check whether C₂ will be intact on applying the above 12 kV across the series combination].

The voltage V₂ across the 2 μF capacitor on applying this voltage V_max across the series combination is given by

            V₂ = V_max C₁/(C₁ + C₂)

Threfore V₂ = V_max×6/(6+2) = 6 V_max/8

Since C₂ can withstand a maximum voltage of 6 kV we have 6 V_max/8 = 6 kV

This gives V_max = 8 kV.

This value of V_max being lower than that obtained (12 kV) on considering the 6 μF capacitor, the correct option is 8 kV.

(2) A uniform electric field of intensity E newton/coulomb directed along the positive x-direction exists in a region of space (Fig.). The x- direction is horizontal. A, B, C and D are points at the corners of a square of side a, with AB and CD  parallel to the x-direction. If the electric potential at point A is V volt, what is the potential (in volt) at the point D?

(a) V

(b) V – aE

(c) V – √2 aE

(d) V + √2 aE

(e) V + aE

Since the electric field acts along the positive x-direction, the potential decreases as we move along the positive x-direction.

[Remember that the electric field is the negative gradient of potential].

While moving from A to D the x-coordinate increases by ‘a’ and hence the potential decreases by aE. Therefore, the potential at D is V – aE [Option (b)].

(3) In the above question what is the potential difference between points A and C?

(a) V

(b) V – aE

(c) V + aE

(d) aE

(e) Zero

Since the electric field acts along the x-direction, the potential will change only if the x-coordinate changes. Points A and C have the same x-coordinates and hence they are at the same potential. Therefore, the potential difference between points A and C is zero.

(4) Two small identical spheres are charged equally and suspended in air by strings of equal length. The strings make a small angle θ with each other (Fig.). When the spheres are immersed in oil of density 800 kg m^–3 the angle between the strings is found to be unaltered. If the density of the material of the spheres is 1200 kg m^–3, what is the dielectric constant of the oil?

(a) 1.5

(b) 2.5

(c) 3

(d) 3.5

(e) 4

The repulsive electrostatic force F  between the spheres in air is given by

             F = (1/4πε₀) (q²/d²) where ε₀ is the permittivity of free space (and air, very nearly), qis the charge on each sphere and d is the distance between the spheres.

            When the spheres are in the oil the electrostatic force F₁ between the spheres is given by

            F₁ = (1/4πε₀K) (q²/d²) where K is the dielectric constant of the oil.

The real weight W of each sphere is given by

            W = Vρg where V is the volume, ρ is the density of the material of the sphere and  g is the acceleration due to gravity.

The apparent weight W₁ of each sphere when immersed in oil is given by

            W₁ = Vρg – Vσg where σ is the density of the oil.

[Note that Vσg is the upthrust or the force of buoyancy due to the oil]

When the spheres are in air, we have (Fig.)

     tan α = F /W = (1/4πε₀) (q²/d²)/ Vρg………………..(i)

When the spheres are in oil, we have

tan α = F₁ /W₁ = (1/4πε₀K) (q²/d²) / (Vρg – Vσg)……(ii)

Dividing Eq. (i) by Eq. (ii) we have

            1 = K(ρ – σ) / ρ

Therefore, K = ρ/(ρ – σ) = 1200/400 = 3

Kerala Engineering Entrance 2012 (KEAM Engg. 2012) Questions on Surface Tension

“It doesn't matter how beautiful your theory is; it doesn't matter how smart you are. If it doesn't agree with experiment, it's wrong.”

– Richard Feynman

Two questions on surface tension were included in the KEAM (Engineering) 2012 question paper. Here are the questions with solution:

(1) If two capillary tubes of radii r₁ and r₂ in the ratio 1 : 2 are dipped vertically in water, then the ratio of capillary rises in the respective tubes is

(a) 1 : 4

(b) 4 : 1

(c) 1 : 2

(d) 2 : 1

(e) 1 : √2

The capillary rise h due to surface tension is relate to the surface tension S as

            S = hrρg/2 cosθ

where r is the radius of the capillary tube, ρ is the density of the liquid, g is the acceleration due to gravity and θ is the angle of contact.

Evidently h is inversely proportional to r.

Therefore, h₁/h₂ = r₂/r₁ = 2 : 1

(2) If the excess pressure inside a soap bubble of radius r₁ in air is equal to the excess pressure inside air bubble of radius r₂ inside the soap solution, then r₁ : r₂ is         

(a) 2 : 1

(b) 1 : 2

(c) 1 : 4

(d) √2 : 1

(e) 1 : √2

The excess pressure inside a soap bubble in air is 4S/r where as the excess pressure inside an air bubble in the soap solution is 2S/r where S is the surface tension (of soap solution) and r is the radius of the bubble.

[In the case of the air bubble in the soap solution there is one liquid surfaoe only and that is why the excess pressure is 2S/r and not 4S/r. Remember that in the case of a soap bubble in air there are two liquid surfaces]. 

As given in the question, we have

            4S/r₁ = 2S/r₂

This gives r₁/r₂ = 2

Or, r₁ : r₂ = 2 : 1

IIT JEE 2012 Questions on Magnetic Field due to Infinitely Long Current Carrying Cylinders

“The world is a dangerous place, not because of those who do evil, but because of those who look on and do nothing.”

– Albert Einstein.

Today we will discuss two questions pertaining to the magnetic field produced by infinitely long current carrying cylinders. The first question is the usual single correct answer type multiple choice question. The second one is an integer answer type question, the answer to which is a single digit integer ranging from 0 to 9. Here are the questions with their solution.   

(1) An infinitely long conducting cylinder with inner radius R/2 and outer radius R carries a uniform current density along its length. The magnitude of the magnetic field B‌ as a function of the radial distance r from the axis is best represented by

‌‌Case (i): r < R/2

Using Ampere’s circuital law we find that the magnetic fiel at points at distance r < R/2 is zero.

[We have ∫B.dℓ = μ₀I.

Since the current passing through the surface enclosed by the path of integration is zero, the value of B is zero at distance r < R/2]

‌‌Case (ii)

For distance r lying between R/2 and R (or, R/2 ≤ r < R) we have

            ∫B.dℓ = μ₀[πr²J – π(R/2)²J] where J is the current density

‌[πr²J – π(R/2)²J is the current passing through the surface enclosed by the path of integration in this case]

Or, B×2πr = μ₀[πr²J – π(R/2)²J]

This gives B = (μ₀J/2)[r – (R²/4r)]

Case (iii)

For r ≥ R we have

            B.dℓ = μ₀[πR²J – π(R/2)²J]

Or, B×2πr = μ₀[πR²J – π(R/2)²J]

This gives B = (μ₀J/2r)[R² – (R²/4r)] = 3μ₀JR²/8r

The graph shown in option (D) indicates the above three cases correctly.

(2) A cylindrical cavity of diameter ‘a’ exists inside a cylinder of diameter 2a as shown in the figure. Both the cylinder and the cavity are infinitely long. A uniform current density J flows along the length. If the magnitude of the magnetic field at the point P is given by N μ₀ aJ/12, then the value of N is

            If the cylinder has no cavity, the magnetic flux density B₁ at the point P is given by

            ∫B₁.dℓ = μ₀πa²J

Therefore, B₁×2πa = μ₀πa²J

Or, B₁ = μ₀aJ/2

Since there exists a cavity, the current is reduced by π(a/2)²J and the magnetic field is reduced by B₂.

The field B₂ is given by

            ∫B₂.dℓ = μ₀π(a/2)²J

Therefore, B₂×2π(3a/2) = μ₀π(a/2)²J

[The circular path of integration in this case has radius a+(a/2) = 3a/2]

Or, B₂ = μ₀aJ/12

The magnitude B of the magnetic field at the point P due to the actual conductor with the cavity is given by

            B = B₁ – B₂ =  μ₀aJ/2– μ₀aJ/12

Or, B = 5μ₀aJ/12

In the question the above field is given as Nμ₀aJ/12

            Therefore N = 5.

AIPMT Main Exam – Two Questions Involving Discharge of a Capacitor Through an Inductor

"Live as if you were to die tomorrow. Learn as if you were to live forever."

– Mahatma Gandhi

Questions appearing in AIPMT Main (Physics) question paper are generally not as simple as those appearing in AIPMT Preliminary question paper. Those who get qualified (by passing the preliminary examination) should bear this in mind and prepare accordingly. I give below two questions involving the discharge of a charged capacitor through an inductor:

(1) A condenser of capacity C is charged to a potential difference of V₁. The plates of the condenser are then connected to an ideal inductor of inductance L. The current through the inductor when the potential difference across the condenser reduces to V₂ is

(1) [C(V₁ – V₂)²/L]^1/2

(2) C(V₁² – V₂²)/L

(3) C(V₁² + V₂²)/L

(4) [C(V₁² – V₂²)/L]^1/2

The above question appeared in AIPMT Main 2010 question paper.

            The initial energy of the charged capacitor is ½ CV₁². When the potential difference across the capacitor reduces to V₂  the energy is ½ CV₂². Therefore, the energy lost by the capacitor is ½ C(V₁² – V₂²). This amount of energy is gained by the inductor.

If I represents the discharge current at the instant when the potential difference across the capacitor reduces to V₂ we have

             ½ C(V₁² – V₂²) = ½ LI²

[Remember that the energy of an inductor carrying a current I is ½ LI² and the energy is store in the magnetic field]

From the above equation we obtain

             I = [C(V₁² – V₂²)/L]^1/2

(2) Capacitance of 6 μF is charged by 6 V battery. Now it is connected with inductor of 5 mH. Find the current in the inductor when 1/3rd of total energy is magnetic.

This question was asked as half the part of a free response (and not a multiple choice) question in the AIPMT Main 2007 question paper.

[Current practice is to ask objective type (multiple choice) questions].

            The initial energy E₁ of the charged capacitor is given by

            E₁ = ½ CV² = ½ ×(6×10^–6)×6²

Since one-third of total energy is mentioned as magnetic in the question, we understand that the energy of the inductor carrying the discharge current I is E₁/3.

Therefore we have

            E₁/3 = ½ LI²  

Or, (1/3) ×[½ ×(6×10^–6)×6²] = ½ ×(5×10^–3)I²  

This gives I = 0.12 A

KEAM (Engineering) 2010 Questions (MCQ) on Electronics

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

–Mahatma Gandhi

Questions on electronics will be generally interesting to most of you. Today we will discuss questions in this section which appeared in Kerala engineering entrance (KEAM - Engineering) 2010 question paper. Here are the questions with their solution:

 

(1) A full wave rectifier with an a.c. input is shown:

The output  voltage across R_L is represented as

  

The rectified output voltage will be a direct voltage but there will be very large amount of ripples. The capacitor C acts as a filter to remove the ripples; but there will still be a small amount of  ripples in the output. Therefore the correct option is (e).

(2) In the given circuit the current through the battery is

(a) 0.5 A

(b) 1 A

(c) 1.5 A

(d) 2 A

(e) 2.5 A

Since the diode D₁ is reverse biased, no current will flow through the D₁ branch. Diodes D₂ and D₃ are forward biased and hence the battery drives currents through the 20 Ω resistor and the series combination of the two 5 Ω resistors.

The current driven through the 20 Ω resistor is 10 V/20 Ω = 0.5 A.

The current driven through the 10 Ω resistor is 10 V/10 Ω = 1 A.

Therefore, total current through the battery is 0.5 A + 1 A = 1.5 A

(3) The collector supply voltage is 6 V and the voltage drop across a resistor of 600 Ω in the collector circuit is 0.6 V, in a transistor connected in common emitter mode. If the current gain is 20, the base current is

(a) 0.25 mA

(b) 0.05 mA

(c) 0.12 mA

(d) 0.02 mA

(e) 0.07 mA

We have I_CR_C = 0.6 V where I_C is the collector current and R_C is the resistance in the collector circuit.

Therefore,  I_C×600 Ω = 0.6 V from which I_C = 0.6/600 A = 10^–3 A = 1 mA.

Since the current gain β is given by

            β = I_C/I_B where I_B is the base current, we have

            I_B = I_C/β = 1 mA/20 = 0.05 mA.

(4) A pure semiconductor has equal electron and hole concentration of 10¹⁶ m^–3. Doping by indium increases n_h to 5×10²² m^–3. Then the value of n_e in the doped semiconductor is

(a) 10⁶ m^–3

(b) 10²² m^–3

(c) 2×10⁶ m^–3

(d) 10¹⁹ m^–3

(e) 2×10⁹ m^–3

According to the law of mass action we have

            n_i² = n_en_h where n_i is the electron concentration as well as the hole concentration in the intrinsic (pure) semiconductor, n_e is the electron concentration in the doped semiconductor and n_h is the hole concentration in the doped semiconductor.

Therefore n_e = n_i²/n_h = (10¹⁶)²/(5×10²²) = 2×10⁹ m^–3