Kerala Medical Entrance (KEAM) 2009 Questions from Electrostatics

Keral Medical Entrance 2009 question paper contained three questions from electrostatics. They are simple except for the first one (given below) which will demand some three dimensional imagination.

(1) The total electric flux through a cube when a charge 8q is placed at one corner of the cube is

(a) ε₀ q

(b) ε₀ /q

(c) 4πε₀ q

(d) q/4πε₀

(e) q/ε₀

You should remember that the total electric flux originating from a charge q is q/ε₀.

[You will definitely remember that the electric field E at a point distant r from a point charge is q/4πε₀r². The electric field is equal to the electric flux through unit area held normal to the direction of the field. If you imagine a spherical surface with the charge q at the centre, the surface area of the sphere is 4πr² and the total flux passing normally through the spherical surface is (q/4πε₀r²) ×4πr² = q/ε₀]

In the question the charge 8q is placed at one corner of the cube. The total electric flux originating from this charge is 8q/ε₀. But only one-eigths of the flux passes through the cube. (You can place seven more cubes with their corners touching the charge to cover the entire volume around the charge).

The electric flux through a cube when a charge 8q is placed at one corner of the cube is therefore (1/8)×(8q/ε₀) = q/ε₀.

(2) A uniform electric field E exists along positive x-axis. The work done in moving a charge 0.5 C through a distance 2 m along a direction making an angle 60º with x-axis is 10 J. Then the magnitude of electric field is

(a) 5 Vm^–1

(b) 2 Vm^–1

(c) √5 Vm^–1

(d) 40 Vm^–1

(e) 20 Vm^–1

The force (F) acting on charge q in electric field E is given by F = qE along the direction of the field (along the positive x-axis here). Since the displacement (d) of the charge is inclined at 60º with the x-axis, the work done (W) is given by

W = qEd cos 60

Therefore, 10 = 0.5×E×2×½, from which E = 20 Vm^–1.

(3) A capacitor of capacitance C is charged to a potential V. If it carries charge Q, then the energy stored in it is

(a) ½ CV

(b) QV

(c) ½ QV²

(d) CV²

(e) ½ QV

The answer is ½ QV which you can obtain from ½ CV² which is the form most of you will usually remember:

½ CV² = ½ CV×V = ½ QV

[The capacitor is charged from zero potential to the potential V. The charge Q is therefore transferred to the capacitor at an average potential of (0+V)/2 = V/2. The work done is therefore QV/2.

To be more rigorous, if the charge on the capacitor at any instant of charging is q and the potential at the instant is v, the work done in transferring an additional charge dq is vdq= (q/C)dq. The total work done in transferring the entire charge Q is ₀ ∫^Q(q/C)dq = Q²/2C = ½ ×Q×(Q/C) = ½ QV].

You will find more questions (with solution) of various entrance examinations at physicsplus.blogspot.com.

Thermodynamics – Two Multiple Choice Questions

Often simple questions on thermodynamics may cause unexpected confusion. The reason for the confusion is usually the lack of your understanding of the fundamentals. If you don’t have any confusion in respect of the following questions, well and good!

(1) A sample of an ideal gas initially having internal energy U₁ and pressure P₁ expands adiabatically and performs work W. Heat energy Q is then added to the gas at constant volume so that its pressure is increased to the initial value P₁. As a result of the above processes, the internal energy of the gas

(a) decreases by Q – W

(b) increases by Q – W

(c) decreases by Q

(d) increases by Q

(e) remains unchanged

Since the gas expands adiabatically and outputs mechanical energy, the internal energy of the gas is decreased. [Remember that during adiabatic process there is no heat transfer between the gas and the surroundings]. On adding heat energy to the gas at constant volume the internal energy of the gas is increased. There is no work involved since the volume is constant (isochoric process).

Evidently the internal energy of the gas increases by Q – W.

[You will obtain the answer from the mathematical statement of the 1^st law of thermodynamics: ∆Q = ∆U + ∆W where ∆Q is the heat energy supplied to the system by the surroundings, ∆W is the work done by the system on the surroundings and ∆U is the increase in the internal energy of the system].

(2) In the given PV diagram, I is the initial state and F is the final state. The gas goes from I to F by (i) IAF (ii) IBF (iii) ICF. The heat absorbed by the gas is

(a) the same in all three processes

(b) the same in (i) and (ii)

(c) greater in (i) than in (ii)

(d) the same in (i) and (iii)

(e) greater in (iii) than in (ii)

This question appeared in Kerala engineering entrance 2009 question paper.

We have ∆Q = ∆U + ∆W

Since the same initial point (I) and the same final point (F) are given for all the three paths, the change in the internal energy (∆U) is the same for all the three processes. During the path IAF the net work done by the gas (∆W) is positive since the work done on the gas during the compression IA (given by the area under the line IA) is less than the work done by the gas during its expansion AF. No work is involved during the process IBF since the volume is constant (isochoric process). During the path ICF the net work done (∆W) by the gas is negative since the work done by the gas during its expansion IC is less than the work done on the gas during its compression CF.

Therefore, the heat absorbed by the gas (∆Q) is greater in (i) than in (ii).

You will find useful multiple choice questions with solution at apphysicsresources as well as at physicsplus.

Kerala Medical Entrance (KEAM) 2009 Questions on Optics

Physics questions in the Kerala Medical Entrance 2009 question paper were generally simple compared to those in the Kerala Engineering Entrance 2009 question paper. Today we will discuss the questions on optics included in the Medical Entrance question paper. Here are those three questions:

(1) A ray of light suffers minimum deviation in equilateral prism P. Additional prisms Q and R of identical shape and of same material as that of P are now combined as shown in figure. The ray will now suffer

(a) greater deviation

(b) no deviation

(c) same deviation as before

(d) total internal reflection

(e) smaller deviation

On combining additional prisms Q and R with P, we obtain a portion of an equilateral prism of the same material. Since the incident ray is such that the deviation produced by the prism P is minimum, the refracted ray will be parallel to the base of the prism P and hence it will pass parallel to the base of the combined prism, suffering the same deviation (minimum deviation) as before. The correct option is (c).

(2) When light is scattered by atmospheric atoms and molecules, the amount of scattering of light of wave length 440 nm is A. The amount of scattering for light of wave length 660 nm is

(a) (4/9) A

(b) 2.25 A

(c) 1.5 A

(d) 0.66 A

(e) A/5

The answer for this question is based on Rayleigh’s scattering formula which says that the amount of light scattered is inversely proportional to the fourth power of the wave length of light. Therefore, in the two cases we have respectively,

A α 1/440⁴ and

x α 1/660⁴ where x is the amount of scattering for light of wave length 660 nm.

Dividing, A/x = 660⁴/ 440⁴ = (3/2)⁴ = 81/16 = 5 nearly.

Therefore, x = A/5.

(3) In the measurement of the angle of a prism using a spectrometer, the reading of first reflected image are Vernier I: 320° 40' Vernier II: 140° 30' and those of the second reflected image are Vernier I: 80° 38'; Vernier II: 260° 24'. Then the angle of the prism is

(a) 59° 58'

(b) 59° 56'

(c) 60° 2'

(d) 60° 4'

(e) 60° 0'

When a parallel beam of light falls symmetrically on the two faces of the prism, the angle between the rays reflected from these faces is 2A where A is the angle of the prism. If you have a clear understanding of the experimental determination of the angle of the prism using this method, you will definitely be able to find the answer to the above question since you will know that the reading of Vernier I was first inecreased from 320° 40' to 360° on rotating the telescope to view the reflected image from the second face. The 360° mark is the same as 0° mark. So after reaching the zero reading the telescope has rotated through 80° 38'.

The difference between Vernier I readings is (360° - 320° 40') + 80° 38' = 39° 20'+ 80°' = 119° 58'. 38

The difference between Vernier II readings is 260° 24' - 140° 30' = 119° 54'.

The mean of the difference between the readings is 119° 56'.

Therefore, the angle of the prism is (119° 56')/2 = 59° 58'

Kerala Medical Entrance (KEAM) 2008 Questions on Nuclear Physics

Try not to be a person of success, but rather a person of virtue.

– Albert Einstein

Here are the two questions from nuclear physics which were included in KEAM (Medical) 2008 question paper:

(1) If the mass defect of ₈O¹⁶ nucleus is 0.128 amu, then the binding energy per nucleon of oxygen is

(a) 8.2 MeV

(b) 7.45 MeV

(c) 7.3 MeV

(d) 7.1 MeV

(e) 8.15 MeV

One atomic mass unit (amu) is equivalent to 931 MeV. Therefore, the total binding energy of the ₈O¹⁶ nucleus is 0.128×931 MeV.

Since there are 16 nucleons in the ₈O¹⁶ nucleus, the binding energy per nucleon of oxygen is (0.128×931)/16 = 7.45 MeV, very nearly.

(2) Two radioactive samples have decay constants 15x and 3x. If they have the same number of nuclei initially, the ratio of number of nuclei after a time 1/6x is

(a) 1/e

(b) e/2

(c) 1/e⁴

(d) 2e/3

(e) 1/e²

The number N of nuclei at time t is given by

N = N₀e^–λt where N₀ is the initial number, e is the base of natural logarithms, and λ is the decay constant.

The required ratio is (N₀e^–15x/6x)/ (N₀e^–3x/6x) = e^–2.5/ e^–0.5 = e^–2 = 1/e²

Three questions from nuclear physics were included in the physics question paper of Kerala Engineering Entrance (KEAM) 2008 examination. You will find those questions with solution here

Two Kerala Engineering Entrance 2005 Questions on Electrostatics

I never did a day’s work in my life. It was all fun.

– Thomas Alva Edison

Here are two multiple choice questions which appeared in Kerala Engineering Entrance 2005 question paper:

(1) A soap bubble is charged to a potential of 16 V. Its radius is then doubled. The potential of the bubble now will be

(a) 16 V

(b) 8 V

(c) 4 V

(d) 2 V

(e) zero

The potential of a bubble of radius R carrying charge Q is Q/4πε₀R so that for a given charge the potential is inversely proportional to the radius. Therefore when the radius is doubled, the potential is halved. The answer is 8 V [Option (b)].

(2) A parallel plate capacitor of capacitance 10 μF is charged to 1μC. The charging battery is removed and then the separation between the plates is doubled. Work done during the process is

(a) 5 μJ

(b) 0.05 μJ

(c) 1 μJ

(d) 10 μJ

(e) 50 μJ

The work done is equal to the increase in the energy of the capacitor.

The initial energy is Q²/2C where C is the initial capacitance and Q is the charge.

Therefore initial energy, Q²/2C = 10^–12/(20×10^–6) = 5×10^–8 J = 0.05 μJ.

When the separation between the plates is doubled, the capacitance is halved (since the capacitance is Kε₀A/d with usual notations) and hence the energy is doubled. The final energy is thus 0.1 μJ.

The increase in energy, which is equal to the work done, is 0.05 μJ.

You will find all posts related to electrostatics on this site by clicking on the label ‘electrostatics’ below this post. More useful questions with solution on electrostatics can be found here as well as here.