Two EAMCET (Medical) 2009 Questions on Photons and Electrons

Today we will discuss two multiple choice questions from modern physics. These questions are from the section ‘electrons and photons’ and they were included in the EAMCET (Medical) 2009 question paper:

(1) In Millikan’s oil drop experiment, a charged oil drop of mass 3.2×10^–14 kg is held stationary between two parallel plates 6 mm apart, by applying a potential difference of 1200 V between them. How many electrons does the oil drop carry?

(1) 7

(2) 8

(3) 9

(4) 10

The electric field ‘E’ between the plates is V/d where V is the potential difference and d is the separation between the plates.

Therefore E = 1200/(6×10^–3) = 2×10⁵ volt per metre.

The electric force on the drop is Eq where ‘q’ is the charge on the drop. Since the drop is stationary, the electric force balances the weight mg of the drop.

Therefore Eq = mg so that

q = mg/E = (3.2×10^–14×10)/( 2×10⁵) = 1.6×10^–18 coulomb.

Since the electronic charge is 1.6×10^–19 coulomb, the number of excess electrons carried by the drop is (1.6×10^–18) /(1.6×10^–19) = 10.

(2) Electrons accelerated by a potential of ‘V’ volts strike a target material to produce ‘continuous X-rays’. Ratio between the de Broglie wave length of the electrons striking the target and the shortest wave length of the ‘continuous X-rays’ emitted is

(1) h/√(2Vem)

(2) (1/c) √(2m/Ve)

(3) (1/c) √(Ve/2m)

(4) he/√(Ve/2m)

The de Broglie wave length λ of the electron is given by

λ = h/p where h is Planck’s constant and p is the momentum of the electron.

The shortest wave length λ_min of the ‘continuous X-rays’ produced by the striking electrons is given by

hc/λ_min = Ve where c is the speed of light in free space and e is the electronic charge.

[Note that we have equated the entire energy of the electron to the energy of the X-ray photon for obtaining minimum wave length (or, the maximum frequency ν_max):

eV = hν_max = hc/λ_min].

Therefore, λ_min = hc/Ve.

The ratio of wave lengths required in the problem is

λ /λ_min = (h/p)/(hc/Ve ) = Ve/pc………..(i)

The kinetic energy of the electron is eV and the momentum p of the electron is related to its kinetic energy as

eV = p²/2m where m is the mass of the electron.

Therefore, p = √(2mVe)

Substituting this in Eq (i), we obtain

λ /λ_min = (1/c) √(Ve/2m)

Kerala Engineering and Medical Entrance Examination (KEAM) 2009 Questions on Digital Circuits

The following question appeared in Kerala Engineering Entrance 2009 Examination question paper:

The truth table for the following logic circuit is

The top AND gate gets inputs (NOT A) and B. The bottom AND gate gets inputs A and (NOT B). The outputs of the two AND gates are respectively [(NOT A) AND B] and [A AND (NOT B)]. These are the inputs for the final OR gate. The output of the final OR gate is [(NOT A) AND B] OR [A AND (NOT B)]. This is the Boolean expression for the XOR gate (exclusive OR gate, also termed EXOR gate). The truth table for the XOR gate is given in option (A).

Even if you don’t know the XOR gate and its truth table, you can substitute values 0 and 1 for the inputs A and B and find the output in each trial. You can easily arrive at the answer given in option (A).

The following question appeared in Kerala Medical Entrance 2009 Examination question paper:

If the two inputs of a NAND gate are shorted, the gate is equivalent to

(a) XOR

(b) OR

(c) NOR

(d) NOT

(e) AND

If both inputs of a two input NAND gate are shorted, both inputs get the same signal. Because of the NOT operation at the output side of the NAND, the output is the complement of the input. The circuit therefore behaves as a NOT gate [Option (d)].

Karnataka CET Questions (MCQ) on Digital Circuits

I give below 4 questions which appeared in Karnataka Common Entrance Test (CET) question papers of the years 2009, 2008, 2007 and 2006 respectively:

(1) In the following combination of logic gates, the outputs A, B and C are respectively

(1) 0, 1, 0

(2) 1, 1, 0

(3) 1, 0, 1

(4) 0, 1, 1

In circuit (A), the output of the input NAND gate is 0 since both inputs are 1. After inversion it becomes 1 and the final OR gate produces output 1.

In circuit (B), the output NAND gate gets 1 and 0 as its inputs and hence the final output is 1.

In the circuit (C), the NOR gate at the input side supplies 0 as one input to the AND gate at the output side. Hence the final output is 0.

Therefore, the correct option is (2).

(2) To get an output Y = 1 from the circuit shown, the inputs A, B and C must be respectively

(1) 0, 1, 0

(2) 1, 0, 0

(3) 1, 0, 1

(4) 1, 1, 0

In all cases the OR gate will provide a high (level 1) input to the AND gate at the output side. The AND gate should have both inputs at level 1 in order to have its output at 1. This is possible only if input C is 1. So the correct option is (3).

(3) The truth table given below is for

(A and B are the inputs, y is the output)

(1) NAND

(2) XOR

(3) AND

(4) NOR

The output y is the complement of AND operation and hence the answer is NAND [Option (1)].

(4) Identify the logic operation performrd by the cicuit given below

(1) NOT

(2) AND

(3) OR

(4) NAND

If you know De Morgan’s theorem, you will immediately obtain the answer as AND. The NOR gate at the output side gets inputs which are complement of A and complement of B. According to De Morgan’s theorem, (NOT A OR NOT B) is the same as [NOT (A AND B)]. Since the output gate is not an OR gate, but a NOR gate, the output is (A AND B).

Therefore, given circuit performs AND operation [Option (2)].

[If you do not know De Morgan’s theorem, you may try substituting values 0 and 1 for A and B to obtain the answer].

Two Questions (MCQ) from Wave Optics

Here are two questions from wave optics. The first question is meant for testing your understanding of coherent sources and interference while the second one is meant for testing your knowledge of the interference pattern produced by Young’s double slit.

(1) Two coherent sources emitting light of wave length λ and amplitude A produce an interference pattern on a screen. The maximum intensity of the interference pattern is I. If the light sources are not coherent but the wave length and amplitude are λ and A respectively, the maximum intensity on the screen will be

(a) I/4

(b) I/2

(c) I

(d) √I

(e) 2√I

At the interference maximum the amplitudes of the light waves get added. Therefore, the resultant amplitude at the interference maximum is A + A = 2A.

The intensity is directly proportional to the square of the amplitude so that we have

I α 4A²

Or, I = k×4A²……………(i)

where k is the constant of proportionality

Since the amplitude of each light wave is A, the intensity (I₀) of light produced by each source is given by

I₀ = k×A²

This gives I₀ = I/4 from (i).

Each source will produce uniform intensity I₀ everywhere on the screen and the resultant intensity everywhere will be 2I₀ (since there are two sources) and we have

2I₀ = 2×I/4 = I/2

The correct option is (b).

(2) The angular width of the interference fringes obtained in a double slit experiment using light of wave length λ is found to be θ. If the entire experimental arrangement is immersed in water having refractive index 4/3, the angular fringe width will be

(a) θ/4

(b) θ/3

(c) θ

(d) 4θ/3

(e) 3θ/4

The angular fringe width is λ/d where d is the separation between the slits. Therefore we have

θ = λ/d.

When the arrangement is submerged in water of refractive index n, the wave length of light is reduced to λ' given by

λ'= λ/n = 3λ/4 since n = 4/3

The angular fringe width now becomes θ'= λ'/d = 3λ/4d = 3θ/4.

You will find some useful multiple choice questions in this section at AP Physics Resources: Multiple Choice Questions on Interference and Diffraction .

“New Year’s day is every man’s birthday.”

– Charles Lamb

Happy New Year…

Two Questions (MCQ) involving Kinetic Energy and Potential Energy

The following question appeared in Kerala Engineering Entrance 2007 question paper:

A simple pendulum is released from A as shown. If m and l represent the mass of the bob and the length of the pendulum, the gain in kinetic energy at B is

(a) mgl/2

(b) mgl/√2

(c) (mgl√3)/2

(d) 2mgl/√3

(e) mgl

From position A, the bob of the pendulum has fallen through a distance l cos 30º (fig.). Therefore, the loss of potential energy is mg l cos 30º = (mgl√3)/2.

Therefore, the gain in kinetic energy by the bob (by the law of conservation of energy) is (mgl√3)/2.

Here is another question which will be useful in the present context:

The bob of a simple pendulum has mass 0.2 kg. The bob is drawn aside so that the string is horizontal and is released. When it swings through 60º its kinetic energy is 0.5 J. The kinetic energy when the sting becomes vertical will be

(a) √3 J

(b) 1/√3 J

(c) √3/2 J

(d) 2/√3 J

(e) 1 J

You can use the figure used with the previous question for working out the present question as well since the angle turned is 60º. The loss of potential energy by the bob on swinging through 60º is mg l cos30º itself. [Or, you may take it as mg l sin60º].

Since the loss of potential energy is equal to the gain of kinetic energy, we have

mg l cos30º = 0.5 J

Or, (mgl√3)/2 = 0.5 J.

The kinetic energy when the string becomes vertical will be mgl since the bob falls through a distance l and loses its entire initial potential energy mgl.

From the above equation mgl = 1/√3 J [Option (b)].

[Note that the mass of the bob given in the question just serves as a distraction].

You will find some useful questions in this section here.

IIT-JEE 2009 Multiple Choice Question (Single Answer Type) on Elastic Collision

The following question which appeared in IIT-JEE 2009 question paper is based on the principle that in an elastic collision between two particles of the same mass, the velocities get interchanged: 

Two small particles of equal masses start moving in opposite directions from a point A in a horizontal circular orbit. Their tangential velocities are v and 2v respectively as shown in the figure. Between collisions, the particles move with constant speeds. After making how many elastic collisions, other than that at A, these two particles will again reach the point A?

(a) 4

(b) 3

(c) 2

(d) 1    

Particle P₁, after traversing one third (AB) of the circular track in the anticlockwise direction, will collide with particle P₂ at position B. This collision occurs when the particle P₂  has traversed two thirds (ACB) of the circular path (since the speed of P₂ is twice that of P₁). 

         Since the collision is elastic and the particles are of equal masses, their velocities are interchanged. P₁ now travels in the clockwise direction with speed 2v and P₂ travels in the anticlockwise direction with speed v. The second collision takes place at position C. (BAC is two thirds of the circular path while BC is one third of the path). The second collision at C results in the reversal of the velocities and P₁ now travels in the anticlockwise direction with speed v where as P₂ travels in the clockwise direction with speed 2v. Therefore, the third collision will occur at A.

Since the third collision at A is not to be counted, the answer is 2 [Option (c)].

*    *    *    *    *    *    *    *    *    *    *    *   *    *    *    *    *

Now suppose we modify the above question as follows:

Two small particles of equal masses start moving in opposite directions from a point A in a horizontal circular frictionless track. Their speeds are v and 3v respectively as shown in the figure. After making how many elastic collisions, other than that at A, these two particles will again reach the point A ?

(a) 4

(b) 3

(c) 2

(d) 1

[You may use the figure shown with the question above, replacing the velocity 2v with 3v]

You can easily arrive at the answer which is 3 [Option (b)].

Apply for Entrance Examination for Admission to Medical/ Agriculture/ Veterinary/ Engineering/ Architecture Degree Courses 2010 (KEAM 2010), Kerala

The Commissioner for Entrance Examinations, Govt. of Kerala, has invited applications for the Entrance Examinations for admission to the following Degree Courses in various Professional Colleges in the State for 2010-11.

(a) Medical: (i) MBBS (ii) BDS (iii) BHMS (iv) BAMS (v) BSMS

(b) Agriculture: (i) BSc. Hons. (Agriculture) (ii) BFSc. (Fisheries) (iii) BSc. Hons. (Forestry)

(c) Veterinary: BVSc. & AH

(d) Engineering: B.Tech. [including B.Tech. (Agricultural Engg.)/B.Tech. (Dairy Sc. & Tech.) courses under the Kerala Agricultural University]

(e) Architecture: B.Arch.

Dates of Exam:

Engineering Entrance Examination (For Engineering courses except Architecture)

19.04.2010 Monday 10.00 A.M. to 12.30 P.M. Paper-I : Physics & Chemistry.

20.04.2010 Tuesday 10.00 A.M. to 12.30 P.M. Paper-II: Mathematics.

Medical Entrance Examination (For Medical, Agriculture and Veterinary Courses)

21.04.2010 Wednesday 10.00 A.M. to 12.30 P.M. Paper-I : Chemistry & Physics.

22.04.2010 Thursday 10.00 A.M. to 12.30 P.M. Paper-II: Biology.

Application form and Prospectus will be distributed from 07.12.2009 to 06.01.2010 through selected branches of Post Offices in Kerala and outside the State. The amount towards the fee of application (Rs. 700/- for general candidates and Rs.350 for SC/ST candidates) is to be remitted in cash at the Post Offices. To find the list of post offices, selected as sales centres, and details of additional fee of Rs.8500/- in the case of candidates opting Dubai as the centre of exam, visit the site http://www.cee-kerala.org/.

Online submission of application is possible in the case of candidates claiming no reservation benefit and those belonging to Non-Keralite category. See details at http://www.cee-kerala.org/.

Last Date for receipt of Application by CEE: 06-01-2010 (Wednesday) - 5 PM

Candidates seeking admission to B.Arch. course should also submit their application to the CEE. There is no state level Entrance Examination for this purpose. These candidates should write the National Aptitude Test for Architecture and should forward the NATA score and mark list of the qualifying examination to the CEE on or before 05-06-2010.

You will find complete details and information updates at http://www.cee-kerala.org/

If you want to see earlier KEAM questions discussed on this site, type in ‘Kerala’ in the search box at the top left of this page and strike the enter key or click on the search button.