Kerala Engineering Architecture Medical (KEAM) 2008 Entrance Examination Questions on Digital Circuits

As usual, simple questions were asked from the section on digital circuits in KEAM 2008 Examination. Here is the question which appeared in the Engineering Entrance question paper:

The combination of the following gates produces

(a) AND gate

(b) NAND gate

(c) NOR gate

(d) OR gate

(e) NOT gate

The first gate is a NAND gate. The second gate also is a NAND gate. But it functions as a NOT gate since its inputs are shorted. So we have a NAND followed by a NOT and hence the combination functions as an AND gate.

Here is the question which appeared in the Kerala Medical Entrance (2008) question paper:

The output Y when all the three inputs are first high and then low, will respectively be

(a) 1, 0

(b) 1, 1

(c) 0, 0

(d) 0, 1

(e) 1, –1

When all the three inputs are high the NAND gate following the AND gate has both inputs at high level. The output Y of the NAND is therefore low. When all the three inputs are low the NAND gate has both inputs low and hence the output is high. The correct option is (d).

You will find AIEEE 2008 and AIPMT 2008 questions on digital circuits with solution here.

All India Pre-Medical/Pre-Dental Entrance Examination (AIPMT) 2008 Questions from Electrostatics

Here are the two questions from electrostatics which appeared in the All India Pre-Medical/Pre-Dental Entrance Examination (AIPMT) 2008 question paper:

(1) A thin conducting ring of radius R is given a charge +Q. The electric field at the centre O of the ring due to the charge on the part AKB of the ring is E. The electric field at the centre due to the charge on the part ACDB of the ring is:

(a) E along OK

(b) E along KO

(c) 3E along OK

(d) 3E along KO

Since the ring is conducting, the charge on it gets distributed uniformly along it. The resultant electric field at the centre of the ring is zero (since a unit positive test charge placed at the centre will be pushed by equal radial forces all around). The field due to the charge on the part ACDB of the ring is therefore equal and opposite to the field due to the charge on the part AKB of the ring. So the answer is E along OK [option (a)].

[It would have been better if it is mentioned that K is the mid point of the arc AKB].

(2) The energy required to charge a parallel plate condenser of plate separation d and plate area of cross-section A such that the uniform electric field between the plates is E, is

(a) ε₀ E²/Ad

(b) ε₀ E²Ad

(c) ½ ε₀E²Ad

(d) ½ ε₀E²/Ad

The energy (U) required for charging a capacitor of capacitance C to V volt is given by

U = ½ CV² where C = ε₀A/d for a parallel plate capacitor with air (or vacuum) as dielectric.

Since E = V/d, we have V = Ed

Therefore, U = (½) (ε₀A/d)(Ed)² = (½) ε₀E²Ad

You can find more questions (with solution) in this section at AP Physics Resources.

Kerala Engineering Entrance 2008 Questions on Electromagnetic Induction

Questions involving electromagnetic induction at your level are usually simple and interesting. Here are the two questions which appeared in Kerala Engineering Entrance 2008 question paper:

(1) If the self inductance of 500 turn coil is 125 mH, then the self inductance of similar coil of 800 turns is

(a) 48.8 mH

(b) 200 mH

(c) 187.5 mH

(d) 320 mH

(e) 78.1 mH

Self inductance is the total magnetic flux linked with the coil when unit current flows through the coil. Since the total flux and the magnetic field are directly proportional to the number of turns in the coil, the self inductance is directly proportional to the square of the number of turns in the coil. Therefore we have

500²/800² = 125/L where L is the self inductance of the 800 turn coil.

This yields L = 320 mH.

(2) The flux linked with a circuit is given by Φ= t³ + 3t – 7. The graph between time (x–axis) and induced emf (y–axis) will be a

(a) straight line through the origin

(b) straight line with positive intercept

(c) straight line with negative intercept

(d) parabola through the origin

(e) parabola not through the origin

The induced emf, V = – dΦ/dt = – 3t² – 3

If V is plotted against t, a parabola which does not pass through the origin will be obtained. So the correct option is (e).

Electrostatics: MCQ on Charged Spherical Drops

Often questions involving the calculation of the potential of a drop obtained by the combination of a number of identical charged spherical droplets are seen in college entrance question papers. If n identical small drops each of radius r carrying charge q coalesce to form a single large drop of radius R, we have

R = n^1/3 r

[You will get this by equating the volumes: n×(4/3)πr³ = (4/3)πR³]

The potential V on the surface of each small drop is given by

V = (1/4πε₀)(q/r)

The total charge on the larger drop is nq. Therefore, the potential V’ on its surface is

V’ = (1/4πε₀)(nq/R) = (1/4πε₀)(nq/n^1/3r) since R = n^1/3 r

Therefore, V’ = n^2/3V

The electric field E on the surface of each small drop is given by

E = (1/4πε₀)(q/r²)

The electric field E’ on the surface of the large drop is given by

E’ = (1/4πε₀)(nq/R²) = (1/4πε₀)(nq/n^2/3r² ) since R = n^1/3 r

Therefore, E’ = n^1/3E

If you can remember the above expressions for the electric potential and field on the larger drop, multiple choice questions involving them can be answered in no time. But it is always more rewarding to remember the basic things so that you can calculate the required quantities in all situations.

The calculation of V’ in the above form itself appeared as a multiple choice question in Kerala Engineering Entrance 2008 question paper.

Now answer the following multiple choice question involving charged drops. You may try them yourself and then check with the given solution. Here are the questions:

(1) Sixty four identical drops each charged by q coulomb to potential V volt coalesce to form a single large drop. The charge and potential on the large drop are respectively

(a) 64q, 64V

(b) 64q, V

(c) 8q, 16V

(d) 16q, 16V

(e) 64q, 16V

The charge on the large drop is the sum of the charges on the small drops and is equal to 64q.

As shown above, the potential on the large drop is V’ = n^2/3V = 64^2/3V =16V [Option (e)].

(2) Two identical soap bubbles A and B are uniformly charged with the same amount of charge. But the charge on A is positive where as the charge on B is negative. (The electrical interaction between the bubbles is negligible). Because of charging, the excess of pressure inside

(a) both bubbles will increase

(b) both bubbles will decrease

(c) both bubbles will remain unchanged

(d) A will increase and that inside B will decrease

(e) B will increase and that inside A will decrease

Because of the repulsive force between like charges, the bubbles will expand and hence the excess of pressure inside both bubbles will decrease.

You will find many useful posts on different branches of Physics at your level at apphysicsresources.blogspot.com and at physicsplus.blogspot.com. The essential equations to be remembered in the section, ‘Electric field and Potential’ can be found here.

Kerala Medical Entrance (KEAM) 2008 Questions from Newton’s Laws of Motion

The following questions which appeared in Kerala Medical Entrance 2008 question paper are of the type popular among question setters:

(1) Two blocks of masses 7 kg and 5 kg are placed in contact with each other on a smooth surface. If a force of 6 N is applied on the heavier mass, the force on the lighter mass is

(a) 3.5 N

(b) 2.5 N

(c) 7 N

(d) 5 N

(e) 6 N

The common acceleration a of the system of masses is given by

a = Driving force/Total mass moved = 6 N/(7+5) kg = 0.5 ms^–2

The force on the lighter mass = Ma = 5×0.5 = 2.5 N.

(2) A body of mass 60 kg suspended by means of three strings P, Q and R as shown in the figure is in equilibrium. The tension in the string P is

(a)130.9g N

(b) 60g N

(c) 50g N

(d) 103.9g N

(e) 109g N

Let us denote the tensions in the three strings by P, Q and R. The concurrent forces P, Q and R keep the common meeting point of the strings in equilibrium so that they can be represented by the three sides of a triangle taken in order as shown in the adjoining figure. From this right angled triangle,

Q/P = tan 30º where Q = weight of the mass M = Mg =60g newton.

[This equation is often written as W/_H = tan θ where W is the weight and H is the horizontal force and is known as tangent law].

Substituting for Q, we have 60g/P = 1/√3, from which P = 60g×√3 = 103.9g N.