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).