If we did all things we are capable of, we would literally astound ourselves.

– Thomas A. Edison

Saturday, March 21, 2009

Electronics- Multiple Choice Questions involving Diodes

I am enough of an artist to draw freely upon my imagination. Imagination is more important than knowledge. Knowledge is limited. Imagination encircles the world.

– Albert Einstein


Here are three questions involving diodes:

(1) In the circuit shown in figure the inputs at A and B are respectively 7 V and 0.8 V. If the potential drops across the forward biased diodes are negligible and the breakdown voltage of the zener diode is 5V, the output voltage is

(a) 0.8 V

(b) 7 V

(c) 3.9 V

(d) 0 V

(e) 5 V

The diode connected to input A will be forward biased and the diode connected to input B will be reverse biased. The input of 7 V will be impressed at the junction between the diodes and the resistor. The zener diode will breakdown and the output voltage will be the breakdown voltage of the zener diode, which is equal to 5 V. (Two volts will be dropped across the resistor).

(2) If the logical 0 is 0 volt and logical 1 is + 5 volt, the circuit shown in the above question can function as

(a) AND gate

(b) NOT gate

(c) EXOR gate

(d) OR gate

(e) none of the above

If both inputs are 0 volt the output will be zero. If at least one input is 5V or all inputs are 5 V, the output will be 5 V and the circuit can function as an OR gate.

(3) The current through the 40 Ω resistor in the circuit shown in the adjoining figure is

(a) 50 mA

(b) 80 mA

(c) 40 mA

(d) 92.3 mA

(e) 44.4 mA

The diode D2 is reverse biased and hence no current flows through its branch. The branches containing diodes D1 and D3 are in parallel and the diodes are forward biased. The effective resistance of these two branches in parallel is (20×20)/(20+20) = 10 Ω. Therefore, the total resistance in series with the 4 volts battery is 40 Ω +10 Ω = 50 Ω.

Therefore,the current through the 40 Ω resistor is 4 V/50 Ω = 0.08 A = 80 mA.

You will find some useful multiple choice questions (with solution) in electronics here as well as here.

Saturday, March 07, 2009

Dates of Kerala Govt. Entrance Examinations (KEAM 2009) for the Engineering and Medical Streams Changed

The Commissioner for Entrance Examinations, Govt. of Kerala, has notified that the dates of the Entrance Examinations conducted by Govt. of Kerala for Admission to Medical/ Agriculture/ Veterinary/ Engineering Degree Courses 2009 (KEAM 2009), have been changed. This is the second time the dates get changed. Here is the renewed time table:

Dates of Exam:

Engineering Entrance Examination:

25.05.2009 Monday 10.00 A.M. to 12.30 P.M. Paper-I : Physics & Chemistry.

26.05.2009 Tuesday 10.00 A.M. to 12.30 P.M. Paper-II: Mathematics.

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

27.05.2009 Wednesday 10.00 A.M. to 12.30 P.M. Paper-I : Chemistry & Physics.

28.05.2009 Thursday 10.00 A.M. to 12.30 P.M. Paper-II: Biology.

You will find complete details at the site http://www.cee-kerala.org/

The details regarding the National Aptitude Test in Architecture (NATA) required for those who apply for Architecture also can be seen at the site.

Sunday, March 01, 2009

All India Pre-Medical/Pre-Dental Entrance Examination (AIPMT) 2008 Questions on Nuclear Physics

Here are two multiple choice questions which appeared in All India Pre-Medical/Pre-Dental Entrance Examination (AIPMT) 2008:

(1) Two radioactive materials X1 and X2 have decay constants 5λ and λ respectively. If initially they have the same number of nuclei then the ratio of the number of nuclei of X1 to that of X2 will be 1/e after a time

(1) λ/2

(2) 1/(4λ)

(3) e/λ

(4) λ

If the initial number of nuclei is N0 we have

N1 = N0e–5λ t and

N2 = N0eλ t where N1 and N2 are the number of nuclei of X1 and X12 at time t.

Therefore N1/N2 = e–4λt

This will be equal to 1/e when t =1/(4λ).

(2) Two nuclei have their mass numbers in the ratio of 1:3. The ratio of their nuclear densities would be

(1) 3:1

(2) (3)1/3:1

(3) 1:1

(4) 1:3

The mass of a nucleus is directly proportional to the number (A) of the nucleons. The volume of the nucleus is (4/3)πR3 where R is the nuclear radius. But, R = R0A where R0 is a constant (equal to 1.1 ×10-15m). So, the volume of the nucleus also is directly proportional to the nucleon number A. Since the density is the ratio of mass to volume, it follows that the density of nuclear matter is independent of the nucleon number A so that the correct option is (c).

You will find some useful multiple choice questions (with solution) on nuclear physics at physicsplus

Sunday, February 08, 2009

Two Questions on Transformers

Questions on transformers at the 12th class level are simple. But unusually simple questions may make you unusually careless in answering. See the following question which appeared in H.P.P.M.T. 2005 question paper:

An ideal transformer has NP turns in the primary and NS turns in the secondary. If the voltage per turn is VP for primary and VS for secondary VS/VP is equal to

(a) 1

(b) NS/NP

(c) NP/NS

(d) (NP/NS)2

In an ideal transformer the magnetic flux linked per turn of the primary and the secondary windings are the same. Therefore, the voltage per turn of the primary is the same as the voltage per turn of the secondary. The required voltage ratio is 1 [Option (a)].

Here is another simple question:

A small transformer with 80% efficiency has turns ratio 10:1. If a dry cell of emf 1.5 volt is connected across the primary, the voltage across the secondary will be

(a) 15 V

(b) 0.15 V

(c) 12 V

(d) 0.12 V

(e) zero

Don’t waste your time thinking of the efficiency and the turns ratio. A transformer requires a varying voltage across its primary to produce a voltage across its secondary. Since the dry cell supplies a steady voltage and therefore a steady current, there cannot be a flux change. The voltage induced across the secondary is zero.

You will find a few multiple choice questions (with solution) on electromagnetic induction here

Monday, January 26, 2009

Kerala Entrance Examinations for Admission to Medical/ Agriculture/ Veterinary/ Engineering/ Architecture Degree Courses 2009 (KEAM 2009)

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 2009-2010.

(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)
20.04.2009 Monday 10.00 A.M. to 12.30 P.M. Paper-I : Physics & Chemistry.
21.04.2009 Tuesday 10.00 A.M. to 12.30 P.M. Paper-II: Mathematics.

Medical Entrance Examination (For Medical, Agriculture and Veterinary Courses)
22.04.2009 Wednesday 10.00 A.M. to 12.30 P.M. Paper-I : Chemistry & Physics.
23.04.2009 Thursday 10.00 A.M. to 12.30 P.M. Paper-II: Biology.

Sale of Application will commence on : 27-01-2009
Last Date for Submission of Application: 26-02-2009
Full details can be had at

http://www.cee-kerala.org/

In addition to the multiple choice questions (of the type expected to appear in KEAM 2009) on this site, you will find many similar useful questions with solution at the site http://physicsplus.blogspot.com. If you want to see earlier KEAM questions only, type in ‘Kerala’ in the search box at the top left of the site and click on the ‘search blog’ box.