Two Questions on Transformers

Questions on transformers at the 12^th 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 N_P turns in the primary and N_S turns in the secondary. If the voltage per turn is V_P for primary and V_S for secondary V_S/V_P is equal to

(a) 1

(b) N_S/N_P

(c) N_P/N_S

(d) (N_P/N_S)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

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.

AIEEE Questions on Oscillations

When you prepare for any examination, it will be very useful to work out the questions which appeared in earlier examinations. Here are two questions on oscillations which appeared in AIEEE 2006 question paper:

(1) Starting from the origin a body oscillates simple harmonically with a period of 2 s. After what time will the kinetic energy be 75% of the total energy?

(1) 1/12 s

(2) 1/6 s

(3) 1/4 s

(4) 1/3 s

This simple harmonic motion can be represented by the equation,

y = A sin ωt where y is the displacement at the instant t, A is the amplitude and ω is the angular frequency.

The instantaneous velocity v of the particle is given by

v = dy/dt = Aω cosωt

The maximum velocity v_max of the particle is evidently Aω and the maximum kinetic energy which is equal to the total energy is ½ mv_max² where m is the mass of the particle. We have

½ mv² = (¾)(½)mv_max²

Therefore, ½ m (Aω cosωt)² = (¾)(½)m(Aω)² from which cosωt = (√3)/2

Therefore, ωt = π/6 so that t = π/6ω = π/(6×2π/T ) = 1/6 s since the period T is 2 s.

(2) The maximum velocity of a particle executing simple harmonic motion with amplitude 7 mm is 4.4 ms^–1. The period of oscillation is

(1) 100 s

(b) 0.01 s

(c) 10 s

(d) 0.1 s

Since the maximum velocity v_max = Aω and the period T = 2π/ω we have

T = 2πA/v_max = 2π×7×10^–3/4.4 = 0.01 s

You will find more questions (with solution) in this section here as well as here.

Apply for All India Engineering/Architecture Entrance Examination 2009 (AIEEE 2009)

Time to apply for AIEEE 2009!

Application Form and the Information Bulletin in respect of the All India Engineering/Architecture Entrance Examination 2009 (AIEEE 2009), which will be conducted on 26-4-2009, are being distributed from 5.12.2008 and will continue till 5.1.2009. Candidates may apply for AIEEE 2009 either on the prescribed Application Form or make application ‘online’. Visit the site http://aieee.nic.in immediately for details. Apply for the exam without delay.

You will find many old AIEEE questions (with solution) on this site. You can access all of them by typing ‘AIEEE’ in the search box at the top left of this page and clicking on the adjacent ‘search blog’ box.

Old AIEEE questions (with solution) can be obtained at the site http://physicsplus.blogspot.com as well by performing a similar search on the site.

Two AIPMT 2008 Questions (MCQ) from Current Electricity

Some of you may understand the principles in physics very well but your capacity for numerical manipulations may be poor. Practice can make you strong in solving questions involving numerical manipulations so that you will not waste your precious time on such questions. Here are two questions which appeared in AIPMT 2008 question paper:

(1) An electric kettle takes 4 A current at 220 V. How much time will it take to boil 1 kg of water from temperature 20º C? The temperature of boiling water is 100º C.

(1) 8.4 min

(2) 12.6 min

(3) 4.2 min

(4) 6.3 min

We have VIt =mcθ where V is the voltage, I is the current, t is the time of flow of the current, m c is the specific heat of water (which is approximately 4200 Jkg^–1 K^–1) and θ is the temperature rise. is the mass of water,

Therefore, 220×4×t = 1×4200×(100 – 20)

This will give t = 381 sec. (nearly) which is approximately 6.3 min.

(2) a galvanometer of resistance 50 Ω is connected to a battery of 3 V along with a resistance of 2950 Ω in series. A full scale deflection of 30 divisions is obtained in the galvanometer. In order to reduce this deflection to 20 divisions, the resistance in series should be

(1) 5550 Ω

(2) 6050 Ω

(3) 4450 Ω

(4) 5050 Ω

Since the current through an ordinary galvanometer is directly proportional to the deflection (remember that in a tangent galvanometer this is not the case) we have

3/(50+2950) = k×30 when the deflection is 30 divisions.

Here k is the proportionality constant (figure of merit of the galvanometer).

If the resistance in series for reducing the deflection to 20 divisions is X we have

3/(50+X) = k×20

On dividing the first equation by the second,

(50+X)/3000 = 3/2 from which X = 4450 Ω.