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.