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.

Apply for Joint Entrance Examination (IIT-JEE 2009) for Admission to IITs and other Institutions

The Joint Entrance Examination for Admission to IITs and other Institutions (IIT-JEE 2009) will be held on April 12, 2009 (Sunday) as per the following schedule:
Paper – 1: 09:00 hrs– 12:00 hrs

Paper – 2: 14:00 hrs– 17:00 hrs

Application materials of the Joint Entrance Examination 2009 will be issued with effect from 19^th November, 2008. On-line submission of Application will also commence on the same day.

Application materials can be purchased from designated branches of Banks and all IITs between 19.11.2008 and 24.12.2008 by paying Rs. 500/- in the case of SC/ST/PD/Female candidates, Rs.1000/- in case of male general/OBC candidates and US$ 100 in the case of candidates appearing in Dubai centre.

The last date for postal request of application materials is 16-12-2008.

The last date for receipt of the completed application at the IITs is 24^th December, 2008.

Online Submission of Application will be available between 19.11.2008 and 24.12.2008 between 8:00 hrs and 17:00 hrs. through the JEE websites of the different IITs.

The JEE websites of the different IITs are given below:
IIT Bombay: http://www.jee.iitb.ac.in

IIT Delhi: http://jee.iitd.ac.in

IIT Guwahati: http://www.iitg.ac.in/jee

IIT Kanpur: http://www.iitk.ac.in/jee

IIT Kharagpur: http://www.iitkgp.ernet.in/jee

IIT Madras: http://jee.iitm.ac.in

IIT Roorkee: http://www.iitr.ac.in/jee

Visit the web site http://www.jee.iitb.ac.in (or any other) for more details.

Two AIPMT 2008 Questions from Thermal Physics

The following questions appeared in the All India Pre-Medical / Pre-Dental Entrance 2008 Examination:

(1) At 10º C the value of the density of a fixed mass of an ideal gas divided by its pressure is x. At 110º C this ratio is

(1) 383x/283

(2) 10x/110

(3) 283x/383

(4) x

We have PV = rT where r is the gas constant for the given mass.

Since the volume V = M/ρ where M is the mass and ρ is the density, we have

P/ρ = rT/M

Since r and M are constants for a given mass of gas, we have

(ρ₁/P₁) /(ρ₂/P₂) = T₂/T₁ where the suffix 1 is for the quantities at temperature 10º C and suffix 2 is for the quantities at temperature 110º.

It is given that (ρ₁/P₁) = x.

Therefore x/(ρ₂/P₂) = 383/283 since the temperatures are in degree kelvin.

From this (ρ₂/P₂) = 283x/383

(2) On a new scale of temperature (which is linear) and called the W scale, the freezing and boiling points of water are 39º W and 239º W respectively. What will be the temperature on the new scale, corresponding to a temperature of 239º C on the Celsius scale?

(1) 117º W

(2) 200º W

(3) 139º W

(4) 78º W

We have 0º C = 39º W and 100º C = 239º W

A temperature difference of 100º C is therefore equivalent to a temperature difference of 200º W. A temperature difference of 1º C is thus equivalent to a temperature difference of 2º W.

Therefore, a temperature 39º C is equal to [39 + (39×2)]º W = 117º W

All India Pre-Medical / Pre-Dental Entrance Examination (AIPMT)-2009

CBSE (Central Board of Secondary Education), Delhi has invited applications for All India Pre-Medical / Pre-Dental Entrance Examination -2009 for admission to 15% of the total seats for Medical/Dental Courses in all Medical/Dental colleges run by the Union of India, State Governments, Municipal or other local authorities in India except in the States of Andhra Pradesh and Jammu & Kashmir, as per the following schedule :-

(i). Preliminary Examination - 5th April, 2009 (Sunday)

(ii). Final Examination - 10th May, 2009 (Sunday)

Candidate can submit their applications for the All India Pre-Medical/Pre-Dental Entrance Examination either offline or online:

Offline:

Offline submission of Application Form may be made using the prescribed application form. The Information Bulletin and Application Form costing Rs.600/- (including Rs.100/- as counseling fee) for General & OBC Category Candidates and Rs.350/- (including Rs.100/- as counseling fee) for SC/ST Category Candidates inclusive of counseling fee can be obtained against cash payment from 22-10-2008 to 01-12-2008 from any of the branches of Canara Bank/ Regional Offices of the CBSE. Details can be obtained by visiting the site http://www.aipmt.nic.in/.

Online:

Online submission of application may be made by accessing the Board’s website http://www.aipmt.nic.in/. from 22-10-2008 (10.00 A.M.) to 01-12-2008 (5.00 P.M.). Candidates are required to take a print of the Online Application after successful submission of data. The print out of the computer generated application, complete in all respect as applicable for Offline submission should be sent to the Deputy Secretary (AIPMT), Central Board of Secondary Education, Shiksha Kendra, 2, Community Centre, Preet Vihar, Delhi-110 301 by Speed Post/Registered Post only. Fee of Rs.600/-(including Rs.100/- as counseling fee) for General and OBC Category Candidates and Rs.350/- (including Rs.100/- as counseling fee) for SC/ST category candidates may be remitted in one of the following ways :

1. By credit card, or

2. Through Demand Draft in favour of the Secretary, Central Board of Secondary Education, Delhi, drawn on any Nationalized Bank payable at Delhi. Instructions for Online submission of Application Form is available on the website http://www.aipmt.nic.in/. Application Form along with original Demand Draft should reach the Board on or before 04-12-2008.

For complete details and information updates visit the web site http://www.aipmt.nic.in/.

MCQ on Newton’s Laws of Motion- KEAM (Engineering) 2008 Questions

The following multiple choice questions appeared in Kerala Engineering Entrance 2008 question paper:

(1) A particle of mass 2 kg is initially at rest. A force acts on it whose magnitude changes with time. The force time graph is shown below

The velocity of the particle after 10 s is

(a) 20 ms^–1

(b) 10 ms^–1

(c) 75 ms^–1

(d) 26 ms^–1

(e) 50 ms^–1

The area under the force-time graph gives the impulse received which is equal to the change in momentum of the particle. Since the initial momentum of the particle is zero, the area under the graph gives the momentum of the particle after 10 s.

The area under the curve is made of rectangles and triangles and is equal to100 newton second. [Note that Ns is the same as kg ms^–1].

The velocity of the particle after 10 s = (100 kg ms^–1)/2 kg = 50 ms^–1

(2) An object of mass 5 kg is attached to the hook of a spring balance and the balance is suspended vertically from the roof of a lift. The reading on the spring balance when the lift is going up with with an acceleration of 0.25 ms^–2 is (g = 10 ms^–2)

(a) 51.25 N

(b) 48.75 N

(c) 52.75 N

(d) 47.25 N

(e) 55 N

The weight of a body of mass ‘m’ in a lift can be remembered as m(g-a) in all situations if you apply the proper sign to the acceleration ‘a’ of the lift. The acceleration due to gravity ‘g’ always acts vertically downwards and its sign is to be taken as positive. Since the lift is moving upwards the sign of its acceleration is negative so that the weight of the object as indicated by the spring balance is m[g-(-a)] = m(g+ a) = 5×10.25 = 51.25 N.

(3) A bullet of mass 0.05 kg moving with a speed of 80 ms^–1 enters a wooden block and is stopped after a distance of 0.4 m. The average resistance force exerted by the block on the bullet is

(a) 300 N

(b) 20 N

(c) 400 N

(d) 40 N

(e) 200 N

The bullet is retarded within the wooden block. The retardation is given by the usual equation of uniformly accelerated motion, v² = u² + 2as. Here v = 0, u = 80 ms^–1 and s = 0.4 m.

Therefore, 0 = 80² + 2a×0.4 from which a = – 8000 ms^–2

The retardation is 8000 ms^–2 and the resistance force exerted by the wooden block is 0.05×8000 = 400 N.

AIPMT 2008 Question on Magnetic force

Most of you might have noted that a magnetic dipole placed in a non uniform magnetic field generally experiences a net force and a torque. If the magnetic field is uniform, the net force will be zero; but there will be a torque (if the magnetic moment vector is not parallel or antiparallel to the magnetic field direction). The following question which appeared in AIPMT 2008 question paper is worth noting:

A closed loop PQRS carrying a current is placed in a uniform magnetic field. If the magnetic forces on segments PS, SR and RQ are F₁, F₂ and F₃ respectively and are in the same plane of the paper and along the directions shown, the force on the segment QP is

(1) √[(F₃ – F₁)² + F₂²]

(2) √[(F₃ – F₁)² – F₂²]

(3) F₃ – F₁ + F₂

(4) F₃ – F₁ – F₂

The net force on the loop in the horizontal direction is F₃ – F₁. Since the force F₂ is in the vertical direction, the reultant of the three forces F₁, F₂ and F₃ is √[(F₃ – F₁)² + F₂²]. Since the net force on the entire loop must be zero in the uniform magnetic field, the force on the segment QP must be the equilibrant of √[(F₃ – F₁)² + F₂²]. Therefore the force on the segment QP has magnitude √[(F₃ – F₁)² + F₂²] and direction opposite [Option (1)].

Now consider a question which is a little more difficult:

A proton of charge q and mass m proceeding along the positive X-direction with speed v encounters a uniform magnetic field of flux density B directed along the negative Z-direction. If the field is confined in the region between x = 0 and x = d and the proton emerges from the field along a direction making an angle of 45º with its initial velocity, the value of d must be

(a) (1/√2)(mv/qB)

(b) √2 mv/qB

(c) mv/qB

(d) 2 mv/qB

(e) mv/2qB

The situation is shown in the adjoining figure. C is the centre of the circular path of the proton in the magnetic field and AB is the radius R drawn from the point A from which the proton exits from the field. We have

R = mv/qB which you get by equating the centripetal force mv²/R to the magnetic force qvB.

Since d = AN = R sin45º where N is the foot of the perpendicular (drawn from A) to the Y-axis, we have

D = R/√2 = (1/√2)(mv/qB)

You will find many useful questions (with solution) on magnetic force at apphysicsresources

Two Kerala Engineering & Medical Entrance 2008 Questions on Resistive Networks

Questions involving resistive networks can be solved are often found in degree entrance examinations. Here is a reasonably good question which appeared in KEAM 2008 (Engineering) question paper:

In the circuit shown, if the resistance 5 Ω develops a heat of 42 J per second, the heat developed in 2 Ω must be about (in J s^–1)

(a) 25

(b) 20

(c) 30

(d) 35

(e) 40

The heat developed per second (power dissipation) in a resistance R is V²/R where V is the voltage across the resistance

[This expression has the forms I²R and VI where I is the current. Even though you can use these forms also to solve the problem, the form V²/R will be more convenient here]

Therefore we have

V²/5 = 42 watt

The power dissipated in the parallel branch containing the resistors 6 Ω and 9 Ω is V²/(6+9) = V²/15 and this must be 14 watt (since the denominator is 3 times)

The total power dissipated in the two parallel branches is thus 42+14 = 56 watt and the effective resistance of the two parallel branches is 5×15/(5+15) = 75/20 Ω.

If I is the current passing through the 2 Ω resistance, we have

I²×2 = W where W is the power dissipated in 2 Ω.

Since the same current passes through the effective resistance (75/20 Ω) of the parallel branches, we have

I²×75/20 = 56

From the above equations (on dividing),

40/75 = W/56 so that W = 40×56/75 = 30 watt nearly

The following MCQ which appeared in KEAM 2008 (Medical) question paper is simple:

In the Wheatstone’s network shown in the figure, the current I in the circuit is

(a) 1 A

(b) 2A

(c) 0.25 A

(d) 0.5 A

(e) 0.33 A

The bridge is balanced (since 2 Ω/4 Ω = 4 Ω/8 Ω). Therefore, the points B and D are equipotential points and there is no current through the diagonal 5 Ω resistor. The circuit thus reduces to 6 Ω and 12 Ω in parallel with the 2 V battery. The parallel combined value of 6 Ω and 12 Ω is 6×12/(6+12) = 4 Ω.

The current, I =2 V/4 Ω = 0.5 A.

IIT-JEE 2008 Matrix-Match type Question----

After the change of pattern with effect from the 2007 exam, the degree of difficulty of IIT-JEE has been considerably reduced, as you can judge from the following Matrix-Match Type question which appeared in the 2008 question paper:

Column I gives a list of possible set of parameters measured in some experiments The variations of the parameters in the form of graphs are shown in Column II. Match the set of parameters given in column I with the graphs given in column II. Indicate your answer by darkening the appropriate bubbles of the 4×4 matrix given in the ORS.

--------------------------Column I-----------------------------------------Column II----
(A) Potential energy of a simple pendulum (y-axis) as a function of displacement (x-axis)

(B) Displacement (y-axis) as a function of time (x-axis) for a one dimensional motion at zero or constant acceleration when the body is moving along the positive x-direction

(C) Range of a projectile (y-axis) as a function of its velocity (x-axis) when projected at a fixed angle

(D) The square of the time period (y-axis) of a simple pendulum as a function of its length (x-axis)

(A) is to be matched with (p) since the potential energy (U) of a simple pendulum is given by

U = mgh = mgℓ(1– cosθ).

Here h is the height of the bob with respect to the mean position of the bob, ℓ is the length of the pendulum and θ is the angular displacement. U is thus related to θ non-linearly as indicated in the graph (p).

(B) is to be matched with (q) and (s). When the acceleration is zero, the velocity is constant and the displacement is directly proportional to the time as indicated in the graph (q). When the acceleration is constant, the velocity is directly proportional to the time and the displacement is non-linearly related to the time as indicated in the graph (s). [Remember s = ut + ½ at²].

(C) is to be matched with (s) since the range R of a projectile is given by

R = (u²sin2θ)/g where u is the velocity of projection and θ is the angle of projection. For fixed angle of projection, the range is directly proportional to the square of the velocity of projection as indicated in graph (s).

(D) is to be matched with (q) since the time period of a simple pendulum is given by

T = 2π√(ℓ/g) so that the square of the time period T is directly proportional to the length ℓ of the pendulum as indicated in graph (q).

You will have to darken the bubbles of the 4×4 matrix as shown

You can find more IIT-JEE questions (with solution) here and at other locations at physicsplus.blogspot.com