Apply for All India Pre-Medical / Pre-Dental Entrance Examination -2010 (AIPMT 2010)

Central Board of Secondary Education (CBSE), Delhi has invited applications in the prescribed form for All India Pre-Medical / Pre-Dental Entrance Examination-2010 (AIPMT 2010) as per the following schedule for admission to 15% of the merit seats for the Medical/Dental Courses.

1. Preliminary Examination: 3^rd April, 2010 (Saturday) 10 AM to 1 PM

2. Final Examination: 16^th May, 2010 (Sunday) 10 AM to 1 PM

For the Preliminary Examination there will be one paper containing 200 objective type questions (four options with one correct answer) from Physics, Chemistry and Biology (Botany & Zoology).

The final examination will consist of one paper containing 120 objective type questions (four options with one correct answer) from Physics, Chemistry and Biology.

The Final Examination is only for those who qualify in the Preliminary Examination.

Candidate can apply for the All India Pre-Medical/Pre-Dental Entrance Examination either offline or online as explained below:

Offline (On prescribed application form):

Offline submission of Application Form may be made using the prescribed application form. The Information Bulletin and Application Form costing Rs.800 for General & OBC Category Candidates and Rs.450/- for SC/ST Category Candidates can be obtained against cash payment from any of the specified branches of Canara Bank/ Regional Offices of the CBSE. Visit the web site www.aipmt.nic.in for details

Online Submission:

Online submission of application may be made by accessing the Board’s website www.aipmt.nic.in. 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 Unit), CBSE, Shiksha Kendra, 2, Community Centre, Preet Vihar, Delhi-110301 by Speed Post/Registered Post in such a way that it should reach the Board on or before 04.12.2009 which is the last date stipulated.

For online submission, the fee of Rs.800/- for General and OBC Category Candidates and Rs.450/- for SC/ST category candidates may be remitted in the following ways :

1. By credit card, or

2. Through Demand Draft in favour of the Secretary, CBSE, Delhi drawn on any Nationalized Bank payable at Delhi.

Detailed instructions for Online submission of application form are available on the website www.aipmt.nic.in.

The last date of receipt of Application Form for both offline and online is 04.12.2009.

In case the application is submitted online, printout of the computer generated form complete in all respects as applicable for offline submission must reach The Deputy Secretary (AIPMT Unit), CBSE, Shiksha Kendra, 2, Community Centre, Preet Vihar, Delhi-110301 on or before the last date. A grace period of 15 days will be allowed from the last date of submission of the application to the candidates belonging to remote areas viz. Mizoram, Assam, Meghalaya, Arunachal Pradesh, Manipur, Nagaland, Tripura, Sikkim, Lahaul and Spiti Districts and Pangi sub-division of Chamba District of Himachal Pradesh, Andaman & Nicobar Islands and Lakshadweep.

For all details and information updates Visit the web site www.aipmt.nic.in.

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Some old AIPMT questions with solution can be seen on this site. You can access them by searching for ‘AIPMT’, making use of the search box provided on this site. Old AIPMT questions with solution can be found at http://physicsplus.blogspot.com also.

EAMCET (Medical) 2009 Questions on Work and Energy

The following questions which appeared in the EAMCET (Medical) 2009 question paper are worth noting:

(1) A block of mass ‘m’ is connected to one end of a spring of spring constant ‘k’. The other end of the spring is fixed to a rigid support. If the mass is released slowly so that the total energy of the system is then constituted by only the potential energy, then ‘d’ is the maximum extension of the spring. Instead, if the mass is released suddenly from the same initial position, the maximum extension of the spring now is (g = acceleration due to gravity)

(1) mg/k

(2) 2d

(3) mg/3k

(4) 4d

The mass m is suspended by means of the spring. Since the spring is extended through a distance d, we have

mg = kd so that k = mg/d

When the mass is suddenly released, suppose the spring extends through an additional distance x. The total extension then is d+x.

The spring mass system momentarily comes to rest in the condition of maximum extension and then tries to return to the initial extension of d, executing simple harmonic oscillations. In the condition of maximum extension (equal to d+x) the gravitational potential energy mg(d+x) of the mass is converted into elastic potential energy of the spring so that we have

mg(d+x) = (½) k(d+x)²

Or, mg(d+x) = ½ (mg/d)(d+x)² since k = mg/d

This gives 2 = (d+x)/d from which x = d

The total extension d+x is therefore equal to 2d [Option (2)]

(2) A particle is projected up from a point at an angle θ, with the horizontal direction. At any time ‘t’, if ‘p’ is its linear momentum, ‘y’ is the vertical displacement and ‘x’ is the horizontal displacement, the graph among the following, which does not represent the variation of kinetic energy of the projectile is

(1) Graph (A)

(2) Graph (B)

(3) Graph (C)

(4) Graph (D)

The kinetic energy of a projectile has to decrease with the increase in its vertical displacement since its gravitational potential energy increases at the cost of its kinetic energy. Therefore graph (A) is incorrect.

[Graphs (B) and (C) are correct since the kinetic energy decreases with the increase in the horizontal displacement x, becomes a minimum at half the horizontal range (corresponding to the maximum height) and then increases. Graph (D) also is correct since the kinetic energy k is given by

k = p²/2m where p is the linear momentum and m is the mass of the particle.

Therefore, k is directly proportional to p², yielding a straight line graph].

Kerala Engineering Entrance 2009 Multiple Choice Questions on Work and Energy

In the KEAM (Engineering ) 2009 question paper three questions were included from the section ‘work, energy and power’. Here are those questions with solution:

(1) A particle is acted upon by a force F which varies with position x as shown in figure. If the particle at x = 0 has kinetic energy of 25 J, then the kinetic energy of the particle at x = 16 m is

(a) 45 J

(b) 30 J

(c) 70 J

(d) 135 J

(e) 20 J

The work done by a variable force acting along the direction of displacement can be found by drawing a force-displacement graph. The area under this graph gives the work done. Since the force has positive and negative values, the area also has positive (+50 units) and negative (– 30 units) values. The net area is +20 units and hence the gain in kinnetic energy during the movement from position x = 0 to the position x = 16 m is 20 J.

The particle has kinetic energy of 25 J at position x = 0. Therefore the kinetic energy at position x = 16 m is 25 + 20 = 45 J.

(2) Two springs P and Q of force constants k_P and k_Q [k_Q = k_P/2] are stretched by applying force of equal magnitude. If the energy stored in in Q is E, then the energy stored in P is

(a) E

(b) 2 E

(c) E/8

(d) E/4

(e) E/2

The potential energy (U) of a spring stretched (by a force F) through distance x is given by

U = ½ kx² where k is the spring constant (force constant) given by k = F/x

This can be rewritten as

U = ½ (F²/k) since x = F/k

The nergy stored in P and Q are respectively given by U_P = ½ (F²/k_P) and U_Q = ½ (F²/k_Q).

Therefore, U_P/U_Q = k_Q/k_P = ½ as given in the question.

This gives U_P = U_Q/2 = E/2.

(3) A rod of mass m and length l is made to stand at an angle of 60º with the vertical. Potential energy of this rod in this position is

(a) mgl

(b) mgl/2

(c) mgl/3

(d) mgl/4

(e) mgl/√2

When the rod is kept inclined at an angle of 60º with the vertical, its centre of gravity is raised (from the ground level) by a height h = (l/2)cos 60º = l/4.

The gravitational potential energy of the rod in this position is mgh = mgl/4.

You will find many useful multiple choice questions on work, energy and power at physicsplus and at AP Physics Resources

AIPMT 2009 - Multiple Choice Questions from Electrostatics

Three questions were included from electrostatics in the All India Pre-Medical/Pre-Dental 2009 Entrance Examination (Preliminary). They are given below with solution. The first question will appear to be a rather difficult and time consuming one. Those who are preparing for AP Physics Exam may make a special note of this question.
(1) Three concentric spherical shells have radii a, b and c (a < b < c) and have surface charge densities σ, − σ and σ respectively. If V_A, V_B and V_C denote the potentials of the three shells, then for c = a + b, we have

(1) V_C = V_B ≠ V_A

(2) V_C ≠ V_B ≠ V_A

(3) V_C = V_B = V_A

(4) V_C = V_A ≠ V_B

The potential V of a spherical shell of radius r having surface charge density σ is given by

V = (1/4πε₀)(4πr²σ/r) = σr/ε₀ where ε₀ is the permittivity of free space.

Potential V_A of the shell A is given by

V_A = σa/ε₀ – σb/ε₀ + σc/ε₀ = (σ/ε₀)[c – (b – a)]

Potential V_B of the shell B is given by

V_B = – σb/ε₀ + (1/4πε₀)(4πa²σ/b) + σc/ε₀

Or, V_B = σ/ε₀ [c – (b² – a²)/b]

Potential V_C of the shell C is given by

V_C = σc/ε₀ – (1/4πε₀)(4πb²σ/c) + (1/4πε₀)(4πa²σ/c)

Or, V_C = σ/ε₀ [c – (b² – a²)/c] = σ/ε₀ [c – (b + a)(b– a)/c]

Since c = a + b we obtain

V_C = σ/ε₀ [c – (b– a)]

The correct option therefore is V_C = V_A ≠ V_B.

(2) Three capacitors each of capacitance C and of breakdown voltage V are joined in series. The capacitance and breakdown voltage of the combination will be

(1) 3 C, V/3

(2) C/3, 3 V

(3) 3 C, 3V

(4) C/3, V/3

This is a very simple question. The effective calacitance C_eff of the combination of three capacitors in series is given by the reciprocal relation,

1/C_eff = 1/C₁ +1/C₂ +1/C₃

Here C₁ = C₂ = C₃ = C so that C_eff = C/3

The breakdown voltage V_eff for the series combination is the sum of the individual breakdown voltages:

V_eff = 3V

(3) The electric potential at a point (x, y, z) is given by V = − x²y − xz³ + 4. The electric field E at that point is:

(1) E = i 2xy + j (x² + y²) + k (3xz − y²)

(2) E = i z ³ + j xyz + k z²

(3) E = i (2xy − z³) + j xy² + k 3z²x

(4) E = i (2xy + z³) + j x² + k 3xz²

The electric field E is the negative gradient of potential:

E = − ∂V/∂r = − (i ∂V/∂x + j ∂V/∂y + k ∂V/∂z)

This gives E = i (2xy + z³) + j x² + k 3xz² as given in option (4).

You will find similar useful multiple choice questions (with solution) in electrostatics at aphysicsresources and at physicsplus

Kerala Medical Entrance (KEAM) 2009 Questions from Electrostatics

Keral Medical Entrance 2009 question paper contained three questions from electrostatics. They are simple except for the first one (given below) which will demand some three dimensional imagination.

(1) The total electric flux through a cube when a charge 8q is placed at one corner of the cube is

(a) ε₀ q

(b) ε₀ /q

(c) 4πε₀ q

(d) q/4πε₀

(e) q/ε₀

You should remember that the total electric flux originating from a charge q is q/ε₀.

[You will definitely remember that the electric field E at a point distant r from a point charge is q/4πε₀r². The electric field is equal to the electric flux through unit area held normal to the direction of the field. If you imagine a spherical surface with the charge q at the centre, the surface area of the sphere is 4πr² and the total flux passing normally through the spherical surface is (q/4πε₀r²) ×4πr² = q/ε₀]

In the question the charge 8q is placed at one corner of the cube. The total electric flux originating from this charge is 8q/ε₀. But only one-eigths of the flux passes through the cube. (You can place seven more cubes with their corners touching the charge to cover the entire volume around the charge).

The electric flux through a cube when a charge 8q is placed at one corner of the cube is therefore (1/8)×(8q/ε₀) = q/ε₀.

(2) A uniform electric field E exists along positive x-axis. The work done in moving a charge 0.5 C through a distance 2 m along a direction making an angle 60º with x-axis is 10 J. Then the magnitude of electric field is

(a) 5 Vm^–1

(b) 2 Vm^–1

(c) √5 Vm^–1

(d) 40 Vm^–1

(e) 20 Vm^–1

The force (F) acting on charge q in electric field E is given by F = qE along the direction of the field (along the positive x-axis here). Since the displacement (d) of the charge is inclined at 60º with the x-axis, the work done (W) is given by

W = qEd cos 60

Therefore, 10 = 0.5×E×2×½, from which E = 20 Vm^–1.

(3) A capacitor of capacitance C is charged to a potential V. If it carries charge Q, then the energy stored in it is

(a) ½ CV

(b) QV

(c) ½ QV²

(d) CV²

(e) ½ QV

The answer is ½ QV which you can obtain from ½ CV² which is the form most of you will usually remember:

½ CV² = ½ CV×V = ½ QV

[The capacitor is charged from zero potential to the potential V. The charge Q is therefore transferred to the capacitor at an average potential of (0+V)/2 = V/2. The work done is therefore QV/2.

To be more rigorous, if the charge on the capacitor at any instant of charging is q and the potential at the instant is v, the work done in transferring an additional charge dq is vdq= (q/C)dq. The total work done in transferring the entire charge Q is ₀ ∫^Q(q/C)dq = Q²/2C = ½ ×Q×(Q/C) = ½ QV].

You will find more questions (with solution) of various entrance examinations at physicsplus.blogspot.com.