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

– Thomas A. Edison

Wednesday, October 21, 2009

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: 3rd April, 2010 (Saturday) 10 AM to 1 PM

2. Final Examination: 16th 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.

Friday, October 16, 2009

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].

Tuesday, October 06, 2009

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 kP and kQ [kQ = kP/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 UP = ½ (F2/kP) and UQ = ½ (F2/kQ).

Therefore, UP/UQ = kQ/kP = ½ as given in the question.

This gives UP = UQ/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