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.