Some of the old AIEEE questions (with solution) can be seen 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 search button. You may perform a similar search for old AIEEE questions (with solution) at the site http://physicsplus.blogspot.com also.
Wednesday, November 24, 2010
Wednesday, November 17, 2010
Central Board of Secondary Education (CBSE), Delhi has invited applications in the prescribed form for All India Pre-Medical / Pre-Dental Entrance Examination-2011 for admission to 15% of the merit positions for the Medical/Dental Courses of India.
The exams will be conducted as per the following schedule:
1. Preliminary Examination …. 3rd April, 2011 (Sunday) 10 AM to 1 PM
2. Final Examination …………... 15th May, 2011 (Sunday) 10 AM to 1 PM
Both examinations will be objective type.
Candidate can apply for the All India Pre-Medical/Pre-Dental Entrance Examination either offline or online. The relevant dates are given below:
Sale of bulletin : 13-12-2010 to 31-12-2010
Online submission : 15-11-2010 to 31-12-2010
Last date of receipt of
(a) Offline application : 07-1-2011
(b) Online application : 07-1-2011
All details can be obtained from the official site www.aipmt.nic.in.
Some of the AIPMT physics questions which appeared in earlier question papers can be seen on this blog. You can access them by searching for ‘AIPMT’, making use of the search box provided on this page.
Old AIPMT questions with solution can be found at http://physicsplus.blogspot.com as well.
Tuesday, November 16, 2010
Questions on magnetic field due to current carrying conductors are generally interesting. Four questions were included from this section in the Karnataka CET 2010 question paper. Here are those questions with solution:
(1) Two thick wires and two thin wires, all of same material and same length, form a square in three different ways P, Q and R as shown in the figure:
With correct connections shown, the magnetic field due to the current flow, at the centre of the loop will be zero in the case of ……
(a) Q and R only
(b) P only
(c) P and Q only
(d) P and R only
In all cases the current in the upper branch produces a magnetic field acting normally into the plane of the figure where as the current in the lower branch produces a magnetic field acting normally outwards (towards the reader). The two fields are thus in opposition. But the currents in the two branches are equal in the case of P and R since the resistances are equal. Therefore, the magnetic field due to the current flow, at the centre of the loop will be zero in the case of P and R only [Option (d)].
(2) Magnetic field at the centre of a circular coil of radius R due to a current I flowing through it is B. The magnetic field at a point along the axis at distance R from the centre is……
(d) √8 B
The magnitude of the magnetic field (Baxis) on the axis of a plane circular coil at a distance x from the centre of the coil is given by
Baxis = (μ0nR2I) /2(R2+x2)3/2 where R is the radius of the coil and I is the current in the coil.
Therefore, the magnetic field (B1) at a point along the axis at distance R from the centre is given by
B1 = (μ0nR2I) /2(R2+R2)3/2 = (μ0nR2I) /(2×2√2 ×R3)
Or, B1 = (μ0nI) /(2R×2√2)
But the magnetic field at the centre of the coil is given by
B = (μ0nI) /2R
(3) A current I is flowing through a loop. The direction of the current and the shape of the loop are as shown in the figure. The magnetic field at the centre of the loop is μ0I/R times….. (MA = R, MB = 2R, ÐDMA = 90º)
(a) 5/16, but out of the plane of the paper
(b) 5/16, but into the plane of the paper
(c) 7/16, but out of the plane of the paper
(d) 7/16, but into the plane of the paper
The magnetic field at the centre of a single turn plane circular coil carrying current I is given by
B = (μ0I) /2r where r is the radius
The loop given in the question consists of three-fourths (DIA) of a circular loop of radius R, one fourth (BIC) of a circular loop of radius 2R and two straight wires AB and CD. The straight wires do not produce any magnetic field at the common centre M of the circular arcs. The circular arcs DIA and BIC produce magnetic fields acting normally into the plane of the loop so that they add up to produce the resultant field at M.
Therefore, magnetic field at M = [(¾)× μ0I /2R] + [(¼ )× μ0I /2(2R)]
Or, magnetic field at M = (μ0I/R) [(3/8) + (1/16)] = (μ0I/R)(7/16)
The correct option is (d).
(4) PQ and RS are long parallel conductors separated by certain distance. M is the mid point between them (see the figure). The net magnetic field at M is B. Now, the current 2A is switched off. The field at M now becomes……
(a) 2 B
(d) 3 B
If the magnitude of the magnetic field produced by the conductor RS (which carries current 1 A) is B, the magnitude of the magnetic field produced by the conductor PQ (which carries current 2 A) must be 2B. But these fields are oppositely directed. That’s why the resultant field at M has magnitude B when both conductors carry currents. When the current 2A is switched off, the field at M becomes B.
[If one of the options were – B, you would pick it out as the correct one].
Monday, November 01, 2010
The Joint Entrance Examination 2011 (JEE 2011) for admission to the under graduate programmes in the IITs at Bhubaneswar, Bombay, Delhi, Gandhinagar, Guwahati, Hyderabad, Indore, Kanpur, Kharagpur, Madras, Mandi, Patna, Rajasthan, Roorkee and Ropar as well as at IT-BHU Varanasi and ISM Dhanbad will be conducted on 10th April, 2011
Here are the Important Dates for IIT JEE 2011:
November 01 to December 15, 2010: Online application process
November 12 to December 15, 2010:
Last date for receiving application forms (Off line or printed copy of Online) at IITs: 17:00 hrs, Monday, December 20, 2010
Date of Exam of IIT JEE 2011:
Sunday, April 10, 2011 Paper 1: 09:00 to 12:00 hrs;
Paper 2 : 14:00 to 17:00 hrs
All details about IIT JEE 2011 can be obtained from the following websites:
IIT Bombay: http://www.jee.iitb.ac.in
IIT Guwahati: http://www.iitg.ernet.in/jee/
IIT Delhi: http://jee.iitd.ac.in
IIT Kharagpur: http://www.iitkgp.ernet.in/jee
IIT Kanpur: http://www.jee.iitk.ac.in
IIT Roorkee: http://jee.iitr.ernet.in/
Make it a habit to visit one of the above sites for information updates if any.
Friday, October 22, 2010
Here are two multiple choice questions from the section, ‘work, energy and power’. The first question appeared in AIPMT 2010 question paper while the second question appeared in AIPMT 2008 question paper.
(i) An engine pumps water through a hose pipe. Water passes through the pipe and leaves it with a velocity of 2 ms–1 .The mass per unit length of water in the pipe is 100 kg/m. What is the power of the engine?
(1) 800 W
(2) 400 W
(3) 200 W
(4) 100 W
The power of the engine is equal to the kinetic energy of water flowing out per second. The mass (M) of water flowing out per second is 2×100 = 200 kg/s
Kinetic energy of water flowing out per second = ½ Mv2 = ½ ×200×22 = 400 W.
(ii) Water falls from a height of 60 m at the rate of 15 kg/s to operate a turbine. The losses due to frictional forces are 10 % of energy. How much power is generated by the turbine? (g = 10 ms–2)
(1) 7.0 kW
(2) 8.1 kW
(3) 10.2 kW
(4) 12.3 kW
The input power to the turbine is the potential energy due to the water falling per second and is equal to mgh = 15×10×60 =9000 watt.
The power generated by the turbine is 10 % less than the above value and is equal to (9000 – 900) watt = 8100 watt = 8.1 kW.