Information Bulletin containing the Application Form for applying for All India Engineering/Architecture Entrance Examination 2011 (AIEEE 2011) will be distributed from 15.12.2010 to 14.1.2011. Candidates can apply for AIEEE 2011 either on the prescribed Application Form or make application ‘online’.

Online submission of the application is possible from 23-11-2010 to 14-1-2011 at the websitehttp://aieee.nic.in

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 sitehttp://physicsplus.blogspot.com also.

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 …. 3^{rd} April, 2011 (Sunday) 10 AM to 1 PM

2. Final Examination …………... 15^{th} 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

(Confirmation page)

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.

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……

(a) B/2

(b) B/4

(c) B/√8

(d) √8 B

The magnitude of the magnetic field (B_{axis}) on the axis of a plane circular coil at a distance x from the centre of the coil is given by

B_{axis} = (μ_{0}nR^{2}I) /2(R^{2}+x^{2})^{3/2} where R is the radius of the coil and I is the current in the coil.

Therefore, the magnetic field (B_{1}) at a point along the axis at distance R from the centre is given by

But the magnetic field at the centre of the coil is given by

B = (μ_{0}nI) /2R

Therefore, we have B_{1} = B/2√2 = B/√8 as given in option (c).

(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 μ_{0}I/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 = (μ_{0}I) /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 theplane of the loop so that they add up to produce the resultant field at M.

Therefore, magnetic field at M = [(¾)×μ_{0}I /2R] + [(¼ )× μ_{0}I /2(2R)]

Or, magnetic field at M = (μ_{0}I/R) [(3/8) + (1/16)] = (μ_{0}I/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

(b) B

(c) B/2

(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].

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 atIT-BHU Varanasi and ISM Dhanbad will be conducted on 10^{th} 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: Sale of Off-line application forms.

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:

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 Mv^{2} = ½ ×200×2^{2} = 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.

You will find similar useful questions in this section here as well as here.

Here are a few questions from optics involving the lens maker’s equation and the concept of magnification in the case of lenses:

(1) A biconvex lens has the same radius of curvature R for its faces. If the focal length of the lens in air is R/2, the refractive index of the material of the lens is

(a) 1.2

(b) 1.33

(c) √2

(d) √3

(e) 2

The refractive index n is related to the focal length f and the radii of curvature R_{1} and R_{2} by the lens maker’s equation,

1/f = (n – 1) (1/R_{1} – 1/R_{2})

By Cartesian sign convention, R_{1} is positive and R_{2} is negative for a biconvex lens so that the above equation becomes

1/f = (n – 1) (1/R_{1} + 1/R_{2})

Substituting for f, R_{1} and R_{2 }we obtain

1/(R/2) = (n – 1) (2/R)

Or, 2/R = (n – 1) (2/R) from which n = 2

(2) A biconvex lens has the same radius of curvature R for its faces. If the focal length of the lens in air is R, what will be its focal length in a liquid of refractive index 1.5?

(a) R/2

(b) R

(c) 2R

(d) 4R

(e) Infinite

Many of you might have noted that an equiconvex lens of refractive index 1.5 has focal length equal in magnitude to the radius of its faces.

Some of you might have noted the converse of the above fact: If the refractive index is 1.5 in the case of an equiconvex (or equiconcave) lens, the magnitude of its focal length will be the same as that of the radius of curvature of its faces

[You can check this by substituting n = 1.5 and R_{1} = R_{2} = R in the lens maker’s equation].

Since the refractive index of the material of the lens is 1.5, its focal length in a liquid of refractive index 1.5 will be infinite.

(3) A converging lens produces a real, magnified and well defined image of a small illuminated square on a screen. The area of the image is A_{1}. When the lens is moved towards the screen without disturbing the object and the screen, the area of the well defined image obtained on the screen is A_{2}. What is the side of the square object?

(a) (√A_{1} +√A_{2})/2

(b) [(A_{1} + A_{2})/2]^{1/2}

(c) (A_{1}A_{2})^{1/2}

(d) (A_{1}A_{2})^{1/4}

(e) [√A_{1 }+√A_{2}]^{1/4}

The two positions (of the lens) for which well defined images of the square are obtained, are the conjugate positions and hence we have

a =√(a_{1}a_{2}) where a_{1 }and a_{2} are the sides of the images in the two cases and a is the side of the square object.

[The linear magnification in the first case is v/u = a_{1}/a.

Since the object distance u and the image distance v are interchanged in conjugate positions, we have, in the second case,

u/v = a_{2}/a.

From the above expressions, 1 = a_{1}a_{2}/a^{2} from which a =√(a_{1}a_{2})].

But a_{1 }= √A_{1 }and a_{2}= √A_{2}

Therefore, a =√(√A_{1}√A_{2}) = (A_{1}A_{2})^{1/4}

You will find similar useful multiple choice questions on optics here as well ashere.

Today we will discuss two questions on magnetic fields due to current carrying wires. These questions appeared in EAMCET (Agriculture-Medicine) 2010 question paper:

(1) A wire loop PQRS is constructed by joining two semicircular coils of radii r_{1} and r_{2} respectively as shown in the figure. If the current flowing in the loop is i, the magnetic induction at the point O is

(1) (μ_{0}i/4)(1/r_{1 }– 1/r_{2})

(2) (μ_{0}i/4)(1/r_{1 }+ 1/r_{2})

(3) (μ_{0}i/2)(1/r_{1 }– 1/r_{2})

(4) (μ_{0}i/2)(1/r_{1 }+ 1/r_{2})

The straight portions PQ and RS do not produce any magnetic field at O. The semicircular portions of radii r_{1} and r_{2 }produce magnetic fields μ_{0}i/4r_{1} and μ_{0}i/4r_{2} respectively at the centre O.

[Remember that the magnetic flux density (magnetic induction) at the centre of a circular loop of radius ris μ_{0}i/2r and hence the magnetic flux density due to a semicircular loop is μ_{0}i/4r].

The field due to the portion of smaller radius r_{1} has greater magnitude μ_{0}i/4r_{1}and is directed perpendicular to the plane of the loop, towards the reader. The field due to the portion of larger radius r_{2} has smaller magnitude μ_{0}i/4r_{2}and is directed perpendicular to the plane of the loop, away from the reader.

The resultant flux density at O is therefore μ_{0}i/4r_{1}– μ_{0}i/4r_{2}, which is equal to (μ_{0}i/4)(1/r_{1 }– 1/r_{2}), as given in option (1).

[The resultant field at O is directed normal to the plane of the loop, towards the reader. This point too can be incorporated in the question to make it a little more difficult].

(2) A wire of length 6.28 m is bent into a circular coil of 2 turns. If a current of 0.5 A exists in the coil, the magnetic moment of the coil is, in Am^{2}

(1)π/4

(2) ¼

(3) π

(4) 4π

The magnetic moment m of a plane circular coil of area A with n turns carrying current i is given by

m = niA

The radius r of the coil is given by

2×2πr = 6.28

Therefore, r = 6.28/4π = ½ and area A = πr^{2} = π/4

Physics questions appearing in AIPMT question papers are generally simple. Here are two questions on gravitation which appeared in AIPMT 2010 question paper:

(1) A particle of mass M is situated at the centre of a spherical shell of the same mass and radius a. The gravitational potential at a point situated at a/2distance from the centrewill be

(1) – 4GM/a

(2) – 3GM/a

(3) – 2GM/a

(4) – GM/a

The gravitational potential V at a point situated at distance a/2from the centre of the shell is equal to the sum of the gravitational potentials due to the particle of mass M and the shell of mass M.

Therefore, V = (– GM/a) + [– GM/(a/2)] = – 3GM/a

[Note that the gravitational potential is a negative quantity and that the potential due to the spherical shell is constant everywhere inside the shell and is equal to the surface value – GM/a].

(2) The radii of circular orbits of two satellites A and B of the earth are 4R and R, respectively. If the speed of satellite A is 3 V, then the speed of satellite B will be

(1) 3V/2

(2) 3V/4

(3) 6V

(4) 12V

The orbital speed v of a satellite is inversely proportional to the square root of the orbital radius r [since v =√(Gm/r)]. Therefore we have

v_{A}/v_{B} = √(r_{B}/r_{A})

Here v_{A} = 3 V, r_{A} = 4 R and r_{B} = R (as given in the question).

Therefore 3 V/v_{B} = √(R/4R) = ½ so that v_{B}= 6 V

You will find some useful questions (with solution) on gravitation here as well as here.

Here are the two questions on electronics which were included in EAMCET 2010 Agriculture-Medicine and Engineering question papers respectively:

(1) In the figures shown below

(1) In both Fig. (a) and Fig. (b) the diodes are forward biased

(2) In both Fig. (a) and Fig. (b) the diodes are reverse biased

(3) In Fig. (a) the diode is forward biased and in Fig. (b) the diode is reverse biased

(4) In Fig. (a) the diode is reverse biased and in Fig. (b) the diode is forward biased

In Fig. (a) the anode of the diode is at a positive potential of 10 V while the cathode is at zero volt and hence it is evidently forward biased. In Fig. (b) the anode of the diode is at a higher negative potential compared to the cathode and hence it is reverse biased [Option (3)].

(2) A transistor having a β equal to 80 has a change in base current of 250 μA. Then the change in collector current is

(1) 20,000 mA

(2) 200 mA

(3) 2000 mA

(4) 20 mA

The current gain β is given by

β = ∆I_{c}/∆I_{b} at constant collector voltage where ∆I_{c} is the change in collector current and ∆I_{b} is the change in base current.