entrancephysics

Apply for All India Engineering/Architecture Entrance Examination 2011 (AIEEE 2011)

2010-11-24T16:27:00.003+05:30

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 website http://aieee.nic.in

The date of Examination is 24^th April 2011.

Visit the site http://aieee.nic.in and www.cbse.nic.in immediately for details and information updates.

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.

Apply for All India Pre-Medical / Pre-Dental Entrance Examination -2011 (AIPMT 2011)

2010-11-17T10:53:00.000+05:30

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.

Old AIPMT questions with solution can be found at http://physicsplus.blogspot.com as well.

Karnataka CET 2010 Questions (MCQ) on Magnetic Field due to Current Carrying Conductors

2010-11-16T11:57:00.004+05:30

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

(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

(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 = (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

Therefore, we have B₁ = 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 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

(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

(b) 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].

It’s Time to Apply for Joint Entrance Examination 2011 (JEE 2011) for Admission to IITs

2010-11-01T16:37:00.001+05:30

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 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:

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/

IT Madras: http://jee.iitm.ac.in

Make it a habit to visit one of the above sites for information updates if any.

AIPMT Questions (MCQ) from Work, Energy & Power

2010-10-22T17:42:00.001+05:30

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.

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

Multiple Choice Questions on Lenses

2010-10-18T14:57:00.001+05:30

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

(d) √3

(e) 2

The refractive index n is related to the focal length f and the radii of curvature R₁ and R₂ by the lens maker’s equation,

1/f = (n – 1) (1/R₁ – 1/R₂)

By Cartesian sign convention, R₁ is positive and R₂ is negative for a biconvex lens so that the above equation becomes

1/f = (n – 1) (1/R₁ + 1/R₂)

Substituting for f, R₁ and R₂we 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

(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 R₁ = R₂ = 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 A₁. 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 A₂. What is the side of the square object?

(a) (√A₁ + √A₂)/2

(b) [(A₁ + A₂)/2]^1/2

(d) (A₁A₂)^1/4

(e) [√A₁+√A₂]^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 =√(a₁a₂) where a₁and a₂ 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 = a₁/a.

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

u/v = a₂/a.

From the above expressions, 1 = a₁a₂/a² from which a =√(a₁a₂)].

But a₁= √A₁and a₂ = √A₂

Therefore, a =√(√A₁√A₂) = (A₁A₂)^1/4

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

EAMCET (Medical) 2010 Questions on Magnetic Fields Due to Current Carrying Conductors

2010-10-04T22:03:00.003+05:30

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 r₁ and r₂ respectively as shown in the figure. If the current flowing in the loop is i, the magnetic induction at the point O is

(1) (μ₀i/4)(1/r₁– 1/r₂)

(2) (μ₀i/4)(1/r₁+ 1/r₂)

(3) (μ₀i/2)(1/r₁– 1/r₂)

(4) (μ₀i/2)(1/r₁+ 1/r₂)

The straight portions PQ and RS do not produce any magnetic field at O. The semicircular portions of radii r₁ and r₂produce magnetic fields μ₀i/4r₁ and μ₀i/4r₂ respectively at the centre O.

[Remember that the magnetic flux density (magnetic induction) at the centre of a circular loop of radius r is μ₀i/2r and hence the magnetic flux density due to a semicircular loop is μ₀i/4r].

The field due to the portion of smaller radius r₁ has greater magnitude μ₀i/4r₁ and is directed perpendicular to the plane of the loop, towards the reader. The field due to the portion of larger radius r₂ has smaller magnitude μ₀i/4r₂ and is directed perpendicular to the plane of the loop, away from the reader.

The resultant flux density at O is therefore μ₀i/4r₁– μ₀i/4r₂, which is equal to (μ₀i/4)(1/r₁– 1/r₂), 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 Am²

(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 = πr² = π/4

Substituting, m = 2×0.5 × π/4 = π/4.

AIPMT Questions (MCQ) on Gravitation

2010-08-26T13:28:00.002+05:30

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/2 distance from the centre will be

(1) – 4GM/a

(2) – 3GM/a

(3) – 2GM/a

(4) – GM/a

The gravitational potential V at a point situated at distance a/2 from 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.

EAMCET 2010 Multiple Choice Questions on Electronics

2010-07-11T22:53:00.002+05:30

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.

Therefore, ∆I_c = β∆I_b = 80×250 μA = 20,000 μA = 20 mA.

Two Questions (EAMCET 2009 and EAMCET 2008) on Energy and Power

2010-05-28T21:47:00.000+05:30

Today we will discuss two multiple choice questions involving energy and power. The first question appeared in the EAMCET 2009 (Engineering) question paper and the second question appeared in the EAMCET 2008 (Engineering) question paper:

(1) A motor of power P₀ is used to deliver water at a certain rate through a given horizontal pipe. To increase the rate of flow of water through the same pipe n times, the power of the motor is increased to P₁. The ratio of P₁ to P₀ is

(1) n : 1

(2) n² : 1

(3) n³ : 1

(4) n⁴ : 1

If the mass of water delivered per second is m when the power is P₀, we have

P₀ α ½ mv² where v is the velocity of water

[We do not equate P₀ to ½ mv² since the efficiency of the motor will not be 100%].

To increase the rate of flow of water through the same pipe n times, the velocity of water has to be made n times. The mass of water delivered per second then is nm. Therefore we have

P₁ α ½ nm (nv)²

From the above we obtain

P₁/ P₀ = n³ [Option (3)].

(2) A river of salty water is flowing with a velocity 2 ms^–1. If the density of water is 1.2 g cm^–3, then the kinetic energy of each cubic metre of water is

(1) 2.4 J

(2) 24 J

(3) 2.4 kJ

(4) 4.8 kJ

The density of the water in the river is 1.2 g cm^–3, which is equal to 1200 kg m^–3. Mass of one cubic metre of the water is 1200 kg and its kinetic energy E is given by

E = ½ mv² = ½ ×1200×2² = 2400 J = 2.4 kJ.

Multiple Choice Practice Questions involving Power in Direct Current Circuits for AP Physics B Exam.

2010-05-10T14:45:00.002+05:30

If you grasp fundamental principles thoroughly, you will be able to answer complicated questions without much difficulty. The following questions meant for AP Physics aspirants are relatively simple but they will definitely help you to test your understanding of fundamental principles.

(1) For transferring 50 coulomb of charge through a 10 Ω resistor, the work required is 150 J. The potential difference across the resistor is

(a) 500 V

(b) 50 V

(d) 5 V

(e) 3 V

We have W = VQ where W is the work done in transferring Q coulombs of charge through a potential difference of V volt.

Therefore, V = W/Q = 150/50 volt = 3 V.

(2) Four resistors R₁, R₂, R₃ and R₄ having resistances 12 Ω, 16 Ω, 26 Ω and 36 Ω respectively are connected in parallel with a battery of negligible internal resistance. The current in the 12 Ω resistor is 0.5 A. What is the electric power dissipated in the 36 Ω resistor?

(a) 0.5 W

(b) 1 W

(d) 18 W

(e) 36 W

Since the resistors are in parallel, the potential difference across them will be the same. (This is true even if the battery has internal resistance).

The potential difference across the 12 Ω resistor is 12×0.5 volt = 6 V.

The potential difference across the 36 Ω resistor also is 6 V.

Therefore, the power dissipated in the 36 Ω resistor = V²/R = 6²/36 = 1 W.

(3) Two 24 V, 16 W electric lamps are connected in series and this series combination is connected across a 12 V battery of negligible internal resistance. The power consumed by each lamp is

(a) 1 W

(b) 2 W

(d) 8 W

(e) 16 W

If the resistance of each lamp is R, we have (from power, P = V²/R)

16 = 24²/R ………..(i)

When the series combination of the lamps is connected across a 12 V battery, each lamp has 6 volt across it. The power (X) consumed by each lamp is therefore given by

X = 6²/R …………(ii)

From equations (i) and (ii) X = 1 W.

(4) A 12 V battery of internal resistance 3 Ω is connected in series with a 1 Ω resistor and a variable resistor. The power dissipated in the variable resistor will be maximum when the current through it is

(a) 0.5 A

(b) 1 A

(d) 3 A

(e) 4 A

According to maximum power transfer theorem maximum power transfer occurs when the external resistance is equal to the internal resistance of the battery. Therefore, the variable resistor and the fixed 1 Ω resistor together must be 3 Ω for maximum power transfer. The current in the circuit then is 12 V/6 Ω = 2 A [Option (c)].

You will find similar useful questions at AP Physics Resources and at physicsplus.

AIPMT 2010 Multiple Choice Questions on Vectors

2010-04-21T15:30:00.005+05:30

The following questions (MCQ) on vectors were included in the AIPMT 2010 question paper:

(1) Six vectors, a through f have the magnitudes and directions indicated in the figure. Which of the following statements is true?

(1) b + e = f

(2) b + c = f

(3) d + c = f

(4) d + e = f

The correct option is (4) as is clear from the adjoining figure

(2) A particle has initial velocity (3î + 4ĵ) and has acceleration (0.4î + 0.3ĵ). Its speed after 10 s is

(1) 10 units

(2) 7 units

(3) 7√2 units

(4) 8.5 units

We have v = u + at with usual notations.

Therefore the velocity after 10 s is given by

v = (3î + 4ĵ) + (0.4î + 0.3ĵ)10 = 7î + 7ĵ

The speed after 10 s is the magnitude of the above velocity vector and is equal to √(7² + 7²) = 7√2 units [Option (3)].

Now see these questions (MCQ) on vectors.

Thermodynamics- AIPMT 2009 Multiple Choice Questions

2010-03-13T21:29:00.002+05:30

Here are two multiple choice questions on thermodynamics which appeared in AIPMT 2009 question paper:

(i) In thermodynamic processes which of the following statements is not true?

(1) In an isochoric process pressure remains constant

(2) In an isothermal process the temperature remains constant

(3) In an adiabatic process PV^γ = constant

(4) In an adiabatic process the system is insulated from the surroundings

The only statement which is not true is statement (1). A process in which the pressure remains constant is called isobaric process (and not isochoric process).

[Note that an isochoric process is one in which the volume remains constant].

(ii) The internal energy change in a system that has absorbed 2 Kcals of heat and

done 500 J of work is:

(1) 6400 J

(2) 5400 J

(3) 7900 J

(4) 8900 J

When a quantity Q of heat is supplied to a system it is used to do an amount of work W by the system and to increase the internal energy of the system by ∆U:

Q = ∆U + W

Note that Q given in kilo calories is to be converted into joule which is the SI unit of heat. Therefore Q = 2×1000×4.2 J = 8400 J (remembering that 1 cal = 4.2 J very nearly).

The increase the internal energy, ∆U = Q – W =8400 – 500 = 7900 J.

Find some more multiple choice questions (with solution) in this section here as well as here.

AIPMT 2009 and AIPMT 2008 Questions on Waves

2010-02-12T20:04:00.001+05:30

Today we will discuss the multiple choice questions on waves included in All India Pre-Medical / Pre-Dental Entrance Examination (AIPMT) 2009 and 2008 question papers. Here are the AIPMT 2009 questions:

(1) The electric field part of of an electromagnetic wave in a medium is represented by

E_x = 0;

E_y = 2.5 N/C cos[(2π×10⁶ rad/s)t – (π×10^–2 rad/m)x];

E_z = 0.

The wave is:

(1) moving along x-direction with frequency 10⁶ Hz and wave length 100 m.

(2) moving along x-direction with frequency 10⁶ Hz and wave length 200 m.

(3) moving along –x-direction with frequency 10⁶ Hz and wave length 200 m.

(4) moving along y-direction with frequency 2π×10⁶ Hz and wave length 200 m.

The electric field variation is in accordance with the equation y = A sin (ωt – kx) which represents a progressive wave proceeding along the positive x-direction. Instead of the usual displacement y we have the electric field E. In place of the angular frequency ω we have 2π×10⁶ (remember ω = 2πn) which means that the linear frequency n is 10⁶ Hz.

Since the propagation constant k = 2π/λ where λ is the wave length, we have

π×10^–2 = 2π/λ from which λ = 200 ms^–1.

So the correct option is (2).

[The units of electric field E (N/C), angular frequency ω (rad/s) and the propagation constant k (rad/m) given in the wave equation in the question should not distract you.

You should note that the negative sign in the wave equation y = A sin (ωt – kx) or y = A sin (kx – ωt) indicates that the wave is propagating along the positive x-direction. A wave propagating along the negative x-direction is represented by y = A sin (ωt + kx).

You can make use of any other form of the wave equation, for instance, y = A sin [2π(t/T – x/ λ)], also to solve the problem].

(2) A wave in a string has an amplitude of 2 cm. The wave travels in the + ve direction of x axis with a speed of 128 m/sec. and it is noted that 5 complete waves fit in 4 m length of the string. The equation describing the wave is:

(1) y = (0.02) m sin (15.7x − 2010t)

(2) y = (0.02) m sin (15.7x + 2010t)

(3) y = (0.02) m sin (7.85x − 1005t)

(4) y = (0.02) m sin (7.85x + 1005t)

The unit of amplitude is metre, shown by its symbol ‘m’ in the equation (which should not distract you).

The equation describing the wave propagating in the +ve x-direction is

y = A sin (kx – ωt)

The amplitude A as given in the question is 2 cm = 0.02 m.

The velocity of the wave, v = ω/k = 128 ms^–1.

Wave length λ = 4/5 m.

But k = 2π/λ = 2π×5/4 = 7.85 m and so ω = kv = 7.85×128 = 1005

So the equation of the wave is

y = (0.02) m sin (7.85x − 1005t)

The correct option is (3).

Here is the AIPMT 2008 question:

The wave described by y = 0.25 sin(10πx – 2πt) where x and y are in metres and t in seconds is a wave traveling along the

(1) negative x-direction with amplitude 0.25 m and wavelength λ = 0.2 m.

(2) negative x-direction with frequency 1 Hz.

(3) positive x-direction with frequency π Hz and wavelength λ = 0.2 m.

(4) positive x-direction with frequency 1 Hz and wavelength λ = 0.2 m.

The given wave equation is in the form y = A sin [2π(x/λ – t/T)].

The negative sign in the equation shows that the wave is propagating along the positive x-direction.

By comparison we obtain 2πx/λ = 10πx from which λ = 0.2 m. Also, 2πt/T =2πt from which the frequency 1/T = 1 Hz. [Option (4)].

You will find a useful post on waves at AP Physics Resources.

Karnataka Common Entrance Test (CET) 2006 Questions on Interference and Diffraction of Waves

2010-01-22T14:52:00.005+05:30

The following three questions appeared in Karnataka CET 2006 question paper:

(1) When a low flying aircraft passes overhead, we sometimes notice a slight shaking of the picture on our TV screen. This is due to

(1) diffraction of signal received from the antenna

(2) interference of the direct signal received by the antenna with the weak signal reflected by the passing aircraft

(3) change of magnetic flux occurring due to the passing aircraft

(4) vibrations created by the passing aircraft.

The microwaves reaching the antenna directly and after reflection at the aircraft interfere. The interference pattern produced is unstable since the aircraft is moving. This results in a shaking picture [Option (2)].

(2) A beam of light of wave length 600 nm from a distant source falls on a single slit 1 mm wide and the resulting diffraction pattern is observed on a screen 2 m away. The distance between the first dark fringes on either side of the central bright fringe is

(1) 1.2 cm

(2) 1.2 mm

(3) 2.4 cm

(4) 2.4 mm

This is a popular question repeatedly asked in various entrance tests with changes in numerical values.

The diffraction bands are distributed symmetrically on either side of the central maximum and the angular separation of the centre of the first dark fringe from the centre of the central maximum is λ/a where λ is the wave length of light and a is width of the slit.

Therefore, the angular separation between the first dark fringes on either side of the central bright fringe is 2λ/a.

The distance (linear separation) between the first dark fringes on either side of the central bright fringe is D×2λ/a where D is the distance between the slit and the screen and we have

D×2λ/a = 2×(2×600×10^–9/10^–3) = 2.4×10^–3 m = 2.4 mm.

(3) If white light is used in the Newton’s ring experiment, the colour observed in the reflected light is complementary to that observed in the transmitted light through the same point. This is due to

(1) 90º change of phase in one of the reflected waves

(2) 180º change of phase in one of the reflected waves

(3) 145º change of phase in one of the reflected waves

(4) 45º change of phase in one of the reflected waves

The mechanism of formation of Newton’s rings is the same as that of the production of colours in thin films and is shown in the figure. Usually the light beam falls normally on the air film included between the upper lens and the lower glass plate, but we have shown the rays slanting to make things clear.

Newton’s rings in the reflected system are formed by the interference of waves 1 and 2 (fig.) where as Newton’s rings in the transmitted system are formed by the interference of waves 3 and 4.

If the phase change of π introduced due to reflection at a denser medium is ignored, the difference between the path lengths of the rays 1 and 2 is the same as the difference between the path lengths of the rays 3 and 4.

In the case of waves 1 and 2 there is an additional path difference of λ/2 because of the phase difference π introduced due to reflection at B. In the case of waves 3 and 4 there is an additional path difference of λ because of the phase difference 2π introduced due to reflections at B and C. If the condition for brightness is satisfied for one colour in the case of the interfering waves in the reflected system, the condition for darkness will be satisfied for the same colour in the case of the interfering waves in the transmitted system. So the colours present in the reflected system will be absent in the transmitted system and vice versa. The basic reason for this is the 180º phase change [Option (2)].

You will find a useful post in this section at physicsplus.

Two Questions (MCQ) on Magnetic Effect of Electric Current

2010-01-05T19:59:00.009+05:30

You will find many questions (with solution) involving magnetic fields on this blog, which can be accessed by trying a search for ‘magnetic field’ using the search box at the top of this page (or, you may click on the label ‘magnetic field’ below this post.

Today we will discuss two multiple choice questions on magnetic field produced by current carrying conductors:

(1) The adjoining figure shows a plane wire loop made of two semicircular portions of radii R₁ and R₂ and two straight portions. The wire loop contains a battery which drives a current I through the loop. What is the magnitude of the magnetic flux density at the common centre ‘O’ of the semicircular portions of this wire loop?

(a) (μ₀I/2π)(1/R₁ + 1/R₂)

(b) (μ₀I/4)(1/R₁ + 1/R₂)

(d) (μ₀I/4π)(1/R₂ – 1/R₁)

(e) Zero

The magnetic field produced at O by the straight portions of the loop is zero. The semicircular portion of smaller radius R₂ produces a magnetic field μ₀I/4R₂which is directed normally into the plane of the figure (away from the reader).

[Note that a single turn plane circular coil of radius R produces a magnetic field μ₀I/2R at its centre].

The semicircular portion of larger radius R₁ produces a smaller magnetic field μ₀I/4R₁which is directed normally outwards (towards the reader).

The resultant magnetic field at the centre O has magnitude (μ₀I/4R₂ – μ₀I/4R₁)= (μ₀I/4)(1/R₂ – 1/R₁), as given in option (c).

[The resultant field is directed normally into the plane of the figure (away from the reader). The above question can be modified to check your understanding of this fact also].

(2) A current I enters a circular coil of radius R, branches into two parts and then recombines as shown in the circuit diagram.

The resultant magnetic field at the centre of the coil is

(a) zero

(b) μ₀I/2R

(d) (¼ )(μ₀I/2R)

(e) (½)(μ₀I/2R)

This question appeared in KEAM 2009 (Engineering) question paper.

The magnetic field produced at the centre by the straight current leads is zero. So it is sufficient to consider the fields produced by the circular portions.

The currents through the branches are inversely proportional to the lengths of the branches while the magnetic fields for a given current are directly proportional to the lengths of the branches. The magnitudes of the fields due to the two branches are therefore equal. Since the fields due to the branches are perpendicular to the plane of the loop and oppositely directed, the net field at the centre is zero. This argument is enough for answering the above question.

If you want to make your argument more rigorous, you will proceed as follows:

Since the longer branch is made of three quarters of the circle and the shorter branch is made of one quarter of the circle, the resistance of the longer branch is 3 times that of the shorter branch. The currents through the shorter and longer branches are therefore given respectively by

I₁ = 3I/4 and

I₂ = I/4

The magnetic field produced at the centre by the shorter branch is directed normally into the plane of the coil (away from the reader) and has magnitude B₁ given by

B₁ = (¼)(μ₀I₁/2R) = 3μ₀I/32R, on substituting for I₁.

The magnetic field produced at the centre by the longer branch is directed normally outwards (towards the reader) and has magnitude B₂ given by

B₂ = (¾)(μ₀I₂/2R) = 3μ₀I/32R, on substituting for I₂.

The fields B₁ and B₂ get canceled and the resultant field at the centre is zero [Option (a)].

[Note that in all situations of branching of current at a circular coil, if the current leads are straight and point to the centre of the coil, the magnetic field at the centre will be zero (the angle shown need not necessarily be 90º)].

Now, find an interesting question in this section here.

2010-01-01T15:04:00.001+05:30

“Be always at war with your vices, at peace with your neighbors, and let each New Year find you a better man.”

– Benjamin Franklin

Happy New Year…

Multiple Choice Questions on Dimensions of Physical Quantities

2009-12-23T16:04:00.006+05:30

The world is a dangerous place, not because of those who

do evil, but because of those who look on and do nothing.

–Albert Einstein

Today we will discuss a few questions on units and dimensions:

(1) Van der Waals equation of state for real gases is [P + (a/V²)](V – b) = RT where P is the pressure, V is the volume of one mole of gas, R is the universal gas constant, T is the temperature and a and b are constants. The dimensional formula of a is

(a) [M⁰L⁶T⁰]

(b) [ML^–1T^–2]

(d) [ML^–2T^–2]

(e) [ML³T^–2]

This is a question popular among question setters and has appeared many times in various entrance test papers. Since a/V² appears as added to P, the dimensions of a/V² must be the same as those of P. Therefore, ‘a’ has the dimensions of PV².

Note that pressure P is force per unit ares and hence the dimensional formula for P is [ML^–1T^–2]. The dimensional formula for V²is [L⁶]. Therefore, the dimensional formula for ‘a’ is [ML⁵T^–2].

(2) The dimensional formula for Stefan’s constant is

(a) [MT^–3K^–4]

(b) [ML²T^–2K^–4]

(d) [MT^–2L⁰]

(e) [MT⁴L⁰]

This question appeared in Kerala Engineering Entrance (KEAM) 2009 question paper. You will find similar questions in other entrance test papers also.

We have E = σT⁴ where E is the energy radiated per second from unit area, σ is Stefan’s constant and T is the temperature of the black body.

Therefore, σ = E/T⁴

Note that energy has dimensional formula [ML²T^–2]. Sine E is the energy radiated per second from unit area, the dimensional formula for E is [MT^–3].

Therefore, the dimensional formula for Stefan’s constant σ is [MT^–3K^–4].

(3) Pick out the correct dimensions in length of the quantity μ₀ε₀from the following:

(a) 2

(b) 1

(d) – 2

(e) zero

You know that the speed ‘c’ of electromagnetic waves in free space is given by

c = 1/√(μ₀ε₀)

Therefore μ₀ε₀ = 1/c² and the dimensional formula for μ₀ε₀ is [L^–2T²]. The quantity μ₀ε₀ therefore has – 2 dimensions in length [Option (d)].

(4) Which of the following is dimensionless?

(a) Stress

(b) Gas constant

(d) Efficiency of heat engine

(e) Thermal conductivity

The correct option is (d) since the efficiency is the ratio of the output power to the input power.

Applications invited for Kerala Entrance Examinations for Admission to Medical/ Agriculture/ Veterinary/ Engineering/ Architecture Degree Courses 2010 (KEAM 2010)

2009-12-05T14:48:00.002+05:30

The Commissioner for Entrance Examinations, Govt. of Kerala, has invited applications for the Entrance Examinations for admission to the following Degree Courses in various Professional Colleges in Kerala for 2010-2011.

(a) Medical (i) MBBS (ii) BDS (iii) BHMS (iv) BAMS (v) BSMS

(b) Agriculture (i) BSc. Hons. (Agriculture) (ii) BFSc. (Fisheries) (iii) BSc. Hons. (Forestry)

(c) Veterinary BVSc. & AH

(d) Engineering B.Tech. [including B.Tech. (Agricultural Engg.)/B.Tech. (Dairy Sc. & Tech.) courses under the Kerala Agricultural University]

(e) Architecture B.Arch.

[Candidates aspiring for B.Arch. course should also submit their application to the CEE. Even though there is no state level Entrance Examination for this purpose, these candidates should write the National Aptitude Test for Architecture and should forward the NATA score and mark list of the qualifying examination to the CEE on or before 05.06.2010].

Dates of Exam:

Engineering Entrance Examination (For Engineering courses except Architecture)

19.04.2010 Monday 10.00 A.M. to 12.30 P.M. Paper-I : Physics & Chemistry.

20.04.2010 Tuesday 10.00 A.M. to 12.30 P.M. Paper-II: Mathematics.

Medical Entrance Examination (For Medical, Agriculture and Veterinary Courses)

21.04.2010 Wednesday 10.00 A.M. to 12.30 P.M. Paper-I : Chemistry & Physics.

22-04.2010 Thursday 10.00 A.M. to 12.30 P.M. Paper-II: Biology.

Sale of Application Form through selected post offices: From 07-12-2009 to 06-01-2010

To see the list of selected post offices, visit http://www.cee-kerala.org/
Last Date for Receipt of Application by CEE: 06-01-2010 (Wednesday), before 5 P.M.

Facility for online submission of application in the case of non-reservation category and Non-Keralite category is available.

Find details at http://www.cee-kerala.org/

In addition to the multiple choice questions (of the type expected to appear in KEAM 2010) on this site, you will find many similar useful questions with solution at the site http://physicsplus.blogspot.com. If you want to see earlier KEAM questions only, type in ‘Kerala’ in the search box at the top left of this page and click on the search button or press the enter key.

EAMCET 2009 (Medical) Questions (MCQ) on Rotational Motion

2009-11-23T16:58:00.004+05:30

The following two questions were included from rotational motion in the EAMCET 2009 (Medical) question paper. (The first question has appeared in many entrance exam question papers. It was included in the EAMCET 2009 Engineering question paper also). Here are the questions with solution:

(1) A rod of length ‘l’ is held vertically stationary with its lower end located at a position P on the horizontal plane. When the rod is released to topple about P, the velocity of the upper end of the rod with which it hits the ground is

(1) √(g/l)

(2) √(3gl)

(3) 3√(g/l)

(4) √(3g/l)

When the rod falls its gravitational potential energy mgl/2 gets converted into rotational kinetic energy of the rod. (Note that ‘m’ is the mass of the rod and initially the centre of gravity of the rod is at a height l/2 with respect to the horizontal plane).

Therefore we can write

½ Iω² = mgl/2 where I is the moment of inertia of the rod about an axis passing through the end (at P) of the rod and perpendicular to the length of the rod and ‘ω’ is the angular velocity of the rod when it hits the horizontal plane.

Here I = ml²/3.

[Usually you will remember the moment of inertia of a rod about a normal axis through its middle as ml²/12. The moment of inertia about a normal axis through one end is obtained by applying the parallel axis theorem: I = ml²/12 + m(l/2)² = ml²/3].

Substituting for I we have

½ (ml²/3)ω² = mgl/2

Since ω = v/l where ‘v’ is the velocity with which the rod hits the ground, we have

½ (ml²/3)(v/l)² = mgl/2

This gives v = √(3gl)

(2) A rigid uniform rod of mass M and length ‘L’ is resting on a smooth horizontal table. Two marbles each of mass ‘m’ and traveling with uniform speed ‘v’ collide with the two ends of the rod simultaneously and inelastically as shown. The marbles get stuck to the rod after the collision and continue to move with the rod. If m = M/6 and v = L ms^–1, then the time taken by the rod to rotate through π/2 is

(1) 1 sec

(2) 2π sec

(3) π sec

(4) π/2 sec

Because of the collision, the rod will rotate about a normal axis through its middle with an angular velocity ω given by

Iω = mvL/2 + mvL/2 where ‘I’ is the moment of inertia of the rod carrying the masses m and m at its ends.

[Note that we have equated the final angular momentum of the system (containing the rod and the masses) to the initial angular momentum. Before the collision the two masses have angular momentum about the central axis. These are shown on the right hand side of the above equation].

Since v = L the above equation gets modified as

Iω = mL²

After the collision, the rod and the masses move together and the total angular momentum is given by

Iω = [(ML²/12) + 2m(L/2)²] ω

[The first term within the square bracket above is the moment of inertia of the rod and the second term is the moment of inertia of the two masses].

From the above two equations, we have

mL² = [(ML²/12) + mL²/2] ω

Since m = M/6 the above equation becomes

M/6 = [(M/12) + (M/12)] ω = (M/6) ω

Therefore ω = 1 radian /sec and the time taken by the rod to rotate through π/2 radian is π/2 sec.

You will find many questions on rotational motion on this site. You can access all of them by clicking on the label ‘rotation’

You will find many useful questions with solution in this section at physicsplus and at AP Physics Resources.

All India Engineering/Architecture Entrance Examination 2010 (AIEEE 2010)

2009-11-17T15:12:00.001+05:30

It’s time to apply for AIEEE 2010.

Information Bulletin containing the Application Form for applying for All India Engineering/Architecture Entrance Examination 2010 (AIEEE 2010) will be distributed from 1.12.2009 and will continue till 31.12.2009. Candidates can apply for AIEEE 2010 either on the prescribed Application Form or make application ‘online’.

Online submission of the application is possible from 16-11-2009 to 31-12-2009 at the website http://aieee.nic.in

The date of Examination is 25^th April 2010.

Visit the site http://aieee.nic.in immediately for details.

You will find many old AIEEE questions (with solution) 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. Or, you may hit the enter key after feeding the search words.

You may perform a similar search for old AIEEE questions (with solution) at the site http://physicsplus.blogspot.com also.

IIT-JEE 2009 Multiple Correct Answer Type Questions on Thermodynamics

2009-11-06T12:56:00.005+05:30

The following multiple correct answer type questions on thermodynamics were included in the IIT-JEE 2009 question paper:

(1) The figure shows the P–V plot of an ideal gas taken through a cycle ABCDA. The part ABC is a semi-circle and CDA is half of an ellipse. Then,

(A) the process during the path A → B is isothermal

(B) heat flows out of the gas during the path B → C → D

(D) positive work is done by the gas in the cycle ABCDA.

Statement (A) is incorrect since an isothermal curve in a PV diagram must be a hyperbola and not an arc of a circle.

Temperature at D is less than that at B (as obtained from PV = nRT with usual notations). Therefore statement (B) is correct.

During the path AB the gas expands, doing work. During the path BC the gas is compressed and work is done on the gas. But these works are unequal since the area under AB is greater than the area under BC. Therefore, statement (c) is incorrect.

The cycle of operation is clockwise and the cyclic curve encloses an area. Therefore, positive work is done by the gas and statement (D) is correct.

Thus the correct options are (B) and (D).

[In the present context you will find a useful post (which will clear your doubts) at AP Physics Resources].

(2) C_V and C_P denote the molar specific heat capacities of a gas at constant volume and constant pressure respectively. Then

(A) C_P – C_V is larger for a diatomic ideal gas than for a monoatomic ideal gas

(B) C_P + C_V is larger for a diatomic ideal gas than for a monoatonic ideal gas

(D) C_P.C_V is larger for a diatomic ideal gas than for a monoatonic ideal gas.

In the case of ideal gases C_P – C_V is the same (equal to R, the universal gas constant) and hence statement (A) is wrong.

C_P and C_V are larger for a diatomic ideal gas than for a monoatonic ideal gas and hence C_P + C_V is larger. Statement (B) is correct.

[For diatomic ideal gas C_P = 7R/2 and C_V = 5R/2. For monoatomic ideal gas C_P = 5R/2 and C_V = 3R/2].

Statement (C) is evidently wrong.

Statement (D) is correct since both C_P and C_V are larger for a diatomic ideal gas than for a monoatomic ideal gas.

Therefore, (B) and (D) are correct options

Apply for IIT-JEE 2010

2009-11-02T20:12:00.003+05:30

You may apply now for the Joint Entrance Examination for Admission to IITs and other Institutions (IIT-JEE 2010). The exam will be conducted on April 11^th, 2010 (Sunday) as per the following schedule:
09:00 – 12:00 hrs: Paper – 1

14:00 – 17:00 hrs: Paper – 2

You can apply for IIT-JEE 2010 using either the on-line facility or the off-line facility. On-line application procedure is available from 1^st November 2009 to 7^th December 2009.

Off-line submission of the application using the application materials purchased from designated branches of banks (see JEE web sites of the IITs) also is possible from 16^th November 2009 to 15^th December 2009.

Fees required for online application are Rs. 900/- (for general category, OBC & DS students) and Rs. 450/- for female (any category), SC/ST and Physically Disabled. The fees required for offline application are Rs. 1000/- and Rs. 500 respectively.

The last date for receipt of the completed application at the IITs is 19^th December, 2009 (before 17:00 hrs).

Find details at the JEE websites of the different IITs given below:
IIT Bombay: http://www.jee.iitb.ac.in

IIT Delhi: http://jee.iitd.ac.in

IIT Guwahati: http://www.iitg.ac.in/jee

IIT Kanpur: http://www.iitk.ac.in/jee

IIT Kharagpur: http://www.iitkgp.ernet.in/jee

IIT Madras: http://jee.iitm.ac.in

IIT Roorkee: http://www.iitr.ac.in/jee

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You can access all posts related to IIT-JEE (including old questions with solution) on this site by making a search for ‘IIT’ using the ‘search blog’ box provided. Similar posts can be seen at http://physicsplus.blogspot.com as well.

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

2009-10-21T19:00:00.002+05:30

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

2009-10-16T19:32:00.005+05:30

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

2009-10-06T15:53:00.005+05:30

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

(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

(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

(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

2009-08-15T21:55:00.006+05:30

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_Band 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_Cof 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

2009-08-04T12:40:00.003+05:30

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) ε₀q

(b) ε₀/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 4πr²and 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

(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

(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.