Wednesday, May 30, 2007

Magnetic Force on Moving Charge- Kerala Medical Entrance 2007 Question

The following MCQ appeared in Kerala Medical Entrance 2007 question paper:

A proton with energy of 2 MeV enters a uniform magnetic field of 2.5 T normally. The magnetic force on the proton is (Take mass of proton to be 1.6×10–27 kg)

(a) 3×10–12 N (b) 8×10–10 N (c) 8×10–12 N (d) 2×10–10 N (e) 3×10–10 N

The velocity ‘v’ of the proton is to be found first using the expression for kinetic energy E (in joule). Note that E = 2 MeV = 2×106×1.6×10–19 joule.

We have E = ½ mv2 from which v = √(2E/m) = [2×2×106×1.6×1019/(1.6×10–27)]1/2 = 2×107 ms–1.

Substituting this value of ‘v’ in the expression for magnetic force (F = qvB), we obtain

F = 1.6×10–19×2×107×2.5 = 8×10–12 N.

Note: In the above question we did not take the relativistic increase of mass of the proton into consideration. Since the energy is 2 MeV only and the proton is is fairly heavy, the relativistic increase of mass will be about 0.2% only and you will obtain the velocity of the proton of 2 MeV energy as 1.995×107 ms–1. So, the magnetic force will be slightly reduced.

You should be aware of the relativistic increase in mass when you deal with questions like the above one, especially if you are preparing for GRE Physics Exam or AP Physics Exam in which you can expect questions of the type given below:

An electron with energy of 2 MeV enters a uniform magnetic field of 2.5 T normally. The magnetic force on the electron is nearly (Take the rest mass of electron to be 9.1×10–31 kg)

(a) 1.17×10–10 N (b) 8×10–10 N (c) 3.35×10–10 N (d) 8×10–12 N (e) 3.14×10–14 N

If you calculate the velocity of the electron without considering the relativistic increase in mass, as we did in the previous question, you will get (nearly) v = 8.39×108 ms–1 and the magnetic force F = 3.35×10–10 N, nearly. But, the velocity is more than the velocity of light in free space and therefore is absurd. So, (c) is not the correct option.

If m0 is the rest mass of the electron and ‘m’ is its mass while moving with kinetic energy E (= 2 MeV), we have

(m – m0)c2 = E, from which m = m0 + E/c2. The energy E is to be substituted in joule in this equation

Therefore, m = 9.1×10–31 + (2×106 ×1.6×10–19)/(3 ×108)2 = 9.1×10–31 +3.55×10–30 = 4.46×10–30 kg. Note that the mass of the electron has become nearly five times its rest mass.

But m = m0/√(1 – v2/c2) so that v = c√[1 – (m0/m)2] = 3×108×√[1 – (9.1×10–31 /4.46×10–30)2] = 2.93×108 ms–1

The magnetic force on the electron is qvB = 1.6×10–19×2.93×108×2.5 = 1.17×10–10 N.

Tuesday, May 29, 2007

MCQ on Magnetic Field due to a Thick Current Carrying Conductor

Usually you will be concerned about the magnetic field outside a current carrying conductor. You will usually take for granted that the conductor is thin and practically there is no need to explore within the conductor, even if it is thick. But for any body interested in physics, every thing deserves to be explored! Further, you should be prepared to answer a question of the type given below:

A very long, thick, cylindrical conductor of radius ‘a’ carries a direct current ‘I’. The magnetic field produced at a point P by this current is plotted against the distance of the point P from the centre of the conductor. Which one of the following graphs gives the correct variation?

The magnetic field at any point outside a long conductor carrying a current I, as you may be remembering, is given by

B = μ0I/2πr, where μ0 is the permeability of free space and ‘r’ is the distance of the point from the centre of the conductor. So, the field outside is inversely proportional to the distance ‘r’.

The magnetic field inside the conductor can be found using the above expression itself by substituting for ‘I’ as the current carried by the cylindrical portion of the conductor having radius ‘r’.

Therefore, in place of ‘I’ you have to substitute (I/πa2)(πr2) = Ir2/a2.

Thus, the field inside, B = μ0(Ir2/a2)/2πr = μ0Ir/a2.

Therefore, the field inside is directly proportional to the distance ‘r’

The correct option is graph (A).

[The magnetic field inside or outside the conductor can be easily found using Ampere’s circuital law: Outside the conductor, since the current enclosed by the closed path of radius ‘r’ is the entire current ‘I’, you will write the law as B×2πr = μ0I and will obtain B = μ0I/2πr.

Inside the conductor, since the current enclosed by the closed path of radius ‘r’ is Ir2/a2, you will write the law as B×2πr = μ0(Ir2/a2) and will obtain B = μ0Ir/2πa2].

Monday, May 28, 2007

MCQ on Angular Momentum of a Projectile

A particle of mass ‘m’ is projected with a velocity ‘v’ making an angle of 45º with the horizontal. When the projectile is at its maximum height, the magnitude of its angular momentum about an axis passing through the point of projection and perpendicular to the plane of its path is

(a) zero (b) mv2/4√2g (c) ) mv3/4√2g (d) mv2/√2g (e) mv3/√2g

You will find questions similar to this on many occasions. So, take a special note of this question.

The velocity of a projectile changes continuously along its path because of the change in the vertical component of velocity under the gravitational pull. If θ is the angle of projection, the horizontal component of velocity, vcosθ remains unchanged throughout the path and at the maximum height, the vertical component of velocity is zero and it has the horizontal velocity vcosθ only. The ‘lever arm’ for angular momentum at the maximum height is the maximum height (v2sin2θ)/2g itself so that the angular momentum is (mvcosθ)×(v2sin2θ)/2g = (mvcos45º )×(v2sin2 45º )/2g = mv3/4√2g

A simplified form of this question appeared in KEAM (Medical) 2007 question paper:

A particle is projected with a speed ‘v’ at 45º with the horizontal. The magnitude of angular momentum of the projectile about the point of projection when the projectile is at its maximum height ‘h’ is

(a) zero (b) mvh2/√2 (c) mv2h/2 (d) mvh3/√2 (e) mvh/√2

Since the maximum height is given as ‘h’, you can write the answer in no time as (mvcos45º)×h = mvh/√2.

Saturday, May 26, 2007

MCQ on Rolling Bodies

The following question on rolling bodies will be worth noting since many among you will be able to correct certain misconceptions:

A disc made of iron rolling along a horizontal surface with a velocity ‘v’ encounters an inclined plane of inclination 30º as shown in the figure. It rolls up the plane (without slipping) and reaches a height ‘h’ before turning back. If a solid sphere made of iron were rolling up with the same initial velocity, the height reached before turning back would be (ignoring the loss of energy against friction)

(a) h (b) 7h/8 (c) 9h/10 (d) 12h/13 (e) 14h/15

The entire initial kinetic energy of the disc (or sphere) gets converted into gravitational potential energy on reaching the maximum height. Since this is a case of rolling, the kinetic energy is partly translational and partly rotational. In the case of the disc we have

½ Mv2+ ½ Iω2 = Mgh, where ‘M’ is the mass, ‘I’ is the moment of inertia, ‘ω’ is the angular velocity and ‘g’ is the acceleration due to gravity.

Since ω = v/R and I = MR2/2, this can be rewritten as

¾ Mv2 = Mgh, from which h = 3v2/4g

In the case of the solid sphere The energy equation, ½ Mv2+ ½ Iω2 = Mgh can be rewritten as

½ Mv2+ ½ (2/5)Mr2(v2/r2) = Mgh

We have used the symbol ‘r’ for the radius of the sphere. This simplifies to

(7/10)Mv2 = Mgx, where ‘x’ is the maximum height reached by the sphere.

Therefore, x = 7v2/10g = (7/10)(4/3)(3/4)v2/g = 14h/15, since h = 3v2/4g.

Whenever you solve problems, you should think of other possibilities. A question setter can modify the above question as follows:

A disc and a solid sphere made of the same material have the same mass but radii in the ratio 3:1. If they have the same initial velocity, the ratio of the maximum heights to which they can roll up an inclined plane of given inclination is (ignoring the work done against friction)

(a) 15/14 (b) 14/15 (c) 45/14 (d) 42/15 (e) 3

We have seen in the previous problem that the height is independent of the radius. The correct option is 15/14 (since we require the ratio h/x).

You should note that the two bodies need not be of the same material as the work done against friction can be ignored.

Friday, May 25, 2007

Kerala Engineering/Medical Entrance Examination 2007 Results Announced

The results of Kerala Engineering/Medical Entrance Examinations conducted in April 2007 by the Commissioner for Entrance Examinations, Govt. of Kerala are announced. The results are available at and at

Thursday, May 24, 2007

NCERT Physics Text Books Online

You may be using various Physics text books for your Class 11 and Class 12 studies, but you should definitely acquire NCERT Physics text books for obtaining high quality content covering your requirement to a very large extent. Generally, NCERT text books in various branches of science are of good quality as I could ascertain from my friends dealing with the subjects. As far as Physics (in Class11 and Class 12) is concerned, I can recommend the NCERT text books to you without any element of doubt.
Fortunately, these text books are freely available online. You will find a link “Quality Books from NCERT” on the lower left side of this page, for this purpose.

Wednesday, May 23, 2007

A Question Involving Inertial Force and Frictional Force

Here is a question which requires your understanding of inertial force and frictional force:

Two bodies A and B of masses ‘M’ and ‘m’ respectively move together under the action of a horizontal force ‘F’ as shown in the figure. The body of mass A is sliding on a horizontal frictionless surface. If the coefficient of friction between A and B is μ, B will not move down under its own weight if F is at least

(a) Mg/μm (b) (M+m)g/μm (c) mg (d) mg/μ (e) (M+m)g/μ

There is frictional force μma between the bodies A and B since there is a normal force equal to the inertial force ma which the body B exerts on A.

The acceleration ‘a’ of the bodies under the action of the force F is given by a = F/(M+m)

Therefore, frictional force between A and B = μma = μmF/(M+m).

[Note that the maximum value of the frictional force is equal to mg since it is a self adjusting force just sufficient to balance the weight mg of the body B and to prevent the relative motion between A and B].

The least value of F is given by μmF/(M+m) = mg so that F = (M+m)g/μ

Now, suppose the force F is less than the least value required to prevent the body B from sliding down. If it slides down with acceleration ‘a1’ and the coefficient of kinetic friction is μk what is the value of F?

[In the question we discussed above, the coefficient of friction (μ) we used is the coefficient of static friction, which is the one having the maximum possible value. When the body slides down, the frictional force is lowered and the coefficient of sliding friction therefore has a smaller value. You may be remembering that the least coefficient is the coefficient of rolling friction]

The frictional force when B slides down is μkma so that the net downward force on B is

mg μkma. The downward acceleration therefore is (mgμkma)/m = (g μka)

Therefore we have g μka = a1

Since a = F/(M+m), we have g μkF/(M+m) = a1, from which F = (M+m)(g–a1)/μk

Tuesday, May 22, 2007

Two Questions on Satellites

In the post (titled ‘Questions on Satellites’) dated August 30, 2006, two questions were discussed. The following question similar to the first question appeared in KEAM 2007 (Engineering) question paper:

A satellite is launched in a circular orbit of radius R around the earth. A second satellite is launched into an orbit of radius 1.01 R. The period of second satellite is longer than the first one (approximately) by

(a) 1.5% (b) 0.5% (c) 3%

(d) 1% (e) 2%

In the question discussed in the earlier post, the orbital radius was 1.02 R instead of 1.01R and the answer was obtained as 3%.

You may work out the present question to obtain the answer as 1.5%. If you have any difficulty, see the earlier post dated August 30, 2006. You may easily do this by clicking on the label satellite below this post.

Now see the following MCQ:

If the orbital radius of an artificial satellite is to be increased by 5%, the orbital speed will have to be

(a) decreased by 5% (b) increased by 5% (c) decreased by 2.5%

(d) increased by 2.5% (e) decreased by10%

The orbital speed is obtained by equating the centripetal force to the gravitainal pull:

mv2/r = GMm/r2 where ‘m’ is the mass of the satellite, ‘v’ is the orbital speed, ‘r’ is the orbital radius, G is the gravitational constant and M is the mass of the earth.

From this v = √(GM/r).

Since G and M are constants, dv/v = – ½ dr/r

Instead of the fractional changes dv/v and dr/r, we can use percentage changes and write this equation as

percentage change in v = – ½ ×percentage change in r

Since the change in ‘r’ is an increment of 5%, the change in ‘v’ will be an increment of – ½ ×5%, which means a decrement of 2.5% [Option (c)].

All the essential points to be remembered in gravitation can be found at AP Physics Resources: Gravitation –Equations to be Remembered

Monday, May 21, 2007

A Question on Varying Currents

The following MCQ which appeared in KEAM 2007 (Engineering) question paper can claim to be one different from the usual type:

In a closed circuit, the current I (in ampere) at an instant of time t (in second) is given by I = 4 – 0.08t. The number of electrons flowing in 50 s through the cross section of the conductor is

(a) 1.25×1019 (b) 6.25×1020 (c) 5.25×1019

(d) 2.55×1020 (e) 4.25×1020

To obtain the electrons flowing in 50 s, you have to calculate the total charge flowing in 50 s.

Total charge Q = ∫Idt = ∫(4–0.08t)dt = [4t–(0.08t2/2)].

Since the limits of integration are 0 and 50 seconds, Q = 4×50–(0.08×502/2) = 100 coulomb.

Since the electronic charge is 1.6×10–19 coulomb, the number of electrons flowing in 50 s is 100/(1.6×10–19) = 6.25×1020.

Sunday, May 20, 2007

Change in Internal Energy of a Heated sphere - Two Questions

If you heat a body, its temperature rises and so its internal energy increases. You have come across the increase in internal energy of a gas on many occasions and have noted the difference between the amounts of heat to be supplied on heating the gas at constant volume and at constant pressure to undergo a given temperature rise. Here are two questions involving the work done by a solid on heating it:

(1) If a solid of volume V is heated at constant pressure so as to have a small temperature rise, the work done by the solid will be directly proportional to

(a) V1/2 (b) V–1/2 (c) V (d) V–1 (e) V0

(2) The temperature of a copper sphere of volume ‘V’m3 and density ‘ρ’ kg m3 increases by a small value ∆T when it absorbs a small quantity ∆Q joule of heat at atmospheric pressure ‘P’ pascal. If the specific heat of copper is ‘C’ J kg–1K–1, and the cubical expansivity of copper is ‘γ’ K–1, the increase in internal energy of the copper sphere is

(a) ∆QPV

(b) ∆Q

(c) ∆Q(1γPV/ρC)

(d) ∆Q(1γP/ρC)

(e) ∆Q(1γP/VρC)

Let us consider the second question first since it will give us the answer for the first question also.

The entire heat energy absorbed by the sphere is not used in increasing its internal energy. A small portion is used to do work in expanding (against the atmospheric pressure). Normally we omit this small amount of energy since the expansion of a solid is small. But you have to calculate it here since you are given all data for calculating it.

The rise in temperature of the sphere ∆T = ∆Q/mC where ‘m’ is the mass of the sphere.

Increase in volume of the sphere ∆V = Vγ∆T = (m/ρ)γ(∆Q/mC) = γ∆Q/ρC. [The density ρ can be assumed to be constant as the temperature rise is small].

Therefore, work done against the atmospheric pressure = P∆V = Pγ∆Q/ρC.

The increase in internal energy of the sphere = ∆Q (Pγ∆Q/ρC) = ∆Q(1γP/ρC).

So, the correct option is (d).

Now the answer for question No.1 is option (e) since the above expression for increase in internal energy does not contain the volume V.

Friday, May 18, 2007

A Question on Lorentz Force

The following MCQ appeared in Kerala Engineering Entrance 2007 question paper. I have seen this question on many occasions with more or less the same options. Here is the question:

A uniform electric field and a uniform magnetic field exist in a region in the same direction. An electron is projected with a velocity pointed in the same direction. Then the electron will

(a) be deflected to the left without increase in speed

(b) be deflected to the right without increase in speed

(c) not be deflected but its speed will decrease

(d) not be deflected but its speed will increase

(e) be deflected to the right with increase in speed

This is a simple question and the most suitable option is (c) since the magnetic field (being parallel to the electron) does not exert any force, but the electric field decelerates the electron (since the charge on the electron is negative).

Now, suppose the question setter had changed the last option as “not deflected but its speed will first decrease and will then increase”. Then, this last option would be the correct option since the electron would be decelerated first and then would be accelerated in the opposite direction. [Remember speed is a scalar quantity and the modified option takes advantage of this].

Wednesday, May 16, 2007

Mozilla Firefox Browser

It is about a couple of months since I have been using Mozilla Firefox browser for my blog posts. I find it very convenient since it senses spelling mistakes in the posts and more importantly it saves a lot of time and effort for me when I deal with superscripts and subscripts. I should have known the possibilities of the Mozilla Firefox browser much earlier!

Tuesday, May 15, 2007

IIT-JEE 2007 Linked Comprehension type Questions (MCQ) on Properties of Matter

Among the IIT-JEE 2007 physics questions there were 4 Linked Comprehension Type questions, each paper carrying two of them. Each Linked Comprehension Type question contained a paragraph based on which three multiple choice questions were asked. Here is the question on properties of matter:

Paragraph for the multiple choice questions:

A fixed thermally conducting cylinder has a radius R and height L0. The cylinder is open at the bottom and has a small hole at its top. A piston of mass M is held at a distance L from the top surface, as shown in the figure. The atmospheric pressure is P0.


(1) The piston is now pulled out slowly and held at a distance 2L from the top. The pressure in the cylinder between its top and the piston will then be

(A) P0 (B) P0/2 (C) P0/2 + Mg/πR2 (D) P0/2 Mg/πR2

The correct option is (A) since the air will enter the cylinder through the hole at the top and will keep the pressure at P

(2) While the piston is at a distance 2L from the top, the hole at the top is sealed. The piston is then released, to a position where it can stay in equilibrium. In this condition, the distance of the piston from the top is

(A)[2P0πR2/(πR2P0 +Mg)](2L)

(B) [(P0πR2–Mg)/πR2P0](2L)

(C)[(P0πR2 +Mg)/πR2P0](2L)

(D) [P0πR2/( πR2P0 –Mg)](2L)

If P is the pressure inside (the region between the piston and the top of the cylinder), the weight of the piston is balanced by the thrust (P0 – P)πR2 arising due to the pressure difference between the top and the bottom of the cylinder. Therefore,

(P0 P)πR2 = Mg from which P = P0 –Mg/πR2 = (πR2P0 Mg)/πR2

Now, substitute this value of P in the relation, P0×2L = P×L’, where L’ is the final distance of the piston from the top. [Note that this equation follows from Boyle’s law].

Thus, P0×2L = [(πR2P0 Mg)/ πR2] × L' from which

L' = [P0πR2/( πR2P0 Mg)](2L).

(3) The piston is taken completely out of the cylinder. The hole at the top is sealed. A water tank is brought below the cylinder and put in a position so that the water surface in the tank is at the same level as the top of the cylinder as shown in the figure. The density of water is ρ. In equilibrium, the height H of the water column in the cylinder satisfies

(A) ρg(L0–H)2 +P0(L0–H)+ L0P0 = 0

(B) ρg(L0–H)2 –P0(L0–H)– L0P0 = 0

(C) ρg(L0–H)2 +P0(L0–H)– L0P0 = 0

(D) ρg(L0–H)2 –P0(L0–H)+ L0P0 = 0

The correct option is obtained by applying Boyle’s law. The initial pressure of air column of length L0 is the atmospheric pressure P0. The final pressure of the air column of length (L0–H) is [(P0+(L0–H)ρg] since the pressure of air inside the cylinder exceeds the atmospheric pressure by the hydrostatic pressure exerted by water column of height (L0–H).

Applying Boyle's law, P0L0 = [(P0 +(L0–H)ρg](L0–H).

Rearranging this you will get the equation in option (c).

Sunday, May 13, 2007

MCQ on Drift Velocity of Electrons

Most of you may be aware of the fact that the drift speed of electrons in a current carrying conductor is very small compared to the speed with which a current flows in a circuit. The drift speed is typically of the order of millimeter per second while the speed with which a current flows in a circuit is nearly equal to the speed of light. Understand that the electron need not move from one end of the conductor to the other end for a current to flow. What happens is that an electric field is established in each region of a conductor (at nearly the speed of light) and the electrons in the region drift under its influence.

Now, consider the following MCQ:

A current of 4.4 A is flowing in a copper wire of radius 1 mm. Density of copper is 9×103 kg m–3 and its atomic mass is 63.5 u. If every atom of copper contributes one conduction electron then the drift velocity of electrons is nearly

(a) 0.1 mm s–1 (b) 0.5 mm s–1 (c) 1 mm s–1 (d) 1.5 mm s–1 (e) 5 mm s–1

The current in the wire is given by

I = nAve where ‘n’ is the number of conduction electrons per unit volume, ‘A’ is the area of cross section of the wire, ‘v’ is the drift velocity of the electrons and ‘e’ is the electronic charge.

From this equation, v = I/nAe.

Since the atomic mass of copper is given as 63.5 g, the number of copper atoms in 0.0635 kg is 6.02 ×1023. The density of copper being 9×103 kg m–3, unit volume of copper contains (9×103/0.0635) ×6.02×1023 = 8.5×1028 atoms.

Therefore, number of conduction electrons per unit volume, n = 8.5×1028 (since each atom contributes one conduction electron).

Therefore, drift velocity, v = I/nAe = 4.4/[8.5×1028×π(1×10–3)2×1.6×10–19] = 0.1×10–3 ms–1(nearly).

The following MCQ appeared in Kerala Engineering Entrance 2007 question paper:

When a current I flows through a wire, the drift velocity of the electrons is ‘v’. When current 2I flows through another wire of the same material having double the length and double the area of cross section, the drift velocity of the electrons will be

(a) v/8 (b) v/4 (c) v/2 (d) v (e) 2v

The drift velocity is given by v = I/nAe.

Since the wires are of the same material, the number density (n) of conduction electrons is the same. The electronic charge (e) is a constant. The current and area of cross section are doubled. Therefore, the drift velocity is unchanged.

Saturday, May 12, 2007

Two Questions (MCQ) from Thermal Physics

The following simple questions are for checking whether you have understood certain fundamental principles in Thermal Physics:

(1) Three bodies A, B, and C with thermal capacities in the ratio 1:2:3 are at temperatures T1, T2, and T3 respectively. When A and B are kept in contact, the common temperature is T. When A, B and C are kept in contact, the common temperature is T itself. Then T is equal to

(a) (T1+ T2 +T3)/3 (b) (T1 T2 +T3)/3 (c) 2(T1 + T2)/3 (d) T2 (e) T3

Simple questions often make even fairly intelligent students commit mistake and this is one such question.

The crucial point you must note is that the common temperature of A and B is unchanged when C also is kept in contact with them. So, the common temperature of A and B must be the same as the temperature of C which is T3.

(2) Equal masses of three liquids A, B,and C with specific heats c1, c2 and c3 at temperatures t1, t2 and t3 (all in degree Celsius) respectively are thoroughly mixed. The resulting temperature is

(a) (c1t1 + c2t2 + c3t3)/3

(b) (c1t1 + c2t2 + c3t3)/(c1 + c2 + c3)

(c) 3(c1t1 + c2t2 + c3t3)/(c1 + c2 + c3)

(d) 3(t1 + t2 + t3)/(c1 + c2 + c3)

(e) (t1 + t2 + t3)/ (c1 + c2 + c3)

The initial heat content is (mc1t1 + mc2t2 + mc3t3) where ‘m’ is the mass of each liquid.

The final heat content is (mc1 + mc2 + mc3)t where ‘t’ is the common temperature after mixing. [We have taken the reference heat energy level at zero degree Celsius].

Equating these two, we obtain t = (c1t1 + c2t2 + c3t3)/(c1 + c2 + c3)

Wednesday, May 09, 2007

Kerala Govt Engineering Entrance 2007-Questions on Alternating Currents

Kerala Government Engineering Entrance 2007 question paper contained four questions from alternating currents out of 72 questions in the physics part. Here are the questions:
(1) A square coil of side 25 cm having 1000 turns is rotated with uniform speed in a magnetic field about an axis perpendicular to the direction of the field. At an instant t the e.m.f. induced in the coil is e = 200 sin100πt. The magnetic induction is
(a) 0.50 T (b) ) 0.02 T (c) ) 10–3 T (d) ) 0.1 T (e) ) 0.01 T
If you remember the production of alternating voltage when a plane coil rotates in a magnetic field, you will be able to answer this question in no time. The induced alternating e.m.f. is
e = naBω sin (ωt) and the maximum emf induced is naBω where ‘n’ is the number of turns, ‘a’ is the area (of the coil), B is the magnetic induction and ‘ω’ is the angular speed. So, we have
1000×(25×10–2)2×B×100π = 200. [The value of ‘ω’ is obtained from the form of the equation for the emf].
This gives B = 0.01 T
(2) When a d.c. voltage of 200 V is applied to a a coil of self inductance 2√3/π H, a current of 1 A flows through it. But by replacing d.c. source with a.c. source of 200 V, the current in the coil is reduced to 0.5 A. Then the frequency of a.c. supply is
(a) 100 Hz (b) 75 Hz (c) 60 Hz (d) 30 Hz (e) 50 Hz
In an LR circuit the direct current is limited by resistance only. On applying a direct voltage of 200V, the current flowing is 1 A which means that the resistance of the coil is 200 Ω.
When the a.c. source is connected, the current is given by I = V/Z where the impedance Z is given by Z = √(R2+L2 ω2)
Therefore 0.5 = 200/√[2002+(2√3/π)2 ω2] = 200/√[2002+(2√3/π)2 (2πf)2], from which f = 50 Hz.
(3) In a LR circuit, the value of L is (0.4/π) henry and the value of R is 30 Ω. If in the circuit, an alternating emf of 200 V at 50 cycles per second is connected, the impedance of the circuit and the current will be
(a) 11.4 Ω, 17.5 A (b) 30.7 Ω, 6.5 A (c) 40.4 Ω, 5 A
(d) 50 Ω, 4 A (e) 35 Ω, 6.5 A
This question is quite simple and straight forward.
Impedance, Z = √(R2+L2 ω2) = √[(302+(0.4/π)2 (100π)2)] since ω = 2πf =2π×50 = 100π.
This gives Z = 50 Ω.
Current I = V/Z = 200/50 = 4 A.
(4) A transformer has an efficiency of 80%. It is connected to a power input of 5 kW at 200 V. If the secondary voltage is 250 V, the primary and secondary currents are respectively
(a) 25 A, 20 A (b) 20 A, 16 A (c) 25 A, 16 A (d) 40 A, 25 A (e) 40 A, 16 A
This question also is quite simple and can be worked out within a minute.
Primary current IP = (5000 watts)/(200 volts) = 25 A.
Secondary current = (5000×0.8 watts)/(250 volts) = 16 A.
[Note that the output power from the secondary side is efficiency times the input power].

Monday, May 07, 2007

Two Questions Involving Friction in Rotational Motion

A coin is placed on a horizontal turn table, at a distance‘r’ from the axis of rotation of the turn table. Starting from rest, the turn table rotates with a constant angular acceleration ‘α’. If the coefficient of friction between the coin and the turn table is ‘μ’, the time after which the coin will begin to slip is

(a) √(μ/α) (b) √(μg/α2) (c) √(μg/αr) (d) √(μgα2/r) (e) √(μg/α2r)

The coin will start slipping when the centrifugal force begins to exceed the frictional force. In the limiting case therefore, mrω2 = μmg where ‘m’ is the mass of the coin and ‘ω’ is the angular velocity of the turn table when the coin begins to slip.

But, ω = ω0 + αt = αt since the initial angular velocity (ω0) is zero.

[Note that the above equation is similar to the equation, v = v0 + at in the case of linear motion].

Substituting this value of ω in the condition for slipping, we have

mr(αt)2 = μmg from which t = √(μg/α2r).

Now, let us modify the above question as follows:

In a region of space where gravitational force is negligible, a coin is kept in contact with a rod AB as shown, at a distance ‘r’ from the end A. If the rod starts from rest and rotates with a constant angular acceleration ‘α’ about an axis XX’ perpendicular to the rod and passing through the end A, the time after which the coin will begin to slip is

(a) √(μα) (b) √(μα/r) (c) √(μ/αr) (d) √(μ/α) (e) negligibly small

This problem is very much different since gravity is negligible. In the previous problem, the normal force that produced friction was the result of the weight of the coin where as here it is produced because of the inertial force produced by the acceleration of the rod. When the rod rotates with an angular acceleration α, the coin presses against the rod with an inertial force equal to ma where ‘m’ is the mass and ‘a’ is the tangential acceleration of the mass.

As in the previous problem, the condition for slipping to start is obtained by equating the centrifugal force to the frictional force:

mrω2 = μma

Therefore, ω = √(μa/r).

But, ω = ω0 + αt = αt (since the initial angular velocity ω0 is zero); a = αr.

Substituting these we obtain αt = √(μα) so that t =√(μ/α) .

Sunday, May 06, 2007

AP (Advanced Placement) Physics Examinations

Many among the visitors of this site might have already noted that the posts here are useful for preparing for the AP (Advanced Placement) Physics Exams. Even though you find multiple choice questions here, you will definitely find the posts useful for answering the free-response section of the exams since the questions are answered with all necessary theoretical details. In many cases the discussions touch even minute details. This is done for helping students who are just above average. The needs of those who appear for the AP Physics Exams will be considered while discussing questions here. You may make use of the facility for comments for communications in this regard.

Saturday, May 05, 2007

Two Multiple Choice Questions on Elastic Collision

You should remember that momentum and kinetic energy are conserved in elastic collisions where as momentum alone is conserved in inelastic collisions. Let us consider two questions (MCQ) involving elastic collision in one dimension:

(1) A system consisting of two identical blocks A and B, each of mass ‘m’, connected by a light spring of force constant ‘k’ is resting on a smooth horizontal surface. A third identical block C of mass ‘m’ moving with a velocity ‘v0’along the direction of the line joining A and B collides elastically with A and compresses the spring. The maximum compression of the spring is

(a) √(mv0/k) (b) √(mv0/2k) (c) √(m/v0k) (d) v0√(m/k) (e) v0√(m/2k)

At the instant of collision, you need consider the two identical colliding masses C and A only. As the collision is elastic, the entire momentum and kinetic energy of C are transferred to A and C comes to rest. The block A then moves towards B, compressing the spring. When the compression is maximum, both A and B move with the same velocity.

Since the momentum is always conserved,

mv0 = (m + m)v where ‘v’ is the common velocity with which the masses A and B move. Therefore, v = v0/2.

The entire kinetic energy of C is transferred to A, which then compresses the spring and pushes B so that we have

½ mv02 = ½ (m + m)(v0/2)2 + ½ kx2, where ‘x’ is the maximum compression.

This gives x = v√(m/2k).

(2) An α-particle of mass ‘m’ suffers a one-dimensional elastic collision with a nucleus of unknown mass at rest. After the collision the α-particle is scattered directly backwards, losing 75% of its kinetic energy. Then, the mass of the nucleus is

(a) m (b) 2m (c) 3m (d) (3/2)m (e) 5m

This MCQ appeared in the Kerala Engineering Entrance 2003 question paper.

The total momentum of the system is equal to the initial momentum p1 of the α-particle. Equating the total initial and final momenta,

p1 = P – p2 where P and p2 are respectively the final momenta of the nucleus and the α-particle. (The negative sign for p2 is because of the backward motion of the α-particle after the collision).

But the magnitude of p2 is p1/2 since the final kinetic energy of the α-particle is ¼ of its initial kinetic energy. [Remember that K.E. = p2/2m and hence the momentum p is directly proportional to the square root of the kinetic energy].

Therefore, P = p1 + (p1/2) = (3/2) p1.

Since the collision is elastic, kinetic energy also is conserved.

Therefore, K.E. lost by the α-particle = K.E gained by the nucleus so that

P2/2M = (3/4) p12 /2m where ‘M’ is the mass of the nucleus.

Substituting for P, [(3/2)p1]2/ 2M = [(3/4)p12]/2m

This gives M = 3m.