## Friday, July 27, 2007

### Multiple Choice Questions on Newton’s Laws of Motion

You will have to apply Newton’s Laws of Motion in different branches of Physics, but you will find questions specifically meant for checking your understanding of these laws in AP Physics Examination, Graduate Record Examination (GRE) and Medical and Engineering Entrance Examinations.

Consider the following MCQ based on impulse:
The force ‘F’ acting on a particle of mass ‘m’ is indicated by a force-time graph (Fig.) . The momentum received by the particle during the time from zero to 8 s is

(a) 24 Ns (b) 20Ns (c) 12Ns (d) 6Ns (e) zero

The area under the force-time graph gives the impulse imparted to the particle. Impulse is a vector quantity and so you must consider its sign while adding the areas. The impulse received from zero to 2 seconds is positive and is equal to the area of the triangle, which is 6 Ns. The impulse received during the time from 2 s to 4 s is the area of the rectangle, which is – 6 Ns. The impulse received during the time from 4 s to 8 s is the area of the larger rectangle, which is 12 Ns.

Hense the net impulse received during the time from zero to 8 seconds is 6 –6 +12 = 12 Ns.

Since the impulse is equal to the change of momentum, the correct option is (c).

Now consider the following question:

A boy caught a ball of mass 200g moving with a speed of 30 ms–1. If the catching process be completed in 0.1 s, the force of impact exerted by the ball on the hands of the boy is

(a) 60 N (b) 40 N (c) 30 N (d) 20 N (e) 10 N

We have force F = dp/dt where ‘dp’ is the change in linear momentum during the time dt.

The initial momentum of the ball is 0.2×30 = 6 kgms–1 and the final momentum is zero so that dp = 6.

Therefore, F = 6/0.1 = 60 N.

The following MCQ which appeared in Karnataka CET 2002 question paper is meant for checking your grasp of the law of conservation of linear momentum:

A projectile is moving at 20 ms–1at its highest point, where it breaks into two equal parts due to an internal explosion. One part moves vertically up at 30 ms–1 with respect to the ground. Then the other part will move at

If the mass of the projectile is ‘m’, the mass of each fragment after the explosion is m/2. The momentum of the part which moved upwards is (m/2)×30 =15m. This is shown as vector OA in the figure.

The momentum of the projectile at the highest point of its path is 20m and its direction is horizontal. This is shown as vector OC.

The momentum of the other part (after the explosion) is (m/2)×v where ‘v’ is its velocity. This is shown as vector OB in the figure.

Since the momentum is conserved, the total final momentum, which is the vector sum of the momentum vectors OA and OB must be equal to the initial momentum, represented by the vector OC.

Evidently, OB = √(OA2 + OC2).

Or, mv/2 = √[(15m)2 + (20m)2] = 25m.

This gives v = 50 ms–1.

## Wednesday, July 25, 2007

### Solution to Practice Questions on Electromagnetic Induction

In the post dated 24th July 2007, two questions on electromagnetic induction were given without solution. The questions with solution are given below as promised in the post:

(1) A search coil in the form of a plane circular copper loop of 10 turns and diameter 2 cm is held within a magnetic field of 20000 gauss, with the plane of the loop perpendicular to the magnetic field. The resistance of the search coil is 0.1 Ω. If the search coil is withdrawn from the magnetic field in a time span of 0.1s, the total charge (in coulomb) and the average current (in ampere) flowing during this time are respectively

(a) 0.1π, 0.5π (b) 0.2π, 2π (c) 0.2π, π

(d) 0.1π, π (e) 0.02π, 0.2π

The charge ‘Q’ induced in the search coil is given by

Q = Flux change/Resistance = nAB/R where ‘B’ is the magnetic flux density and ‘n’, ‘A’ and ‘R’ are respectively the number of turns, area and the resistance of the search coil.

Therefore, Q = [10× π(0.01)2×2]/0.1 = 0.02 π coulomb.

The average current ‘I’ in the coil is given by

I = Q/t = 0.02 π/0.1 = 0.2 π ampere.

(2) A triangular copper loop ABC is moving in its own plane with velocity ‘v’ in a uniform magnetic field acting perpendicular to the plane of the loop. If the direction of motion is perpendicular to the side AB of the loop, an electric field is induced

(a) in AB only

(b) in AC as well as BC, but not in AB

(c) in BC only

(d) in AB, AC and BC

(e) in no side of the loop There is no change in the magnetic flux linked with the loop and hence the net emf acting in the whole loop is zero. But, since all the conductors AB. AC and BC are moving in a magnetic field, cutting the magnetic field lines, the free charge carriers in them experience magnetic force (Lorentz force) and get shifted towards their ends . So, there are electric fields in all these conductors (generally). In other words, there are induced voltages across AB, AC and BC, but the net voltage in the whole loop is zero since the voltage across AB is equal and opposite to the net voltage across the series combination of AC and BC.

The correct option is (d).

A special case is the possibility of the triangle to be right angled at A or B. If the triangle is right angled at A, the electric field induced across AC is zero. If the triangle is right angled at B, the electric field induced across BC is zero. But the most suitable choice in the above question is still (d).

## Tuesday, July 24, 2007

### Two Practice Questions (MCQ) on Electromagnetic Induction

Try to work out the following questions on electromagnetic induction:
(1) A search coil in the form of a plane circular copper loop of 10 turns and diameter 2 cm is held within a magnetic field of 20000 gauss, with the plane of the loop perpendicular to the magnetic field. The resistance of the search coil is 0.1 Ω. If the search coil is withdrawn from the magnetic field in a time span of 0.1s, the total charge (in coulomb) and the average current (in ampere) flowing during this time are respectively
(a) 0.1π, 0.5π (b) 0.2π, 2π (c) 0.2π, π
(d) 0.1π, π (e) 0.02π, 0.2π

(2) A triangular copper loop ABC is moving in its own plane with velocity ‘v’ in a uniform magnetic field acting perpendicular to the plane of the loop. If the direction of motion is perpendicular to the side AB of the loop, an electric field is induced
(a) in AB only
(b) in AC as well as BC, but not in AB
(c) in BC only
(d) in AB, AC and BC
(e) in no side of the loop
If you have mastered the basic points in electromagnetic induction, you will be able to answer the above questions in a couple of minutes. Try. I’ll be back shortly with the solution.

## Tuesday, July 17, 2007

### IIT-JEE 2007 Assertion-Reason Type MCQ on Lenz’s Law

The following Assertion-Reason Type MCQ which appeared in IIT-JEE 2007 question paper is meant for gauging your grasp of Lenz’s law. Here is the question: STATEMENT-1

A vertical iron rod has a coil of wire wound over it at the bottom end. The rod goes through a conducting ring as shown in the figure. The ring can float at a certain height above the coil

because

Statement-2

In the above situation, a current is induced in the ring which interacts with the horizontal component of the magnetic field to produce an average force in the upward direction.

(a) Statement-1 is True, Statement-2 is True; Statement-2 is a correct

explanation for Statement-1

(b) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct

explanation for Statement-1

(c) Statement-1 is True, Statement-2 is False

(d) Statement-1 is False, Statement-2 is True

Statement-1 is true. But statement-2 false[Option (c)].

When the magnetic flux linked with the ring increases because of the increase of current in the coil, the ring will be repelled in accordance with Lenz’s law, so as to oppose the increase of flux through the ring. When the current through the coil decreases, the flux linked with the ring decreases. In this case, the ring will be attracted towards the coil, so as to oppose the decrease of flux through the ring. Since the coil is fed with an alternating current, these actions are repeated successively and the ring can therefore float above the coil.

The ring floats because of Lenz’s law. Basically it is the result of the interaction of the magnetic field produced by the induced current with the magnetic field produced by the coil, but the magnetic field produced by the coil is along the vertical (through the iron rod) and is not horizontal. The vertical field produces the flux and the change of flux through the ring and this is responsible for the floating of the ring.

Statement-2 is therefore false.

## Sunday, July 15, 2007

### KEAM (Engineering) 2007 Question on Polarisation

In the post dated 3rd May 2007, four questions on Optics which appeared in Kerala Engineering Entrance 2007 test paper were discussed. Here is an additional question which appeared in the same paper:

If the polarising angle of a piece of glass for green light is 54.74º, then the angle of minimum deviation for an equilateral prism made of same glass is (given tan 54.74º = 1.414)

(a) 45º (b) 54.74º (c) 60º (d) 90º (e) 30º

Brewster’s law gives you the refractive index ‘n’ in terms of the polarising angle ‘θ’:

n = tan θ

Therefore, the refractive index of the given equilateral prism is 1.414, which is √2

Since n = sin[(A+D)/2] /sin(A/2) where ‘A’ is the angle of the prism and ‘D’ is the minimum deviation, we have

√2 = sin[(60º +D)/2]/sin 30º

This gives sin(60º +D)/2 = 1/√2 so that D = 30º

Now that you have worked out a problem on polarisation, note the following points:

(i) Transverse waves only can be polarized (Longitudinal waves cannot be polarised).

(ii) The proof for the transverse nature of light is the exhibition of polarisation by light.

## Wednesday, July 11, 2007

### Solution to the Question on Two Dimensional Motion

In the post dated 10th July 2007, a questions on two dimensional motion was given without solution. The question with solution is given below as promised in the post:

A steel sphere A projected up with a velocity of 10 ms–1 at an angle of 60º with the horizontal, collides elastically with an identical steel sphere B located at the highest point of the trajectory of A. The sphere B is the bob of a simple pendulum. After the collision, if the sphere B just moves along a vertical circle, what is the length of the pendulum? ( Take g = 10 ms–2)

(a) 50 cm (b) 55 cm (c) 60 cm (d) 65 cm (e) 70 cm

At the highest point of the trajectory, the velocity of the projectile is horizontal and is equal to ucosθ = 10 cos 60° = 5 ms–1. Since the collision is elastic and the spheres are identical, the sphere A transfers its entire momentum to sphere B. Sphere B therefore moves with the same horizontal velocity of 5 ms–1.

If the sphere is to just move along a vertical circle, the minimum velocity required at the lowest point of the circle is (5gR) where ‘R’ is the radius of the circle (which is the length of the pendulum here). Therefore, we have

(5gR) = 5 ms–1.

On substituting for the acceleration due to gravity, g (=10 ms–1), R = 0.5 m = 50 cm.

Now, consider the following MCQ:

A ball rolling along a horizontal floor with a velocity of 7 ms–1 reaches the edge of a long stair case from where it moves down as a projectile. If the width and the height of the steps of the stair case are 25 cm and 10 cm respectively, the ball will strike the edge of the nth step from the top, where n is equal to

(a) 10 (b) 12 (c) 14 (d) 16 (e) 18

Since the horizontal velocity is unchanged throughout the trajectory of the ball, we have

7 t = 0.25 n where ‘t’ is the time taken to hit the edge of the nth step after leaving the top of the stair case.

Therefore, t = 0.25 n / 7

The vertical fall of the ball during this time is ½ g t2 and we have

½ g t2 = 0.1 n

Substituting for t and g, ½ ×9.8×0.0625 n2/49 = 0.1n

Therefore, n = (49×0.1)/(4.9×0.0625) =16

## Tuesday, July 10, 2007

### Multiple Choice Questions on Two Dimensional Motion (Projectiles)

Two dimensional motion is the combination of two independent one dimensional motions at right angles. The following MCQ can be easily answered if you are aware of this.
A boy, standing on the top of a building, throws a stone up with a velocity of 16 ms–1 in a direction making an angle of 30º with the horizontal. If the height of the building is 9.8 m, the velocity with which the stone will strike the ground will be approximately

(a) 17 ms–1 at angle tan–1(2/3) with the horizontal

(b) 21 ms–1 at angle tan–1(–2/3) with the horizontal

(c) 25 ms–1 at angle tan–1(–3/2) with the horizontal

(d) 27 ms–1 at angle tan–1(–3/2) with the horizontal

(e) 29 ms–1 at angle tan–1(–2/3) with the horizontal

The horizontal and vertical components of the velocity when the stone strikes the ground at C are to be found. The magnitude and direction of the velocity can then be obtained. The horizontal component of velocity (ux) is the same throughout the trajectory (since the gravitational pull is vertically downwards) and is given by

ux = vx = u cosθ = 16 cos30° = 8√3 ms–1.

The initial vertical component of velocity (at the point of projection A) is uy = u sinθ = 16 sin30° = 8 ms–1. The vertical component (vy) of the final velocity (at the point C) is given by

vy 2 = 82 + 2×(–9.8)×(–9.8), making use of the standard relation v2 = u2 + 2as.

[We have taken u which is vertically upwards as positive. The acceleration (due to gravity) and the displacement are downwards and hence should be negative]

Therefore, vy2 = (256.08) so that vy = ± 16 ms–1.

You have to take the negative value since vy is downwards.

Therefore, vy = – 16 ms–1.

The magnitude of the final velocity at the point C is

v = (vx2 + vy2) = (83)2 + 162) = 21 ms–1, approximately.

The velocity ‘v’ is at an angle of tan–1(vy/vx) = tan–1(–16/83) with respect to the horizontal.

So, the angle is tan–1(–2/3) and the correct option is (b).

Now, suppose in the above question, you were asked to find just the speed ( or the magnitude of the velocity) with which the stone strikes the ground at C. Your answer will be 21 ms–1 as you have already found out. But you can find it out a little more easily using the energy conservation method as follows:

The kinetic energy ½ mv2 of the stone at the moment of hitting the ground is equal to the sum of the initial kinetic energy ½ mu2 and the initial gravitational potential energy mgh (with respect to the ground). Therefore we have

½ mv2 = ½ m×162 + m×9.8×9.8, from which v = 21 ms–1

The following question will be quite simple if you remember the basic things in circular motion and elastic collisions ( and of course projectiles).

A steel sphere A projected up with a velocity of 10 ms–1 at an angle of 60º with the horizontal, collides elastically with an identical steel sphere B located at the highest point of the trajectory of A. The sphere B is the bob of a simple pendulum. After the collision, if the sphere B just moves along a vertical circle, what is the length of the pendulum? ( Take g = 10 ms–2)

(a) 50 cm (b) 55 cm (c) 60 cm (d) 65 cm (e) 70 cm

Many among you will be able to work this out easily. Once you have prepared thoroughly for an examination like the AP Physics Exam or a Medical and Engineering entrance Exam, you will be able to get the answer in less than a couple of minutes!

I will be back with the solution shortly, but try your best.

## Friday, July 06, 2007

### Another Question on Electromagnetic Induction

The following question is in continuation of yesterday’s question on electromagnetic induction. In fact I had tried to post it yesterday itself but there were some technical problems.

Here is the question which is meant for checking whether you have understood Lenz’s law clearly:

(1) A current carrying solenoid and a conducting ring are arranged coaxially, the ring being at a distance ‘d’ from one end of the solenoid. By varying the output voltage of a variable voltage source feeding the solenoid, the current ‘I’ in the solenoid can be varied. Pick out the correct statement:

(a) When the current ‘I’ increases, there is an attractive force between the ring and the solenoid

(b) When the current ‘I’ decreases, there is a repulsive force between the ring and

the solenoid

(c) When the current ‘I’ increases, there is a repulsive force between the ring and the solenoid

(d) When the current ‘I’ increases, there no force between the ring and the solenoid

(e) When the current ‘I’ decreases, there no force between the ring and the solenoid

When the current in the solenoid increases, the magnetic flux linked with the ring will increase. This increment in magnetic flux produces an induced current in the ring. By Lenz’s law, the system has to oppose this. The ring and the solenoid will try to move away so that the magnetic field at the ring and the flux through the ring are reduced. So. there is a repulsive force in this case [Option (c) is the correct one].

When the current in the solenoid decreases, the solenoid and the ring will have to come closer to oppose the decrement in flux. In that case there will be an attractive force. [ So, option (b) is incorrect]

You can find additional typical multiple choice questions (with solution) on electromagnetic induction by clicking on the label ‘electromagnetic induction’ either below this post or on the side of this page (or any page).

## Thursday, July 05, 2007

### A Question on Electromagnetic Induction

The following questions will be easy for you if you have understood Lenz’s law clearly: The variation of induced emf (E) with time (t) in a coil if a short bar magnet is moved along the axis with a constant velocity is best represented as This question appeared in IIT Screening 2004 question paper.
When the magnet is moved towards the coil, the magnetic flux linked with the coil goes on increasing and the magnitude of the induced emf goes on increasing as long as the rate of increase of flux is increasing. The rate of increase of flux then has to decrease and hence the magnitude of the induced emf decreases. The rate of increase of flux then becomes zero. Thereafter, the rate of decrease of flux becomes greater and greater, becomes a maximum and then becomes smaller and smaller and finally reduces to negligibly small value. The induced emf therefore follows these variations in the opposite direction (since it is the rate of decrease of flux that occurs now).
The variation of emf is therefore given by curve (B).

## Tuesday, July 03, 2007

### Solution to MCQ on Magnetism

In the post dated 2nd July 2007, two questions on magnetism were given without solution. These questions with solution are given below as promised in the post:

(1) A thin, non-conducting rod of length ‘L’ is uniformly charged to have a linear charge density ‘λ’ from its mid point to one end (The other half of the rod is uncharged). The rod is rotated about an axis perpendicular to its length and passing through the other end with angular velocity ‘ω’. The magnetic dipole moment of the rotating rod is

(a) zero (b) λωL3/3 (c) λωL3/6

The magnetic dipole moment of a current loop is iA where ‘i’ is the current and ‘A’ is the area of the loop. The rotation of the rod OP with linear charge density λ from its middle to the end P is equivalent to many current loops of varying area. Considering the current loop made by the charge λdr on a small length dr of the rod at distance ‘r’ from the end O, the magnetic moment contributed by it is (λdr/T)×πr2 where T is the period of rotation of the rod.

Since T = 2π/ω, the above magnetic moment is ½ (λωr2dr).

The magnetic dipole moment (m) due to the rotation of the entire rod is obtained by integrating the above between the limits L/2 and L.

Therefore, m = [½(λωr2dr) = ½(λω)×[L3/3 – (L/2)3/3] = 7λωL3/48.

(2) A magnetic needle oscillates in a horizontal plane with a period ‘T’ at a place where the angle of dip is 30º. When the same needle is made to oscillate in a vertical plane coinciding with the magnetic meridian, its period will be

(a) T (b) T/√3 (c) T×√3 (d) T×31/4 (e) T×31/4

When the magnetic needle is made to oscillate in the horizontal plane, the restoring force required for the motion is provided by the horizontal component (Bh) of the earth’s magnetic field B so that we have

T = 2π√(I/mBh) where ‘I’ is the moment of inertia of the needle (about the axis of angular oscillation, which is through its centre of mass and is perpendicular to its length.) and ‘m’ is its magnetic dipole moment.

When the needle is made to oscillate in the vertical plane, the restoring force required for the motion is prvided by the vertical component (Bv) of the earth’s magnetic field B so that we have the new period

T’ = 2π√(I/mBv).

Dividing the first equation by the second, T/T’ = √(Bv/Bh) from which

T’ = T/√(Bv/Bh).

But, Bv/Bh = tanθ where ‘θ’ is the angle of dip (which is 30º here) so that

T’ = T/√(tan30º) = T/√(1/√3) = T×31/4.

## Monday, July 02, 2007

### Multiple Choice Questions on Magnetism

I give you two questions (MCQ) from magnetism without solution. See whether you can work out these within five minutes. I’ll be back shortly with the solution.

(1) A thin, non-conducting rod of length ‘L’ is uniformly charged to have a linear charge density ‘λ’ from its mid point to one end (The other half of the rod is uncharged). The rod is rotated about an axis perpendicular to its length and passing through the other end with angular velocity ‘ω’. The magnetic dipole moment of the rotating rod is

(a) zero (b) λωL3/3 (c) λωL3/6

(d) 5λωL3/46 (e) 7λωL3/48

(2) A magnetic needle oscillates in a horizontal plane with a period ‘T’ at a place where the angle of dip is 30º. When the same needle is made to oscillate in a vertical plane coinciding with the magnetic meridian, its period will be

(a) T (b) T/√3 (c) T×√3 (d) T×31/4 (e) T×31/4