## Tuesday, December 03, 2013

### Apply Online for AIPMT 2014

Now is the right time for you to apply online for the All India Pre-Medical/Pre-Dental Entrance Test (AIPMT) 2014. Instead of the National Eligibility Cum Entrance Test (NEET) conducted last year for admitting students to MBBS and BDS courses, the earlier AIPMT has been reintroduced. The Entrance Test which will be conducted from 10 am to 1 pm on 4th May, 2014 (Sunday) will have one paper with 180 objective type questions from Physics, Chemistry and Biology (Botany & Zoology).
The last date for submission of online application without late fee is 31-12-2013.
The last date for submission of online application with late fee is 31-1-2014.
Visit the AIPMT site at http://aipmt.nic.in immediately to get all details.

## Monday, October 07, 2013

### Multiple Choice Questions Involving Electromagnetic Induction

“There is only one corner of the universe that you can be certain of improving, and that’s your own self.”
Aldous Huxley

Many questions on electromagnetic induction have been discussed on this site earlier. You can access those questions by clicking on the label, ‘electromagnetic induction’ below this post. Today we shall discuss a few more questions in this section.

(1) An object of mass M (Fig.) carrying a charge +Q is released from rest at the top of a smooth inclined plane of inclination θ. A magnetic field of flux density B acts  throughout the region and is directed normally into the plane of the figure. Which one among the following statements is correct?
(a) The acceleration of the object down the incline depends on Q but does not depend on B.
(b) The acceleration of  the object down the incline depends on B but does not depend on Q.
(c) The acceleration of the object down the incline depends on both Q and B.
(d) The acceleration of the object down the incline does not depend on Q and B.
(e) All the above statements are incorrect.
The component of the gravitational force along the incline is Mgsinθ which can produce in the object an acceleration gsinθ down the incline. When the object moves down along the inclined plane, the magnetic force (Lorentz force) acting on the object is directed perpendicular to the inclined plane such that it tries to press the object against the inclined plane.
[Apply Fleming’s left hand rule to arrive at the direction of the magnetic Lorentz force].
Since the magnetic force on the object is perpendicular to the inclined plane, it has no component along the plane. The acceleration of the object down the incline is gsinθ itself and is independent of Q and B [Option (d)].
(2) In the above question suppose the inclined plane is rough and the object is able to slide down. Which one among the following statements is correct?
(a) The acceleration of the object down the incline depends on Q but does not depend on B.
(b) The acceleration of the object down the incline depends on B but does not depend on Q.
(c) The acceleration of the object down the incline depends on both Q and B.
(d) The acceleration of the object down the incline does not depend on Q and B.
(e) All the above statements are incorrect.
In this case the frictional force acting between the object and the inclined plane reduces the acceleration of the object. The frictional force F depends on the normal force N acting on the object.
[Remember that F = μN where μ is the coefficient of friction]
But the normal force (which is Mgcosθ + magnetic Lorentz force) depends on both Q and B since the magnetic Lorentz force is QvB where v is the velocity of the object.
Therefore, the acceleration of the object down the incline depends on both Q and B [Option (c)].

(3) A plane rectangular coil of length 2L and breadth L has N turns of insulated copper wire in it. Initially (at time t = 0) the plane of the coil is perpendicular to a uniform magnetic field of flux density B . The coil rotates with angular velocity ω such that the axis of rotation of the coil is at right angles to the magnetic field (Fig.). If the voltage induced in the coil is Vmaxsinωt, the angular velocity of the coil is

(a) Vmax / 2NL2B

(b) 2Vmax t / NL2B

(c)Vmax/NL2B

(d) NL2B/ Vmax

(e) 2NL2B/ Vmax

The magnetic flux φ linked with the coil at the instant t is given by
φ = NAB cos ωt where A is the area of the coil.

The voltage V induced in the coil is is given by

V = dφ /dt = NABω sin ωt

It is given in the question that V = Vmaxsinωt

Therefore, Vmax = NABω

This gives ω = Vmax /NAB
Since the area A = 2L2, we have ω = Vmax / 2NL2B

(4) In the transformer shown in the adjoining figure, the secondary winding has taps A, B, C, D and E. The number of turns between taps A and B is 100. The number of turns between taps B and C is 50, the number of turns between C and D is 40 and that between D and E is 10. The primary has 2000 turns and is connected to 220 volt A. C. mains. Between which taps will you obtain 5.5 volt output?

(a) Between A and B

(b) Between A and B as well as B and E

(c) Between B and C as well as C and E

(d) Between C and D

(e) Between D and E

Since the primary voltage is 220 V and the number of turns in the primary is 2000, the induced voltage per turn is 220/2000 = 0.11 volt.

Therefore, the number of turns required to produce a voltage of 5.5 volt is (5.5)/(0.11) = 50.

Since there are 50 turns between taps B and C as well as between taps C and E, the correct option is (c).

[You can work this out this way also:

Considering the entire secondary of the transformer, the secondary to primary turns ratio is 200/2000 = 1/10. Therefore, the secondary voltage (between the ends A and E) is 220/10 = 22 V. Since the voltage across the entire secondary containing 200 turns is 22 volt, we will get 5.5 volt across 50 turns (between taps B and C as well as between taps C and E)]

## Thursday, August 22, 2013

### JEE (Advanced) 2013 Questions Involving Two Dimensional Motion

"It is unwise to be too sure of one's own wisdom. It is healthy to be reminded that the strongest  might weaken and the wisest might err”
– Mahatma Gandhi

Today we shall discuss three questions which appeared in JEE (Advanced) 2013 question paper. JEE (Advanced) is the examination conducted for admitting students to under graduate engineering programmes in IITs, IT-BHU and ISM Dhanbad (in place of the earlier IIT-JEE).

(1) A particle of mass m is projected from the ground with an initial speed u0 at an angle α with the horizontal. At the highest point of its trajectory, it makes a completely inelastic collision with another identical particle, which was thrown vertically upwards from the ground with the same initial speed u0. The angle that the composite system makes with the horizontal immediately after the collision is

(a) π/4

(b) π/4 + α

(c) π/4 α

(d) π/2

At the highest point (P in fig.) of the trajectory, the velocity of the first particle is u0 cos α.
[The vertical component of the velocity of projection of the first particle is zero at P. The horizontal component (u0 cos α) of the velocity of projection is retained through out the trajectory]
The momentum of the first particle at the highest point P is mu0 cos α and is directed horizontally rightwards.
The velocity (v) of the second particle at the point P is given by the equation of linear motion, v2 = u02 – 2gH. This gives
v = √(u02 – 2gH)
But H = (u02 sin2α)/2g
Therefore we have
v = √[u02 – 2g(u02 sin2α)/2g] = √[u02(1 – sin2α) = u0 cos α
Therefore, the momentum of the second particle at the point P is mu0 cos α and is directed vertically upwards.
The momentum of the composite particle immediately after the inelastic collision is the resultant of the momenta of the two particles. The resultant of the horizontally rightward momentum mu0 cos α and the vertically upward momentum mu0 cos α is inclined at 45º with respect to the horizontal.
Therefore, the correct option is (d).
*          *          *          *          *          *          *          *          *          *          *          *
Two more questions are discussed below. These questions too are single correct answer type multiple choice questions and are based on a given paragraph.
Here is the paragraph and the two questions related to it:

A small block of mass 1 kg is released from rest at the top of a rough track. The track is a circular arc of radius 40 m. The block slides along the track without toppling and a frictional force acts on it in the direction opposite to the instantaneous velocity. The work done in overcoming friction up to the point Q, as shown in the figure below, is 150 J. (Take the acceleration due to gravity, g = 10 ms–2)
(i) The speed of the block when it reaches the point Q is
(a) 5 ms–1
(b) 10 ms–1
(c) 10√3 ms–1
(d) 20 ms–1
The block reaches the point Q after falling through a height h equal to R sin30º = R/2. Therefore it loses gravitational potential energy mgh. Out of this energy 150 joule is used up against friction and the remaining portion is transferred to the block as kinetic energy ½ mv2 where v is the speed of the block at the point Q. Therefore, we have
mgh – 150 = ½ mv2
Here m = 1 kg, g = 10 ms–2 and h = R/2 = 40/2 = 20 m
Therefore, 10×20 – 150 = v2/2
This gives v = 10 ms–1
(ii) The magnitude of the normal reaction that acts on the block at the point Q is
(a) 7.5 N
(b) 8.6 N
(c) 11.5 N
(d) 22.5 N

With reference to the adjoining figure, if the normal reaction acting on the block at the point Q is N, we have

N mg cos 60 = mv2/R

[mv2/R is the centripetal force required for the circular motion of the block]
Therefore N = mv2/R + mg cos 60 = (1×102)/40 + 1×10× ½  = 7.5 newton.

## Wednesday, July 24, 2013

### NEET 2013 Questions (MCQ) from Work and Energy

`“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 `

The following questions on work and energy appeared in the National Eligibility Cum Entrance Test (NEET) 2013 which replaced AIPMT for admitting students to MBBS and BDS courses:.

(1) A uniform force of (3 i + j) newton acts on a particle of mass 2 kg. Hence the particle is displaced from position (2 i + k) metre to position (4 i + 3 j k) metre. The work done by the force on the particle is

(1) 6 J

(2) 13 J

(3) 15 J

(4) 9 J

Work W is the scalar product (dot product) of force F and displacement s.

Or, W = F.s

Since the particle is displaced from position (2 i + k) metre to position (4 i + 3 j k) metre, the displacement is given by

s = (4 i + 3 j k) – (2 i + k) = (2 i + 3 j – 2 k)

Therefore work W = F.s = (3 i + j) . (2 i + 3 j – 2 k) = 6 + 3 = 9 joule [Option (4)].

(2)The upper half of an inclined plane of inclination θ is perfectly smooth while lower half is rough. A block starting from rest at the top of the plane will again come to rest at the bottom, if the coefficient of friction between the block and lower half of the plane is given by

(1) μ = 2/tanθ

(2) μ = 2 tanθ

(3) μ = tanθ

(4) μ = 1/tanθ

This question is popular among question setters (See AIEEE 2005 question paper).

The component of gravitational force along the incline is mg sin θ where m is the mass of the object and g is the acceleration due to gravitaty.

The work done by the gravitational force when the object moves along the incline is mgL sinθ since the force mg sinθ acting along the incline moves the object through the length L of the incline.

The work done by the gravitational force imparts kinetic energy to the object. But the entire kinetic energy is used up in doing work against the frictional force acting along the lower half of the incline. Therefore, we have

mgL sinθ = μmg cosθ (L/2)

[Note that the frictional force is μmg cosθ since the normal force exerted (by the incline) on the object is mg cosθ and the coefficient of friction is μ]

From the above equation we obtain μ = 2 sinθ/cosθ = 2 tanθ [Option (2)].

[You may argue in an equivalent manner like this too:

The object falls through a height L sinθ, thereby losing gravitational potential energy mgL sinθ. The entire gravitational potential is used up in doing work against the frictional force acting along the lower half of the incline. Therefore, we have

mgL sinθ = μmg cosθ (L/2) from which μ = 2 tanθ

*          *          *          *          *          *          *          *          *          *          *          *

You can work out the above problem equally well by using the equation of linear motion,

v2 = u2 + 2as

Considering the uniformly accelerated motion of the object down the upper smooth half of the inclined plane, the final velocity ‘v’ of the object is given by

v2 = 0 + 2 g sinθ (L/2) ………..(i)

Considering the uniformly retarded motion of the object down the lower rough half of the inclined plane, we have

0 = v2 + 2(g sinθ – μg cosθ) (L/2) ………(ii)

Substituting for v2 from Eq. (i), we have

2 gL sinθ = μgL cosθ, from which μ = 2 tanθ]