Saturday, June 30, 2007

MCQ on Carnot Engine

A brief discussion on heat engine and refrigerator was made in the post dated 20th February, 2007 in which you will find a few typical multiple choice questions. You can find the post by clicking on the label ‘Carnot engine’ or ‘thermodynamics’ below this post or on the side of this page.

Here is another MCQ on Carnot engine (reversible heat engine):

A Carnot engine converts 1/5th of the heat, which it absorbs from the source, into work. When the temperature of the sink is reduced by 100º C, its efficiency is doubled. The temperature of the source is

(a) 400 K (b) 450 K (c) 500 K (d) 600 K (e) 700 K

If T1 and T2 are the temperatures of the source and sink respectively, the efficiency (η) is given by η = (T1 – T2)/T1.

Therefore, initially we have 1/5 = (T1 – T2)/T1 so that T2 = 4T1/5.

In the second case (when the temperature of the sink is reduced), we have

2/5 = [T1 – (T2 – 100)]/T1.

Substituting for T2 (= 4T1/5) in the above equation, we obtain T1 = 500 K.

Wednesday, June 27, 2007

Net Magnetic Force and Torque on a Current Loop- Two Multiple Choice Questions

If a plane current carrying coil is placed in a uniform magnetic field, generally there will be a torque on the coil, but the net force on the coil will be zero. But, if the current loop is placed in a non-uniform magnetic field, there will be a torque and a net force on the coil. The torque on the loop will be zero only if the plane of the loop is perpendicular to the direction of the magnetic field B. This follows from the expression for torque(τ) given by

τ = n IABsinθ, where ‘n’ is the number of turns, ‘A’ is the area of the loop, ‘I’ is the current and ‘θ’ is the angle between the magnetic field vector B and the area vector A. Note that the area vector is directed normal to the plane of the area. So, if the plane of the current loop is perpendicular to the direction of the magnetic field, the torque is zero. The torque is maximum (equal to nIAB) when the plane of the loop is parallel to the field B.

Now, consider the following MCQ:

A thin copper wire of length ‘L’ is bent to form a single turn plane circular loop and is suspended in a uniform magnetic field ‘B’ which is directed parallel to the plane of the loop. The torque acting on the loop when a current ‘I’ passes in it is 0.1 N. If the same wire were bent to form a plane circular loop of 10 turns, the torque would be

(a) 0.01 N (b) 0.1 N (c) 1 N (d) 0.001 N (e) 10 N

The torque is maximum (equal to nIAB) since the plane of the coil is parallel to the magnetic field. In the first case, since the number of turns (n) is one,

torque, τ = IAB = I×π(L/2π)2×B since the radius of the single turn coil is L/2π.

Therefore, τ = IL2B/4π

When the coil is of ‘n’ turns (with the wire of the same length L),

torque, τ’ = nIA’B = nI×π(L/2πn)2×B = IL2B/4πn.

The torque thus reduces to (1/n)th of the torque on the single turn coil. Since n = 10, the correct option is 0.01 N.

Let us now consider the following question which is popular among question setters:

A very long straight wire carrying a current ‘I’ is arranged to be coplanar with a rectangular loop of sides ‘a’ and ‘b’ carrying a current ‘i’ as shown. If the nearer parallel side of the loop is at distance ‘r’ from the straight wire, the magnitude of the net magnetic force on the loop is

(a) (μ0I iab)/[2πr(r+b)]

(b) (μ0I iab)/[2πr(r+b/2)]

(c) (μ0I iab)/2πr

(d) (μ0I iab)/[2π(r+b)]

(e) (μ0I iab)/[πr(r+b)]

The magnetic forces on the top and bottom sides of the loop are equal and opposite and hence they will get canceled. The force on the left side of the loop is attractive (towards the straight wire) while the force on the right side of the loop is repulsive (away from the straight wire). But they cannot get canceled since the attractive force is greater because of the proximity of the left side to the straight wire.

The magnetic force on the left side = ia(μ0I/2πr) since (μ0I/2πr) is the magnetic flux density produced by the straight wire at the distance ‘r’.

Similarly, magnetic force on the right side = ia[μ0I/2π(r+b)]

Therefore, the net magnetic force on the loop is

ia(μ0I/2π)[1/r – 1/(r+b)] = 0Iiab)/[2πr(r+b)]

You will find many useful questions (with solution) on magnetic force at apphysicsresources

Thursday, June 21, 2007

Two KEAM (Engineering) 2007 Questions from Electrodynamics

The following multiple choice questions appeared in Kerala Engineering Entrance 2007 question paper:

(1) A conducting rod of 1 m length and 1 kg mass is suspended by two vertical wires through its ends. An external magnetic field of 2 T is applied normal to the rod. Now the current to be passed through the rod so as to make the tension in the wires zero is [Take g = 10 m s–2]

(a) 0.5 A (b) 15 A (c) 5 A (d) 1.5 A (e) 2.5 A

The magnetic force (F) on the rod is given by

F = ILB sinθ where ‘I’ is the current, ‘L’ is the length of the rod, ‘B’ is the magnetic field andθ’ is the angle between the magnetic field and the rod. Since θ = 90º,

F = ILB = I×1×2 = 2I.

The tension in the wires will be zero when the magnetic force balances the weight of the rod.

Therefore, 2I = mg = 1×10 from which I = 5 A.

[It would have been better to mention the magnetic field to be horizontal, in addition to being normal to the rod]

(2) A galvanometer of resistance 20 Ω shows a deflection of 10 divisions when a current of 1 mA is passed through it. If a shunt of 4 Ω is connected and there are 50 divisions on the scale, the range of the galvanometer is

(a) 1 A (b) 3 A (c) 10 mA (d) 30 A (e) 30 mA

This question basically pertains to the conversion of a galvanometer in to an ammeter. Since there are 50 divisions on the scale, the current (Ig) required for full scale deflection is 5 mA.

When the shunt ‘S’ is connected, the maximum current (‘range of galvanometer’, as mentioned in the question) that can be passed into the system (I) is given by

IgG = (I–Ig)S where ‘G’ is the resistance of the galvanometer.

Therefore, I = Ig(G+S)/S = 5×10–3(20+4)/4 = 30×10–3 A = 30 mA.

[You can solve this problem in no time like this: Since the shunt is 1/5th of the resistance of the galvanometer, it will pass five times the current through the galvanometer. So, the total (maximum) current that can be passed in to the system is six times Ig, which is 6×5 mA = 30 mA]

Sunday, June 17, 2007

Two Questions on Direct Current Circuits

I include the following unusually simple MCQ just because it appeared in IIT Screening 2005 question paper. The interesting thing is that some of the students with top ranks in the class may omit such trivial questions because of sheer suspicion. Here is the question:

Find the current through 2Ω

(a) 0

(b) 1 A

(c) 2 A

(d) 4 A

Both poles of the 10V battery are connected to one end of the 2Ω resistor. Similarly, both poles of the 20V battery are connected to the other end of the 2Ω resistor. So, there are no continuous paths for the currents to flow through the 2Ω resistor and the current is zero.

Now, suppose the above circuit is modified as shown in the adjoining figure in which the points A and B are grounded. Assuming that the batteries have negligible internal resistance, what will be the current through the 2Ω resistor?

The two batteries are now connected (through the ground connection) in series with the 2Ω resistor. But, the battery voltages are in opposition so that the net voltage across the 2Ω resistor is 20V – 10 V = 10 V. Therefore, the current is 10V/2Ω = 5A.

"Setting an example is not the main means of influencing others; it is the only means"Albert Einstein

Friday, June 15, 2007

MCQ on Moment of Inertia of a Rod

Moment of inertia is a scalar quantity. So you can add and subtract moment of inertia as you do in the case of scalar quantities, but you should remember that the moment of inertia depends on the axis chosen. You cannot add (or subtract) the moments of inertia of bodies about different axes, but you can add (or subtract) moments of inertia if the axis is the same. Now consider the following MCQ which will make things clear:

A thin uniform rod AB of mass 12M and length L has moment of inertia I about an axis passing through its mid point O perpendicular to its length. If a portion CB of length (L/4) is cut off from the end of the rod, the moment of inertia of the remaining portion AC about the original axis through O will be

(a) (9/16)I

(b) (¾)I

(c) (13/15)I

(d) (41/48)I

(e) (51/64)I

You have to subtract the moment of inertia of the portion CB from the moment of inertia of the original rod AB to obtain the moment of inertia of the remaining portion AC ( since all moments of inertia are about the same axis passing normally through O).

Moment of inertia (I) of original rod AB = 12ML2/12 = ML2 (The mass of the rod in this problem is given as 12M to make the terms simple).

Moment of inertia of the portion CB of mass 3M and length (L/4) is (by applying the parallel axis theorem) 3M(L/4)2/12 + 3M(3L/8)2 = (7/16)ML2

[The first term is the moment of inertia of portion CB about an axis through its centre and the second term is the product of its mass and the square of the distance between the two parallel axes].

Therefore, moment of inertia of the remaining portion AC = ML2(7/16)ML2 = (9/16)ML2 = (9/16)I.

[Alternatively, you can treat the portion AC of the rod to be made of the portions AO and OC and add their moments of inertia about O to arrive at the answer, remembering that the moment of inertia of a thin rod of mass ‘m’ and length ‘l’ about a normal axis through one end is ml2/3. The moment of inertia of the portion AC of the rod is therefore 6M(L/2)2/3 + 3M(L/4)2/3 = ML2/2 + ML2/16 = (9/16)ML2 = (9/16)I].

Wednesday, June 13, 2007

IIT-JEE 2007 Assertion-Reason Type MCQ on Elastic Collisions

The following simple Assertion-Reason Type multiple choice question which appeared in IIT-JEE 2007 question paper is quite simple and is meant for checking whether you have mastered the fundamentals of elastic and inelastic collisions:

STATEMENT-1

In an elastic collision between two bodies, the relative speed of the bodies after collision is equal to the relative speed before the collision

because

Statement-2

In an elastic collision, the linear momentum of the system is conserved.

(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

The important points you have to remember here are:

(i) momentum is conserved in elastic as well as inelastic collisions

(i) in elastic collisions, kinetic energy also is conserved.

[Remember, we are talking of kinetic energy. Note that the total energy is conserved in all collisions in accordance with the law of conservation of energy].

In an elastic collision, the relative speed of the bodies after collision is equal to the relative speed before the collision and this is a consequence of the conservation of kinetic energy.

Therefore, statement-1 is true. Statement-2 also is true, but it is not the reason for statement-1. So, the correct option is (b).

[Note that the ratio of the relative speed of the bodies after collision to the relative speed before collision is called the coefficient of restitution, e. The value of ‘e’ lies between 0 and 1. For perfectly elastic collisions, e =1 and for perfectly inelastic collisions, e = 0]

Sunday, June 10, 2007

Force, Momentum and Circular Motion

You might have noted that the angle between the force acting on a body and the momentum of the body can be anything between zero and 2π. (But the change of momentum is always in the direction of force). Here is a simple question which you should be able to answer in a minute. If you find it difficult you should work harder to understand basic points in mechanics thoroughly.

A particle moves in a plane such that the rectangular components of its momentum vary simple harmonically with the same period and amplitude, but with a constant phase difference of π/2. The angle (in radian) between the momentum of the particle and the force acting on it is

(a) π (b) π (c) zero (d) π/2 (e) varying between zero and 2π

The particle is forced to move simple harmonically along two mutually perpendicular directions. In other words, this is a case of the superposition of two simple harmonic motions of the same perod and amplitude at right angles to each other. Since the phase difference is π/2, the resultant motion is uniform circular motion. The angle between the resultant momentum of the particle and the force acting on the particle is therefore π/2.

Now consider the following MCQ which appeared in IIT-JEE 2007 question paper:

A particle moves in the X-Y plane under the influence of a force such that its linear momentum is p(t) = A[î cos(kt) – ĵ sin(kt)], where A and k are constants. The angle between the force and the momentum is

(a) 0º (b) 30º (c) 45º (d) 90º

Simply by noting that the momentum vector p has simple harmonically varying components A cos(kt) and A sin(kt) in the X and Y directions respectively, you can conclude that this is a case of the superposition two simple harmonic motions of the same frequency and amplitude at right angles, with a constant phase difference of π/2. [The phase difference is π/2 since one is a sine function while the other is a cosine function].

So, this is uniform circular motion and the angle between momentum and force, as usual, is π/2.

Now, if you want to give a quantitative proof, you may proceed as follows:

The time rate of change of momentum is force. Therefore, force

F = dp/dt = –Ak[î sin(kt) + ĵ cos(kt)]

The angle θ between the force vector F and the momentum vector p is given by

Cosθ = F.p/ Fp, where the bold face letters represent vectors.

But F.p = –Ak[î sin(kt) + ĵ cos(kt)] . A[î cos(kt) – ĵ sin(kt)]

= –A2k sin(kt)cos(kt) + A2k cos(kt)sin(kt) = 0

Since cosθ = 0, θ = π/2.

Friday, June 08, 2007

IIT-JEE 2007 Linked Comprehension Type MCQ on Optics

Physics is a subject with great scope for setting questions with numerous variations conforming to the prescribed syllabus. Often the questions may look lengthy and too much descriptive. Don’t be scared by lengthy questions. Most of them will be simple as for instance, the following linked comprehension type multiple choice questions which appeared in IIT-JEE 2007 question paper:

Paragraph for Question Nos. 1 to 3:

The figure shows a surface XY separating two transparent media, medium-1 and medium-2. The lines ab and cd represent wavefronts of a light wave traveling in medium-1 and incident on XY. The lines ef and gh represent wavefronts of the light wave in medium-2 after refraction.

(1) Light travels as a

(a) parallel beam in each medium

(b) convergent beam in each medium

(c) divergent beam in each medium

(d) divergent beam in one medium convergent beam in the other medium

Light rays have to be perpendicular to the wave front. Since the wave fronts are shown as parallel, the rays are parallel. So, the correct option is (a).

(2) The phases of the light waves at c, d, e and f are Фc, Фd, Фe and Фf respectively.

It is given that ФcФf .

(a) Фc cannot be equal to Фd

(b) Фd can be equal to Фe

(c) (Фd Фf ) is equal to (Фc Фe)

(d) (Фd Фc ) is not equal to (Фf Фe)

All points on a wave front will have the same phase. The lines cd and ef represent the wave fronts of the same light wave at two instants. Therefore, points ‘c’ and ‘d’ have the same phase. Points ‘e’ and ‘f’ also have the same phase, but these points have the same phase lag with respect to points ‘c’ and ‘d’. So, the correct option is (c).

(3) Speed of light is

(a) the same in medium -1 and medium -2

(b) larger in medium -1 than in medium -2

(c) larger in medium -2 than in medium -1

(d) different at b and d

The angle of refraction is evidently less than the angle of incidence at XY and hence the speed of light is larger in medium-1 [Option (b)].

Wednesday, June 06, 2007

Mobile Phones for Internet Browsing & Blogging

Mobile phones with multimedia and internet browsing capabilities are becoming cheaper and this is widening the possibilities of blogs for educational purposes. Youngsters find it very convenient to use mobile phones for internet browsing and already a large percent of youngsters prefer mobile phones to personal computers in many parts of the world. For students and youngsters who restrict this facility to healthy and judicious use, there is ample scope for self development. The teaching-learning process will get a boost and the teachers as well as students will find the new generation mobile phone an important tool in their academic activities.

Monday, June 04, 2007

MCQ from Electrostatics- Charged Spherical Shells

Two thin concentric spherical shells of copper having radii R1 and R2 (R2 > R1) carry charges Q1 and Q2 respectively. The potential at a point P (Fig) distant ‘r’ from the common centre and located in between the shells is

(a) (1/4πε0)(Q1/r)

(b) (1/4πε0)(Q1 + Q2)/r

(c) (1/4πε0)[(Q1/R1)+ (Q2/R2)]

(d) (1/4πε0)[(Q1/R1) + (Q2/r)]

(e) (1/4πε0)[(Q1/r) + (Q2/R2)]

Potentials due to the two charged shells get added to give the net potential at P. The potential at P due to the smaller shell of radius R1 is (1/4πε0)(Q1/r) since the point P is outside the shell at distance ‘r’ from its centre. The potential at P due to the larger shell of radius R2 is (1/4πε0)(Q2/R2) since the point P is inside this shell and the potential everywhere inside is the same as the potential at the surface of the shell. The net potential at P is obtained by adding these two potentials.

Therefore, the net potential at P = (1/4πε0)[(Q1/r) + (Q2/R2)]

[This question can be modified by making one of the charges or both charges negative. You will then have to substitute the charges with the proper sign in the above equation].

What is the electric potential at a point inside the smaller shell?

The point is then inside both shells and both produce constant potentials (1/4πε0)(Q1/R1) and (1/4πε0)(Q2/R2) respectively. The net potential is therefore (1/4πε0)[(Q1/R1)+ (Q2/R2)]

What is the electric potential at a point outside the larger shell?

The point is then outside both shells and they produce potentials (1/4πε0)[(Q1/r) and (1/4πε0)[(Q2/r) respectively so that the net potential is (1/4πε0)(Q1 + Q2)/r. This is a very simple case since you can treat the entire charge Q1 + Q2 to be located at the common centre of the shells.

Now, one more question: What is the electric field at a point P in between the two shells?

The larger shell does not produce any field at P since it is inside and there is no potential gradient inside. The field at P is therefore due to the smaller shell only and is equal to (1/4πε0)[(Q1/r2)

You can find more posts on electrostatics by clicking on the label 'electrostatics' below this post or by performing a search for 'electrostatics' making use of the blog search box on this page.

Saturday, June 02, 2007

Two Multiple Choice Questions on Vectors

A methane molecule CH4 may be fitted in a cube of side 2a such that the C atom lies at the body centre and four H-atoms at non-adjacent corners of the cube as shown in figure. The angle between any two C-H bonds is
(a) 120º
(b) cos–1(1/3)
(c) 150º
(d) sin–1(1/3)
(e) cos–1(1/3)
This MCQ appeared in Kerala Engineering Entrance 2006 question paper. You are required to find the angle between the lines CH1 and CH2. Treat these lines as vectors. The coordinate axes are marked (by the question setter) in the figure with H3 as the origin.
The vectors CH1 and CH2 have X, Y and Z components of the same magnitude ‘a’ since the carbon atom C is at the centre of the cube of side 2a. These vectors can be written as
CH1 = a i +a j +a k
CH2 = +a i a j +a k
[Note that bold face letters represent vectors and i, j, k are unit vectors in the X, Y and Z directions respectively].
The magnitudes CH1 and CH2 of these vectors are the same, equal to a√3.
The angle between the vectors CH1 and CH2 is given by
Cos(CH1, CH2) = (CH1 . CH2)/ (CH1)(CH2) = (a2 a2 +a2)/ (a√3)( a√3)
= –(1/3)
The angle between the vectors, which is the bond angle required, is therefore cos–1(1/3).
Now, consider the following question. This is a simple one, but similar questions are often seen in entrance test papers.
Angle (in rad) made by the vector j + √3 k with the Y-axis is
(a) π/8
(b) π/6
(c) π/4
(d) π/3
(e) π/2
You should remember that the direction cosines (the cosines of the angles between coordinate axes and   the vector itself) are given by Cos(A,x) = Ax/A, Cos(A,y) = Ay/A and Cos(A,z) = Az/A.
Since the Y-component has magnitude 1 and the vector has magnitude 2, the cosine of the angle between the Y-axis and the given vector is ½. The angle therefore is π/3.

Friday, June 01, 2007

EAMCET 2005 Question on Friction

The following MCQ appeared in EAMCET 2005 question paper:

A cubical block of mass ‘m’ rests on a rough inclined surface. ‘μ’ is the coefficient of static friction between the block and the surface. A force mg acting on the cube at an angle ‘θ’ with the vertical side of the cube pulls the block. If the block is to be pulled along the surface, then the value of cot(θ/2) is

(a) less than μ (b) greater than μ (c) equal to μ (d) not dependent on μ

The pulling force mg applied at an angle θ with the vertical can be resolved into a vertical (upward) component mgcosθ and a horizontal component mgsinθ as shown in the figure. It is the horizontal component which moves the block along the surface.

But the vertical component is effective in reducing the frictional force since it reduces the normal force from mg to (mg – mg cosθ) so that the frictional force is μ(mg – mg cosθ) = μmg(1cosθ). So, the condition for the block to move along the surface is

mg sinθ > μmg(1cosθ).

Or, sinθ > μ(1cosθ).

But, sinθ = 2 sin(θ/2)cos(θ/2) and (1cosθ) = 2sin2(θ/2).

Substituting these in the above condition, we obtain

cos(θ/2) > μ sin(θ/2), from which cot(θ/2) > μ

Do you know why it is easier to pull a lawn mower than to push it?

You have seen in the above problem that the upward component of the pulling force opposes the weight of the block and therefore reduces the normal force (N) on the surface. The frictional force μN is therefore reduced. If a pushing force is applied to move the block, the vertical component of the pushing force will be downwards and it will add with the weight of the block to produce a greater normal force and therefore a greater frictional force.