User:Thewinster/sandbox

Inductors in parallel (1)
Name the three dots as l, m and n. Redraw the circuit using these three points as reference and you’ll find that the three inductors have their ends joined to the same potential difference. Hence the effective inductance in between A and D is [1/3 + 1/3 + 1/3]-1 = 1 H.

This is a very common circuit diagram to represent elements in parallel. Please keep it in mind. What if we replace the inductors with three capacitors of 3μ each? What would be the equivalent capacitance between the points A and D?

Projectile (2)
Relate everything with the speed. Assume the speed in the beginning to be the V.

Then, $$\frac{mV^2}{2}=E$$. When the ball is at its highest point, its vertical velocity would have decreased to zero (otherwise, won’t it go a little bit higher? Think about it. As a food for thought, after the highest point, the velocity changes its direction and its magnitude increases. No wonder the projectile descends after the highest point!)

Then, at the highest point of the trajectory the speed of the projectile would be its unchanged horizontal velocity - $$\frac{V}{\sqrt2}$$. Find the kinetic energy using this as the speed of the ball. It will be half of E.

Falling from a height (3)
There is one thing that you need to agree with, the direction of velocity vector of both the balls is the same. If their speeds on reaching the ground are same as well, then we know the answer would be B.

Use the equation of motion $$v^2=u^2 + 2as$$. The balls have the same value of initial speed, same acceleration and most importantly – the same displacement. Yes, s is nothing but displacement and we know that whenever they reach the ground their displacements will be same. We happily conclude that their final speeds would be the same as well.

Speed same, direction of velocity same. This implies their velocities on reaching the ground are same.

By the way, will the time required to reach the ground will be the same for them? You may put forward worded arguments or use equations.

Bullet in a block (4)
This question requires a little assumption that the acceleration of the bullet is the same throughout its motion. Then, we use the first data to find the acceleration.

We know the initial and final speeds, we know the displacement. So, to find the acceleration $$v^2=u^2 +2as$$ will do. We get $$a=\frac{-v^2}{8}$$, the negative sign signifying that acceleration is opposite in direction than velocity.

Use the equation again, $$0=\left(\frac{v}{2}\right)^2 - \frac{2sv^2}{8}$$ the further displacement require to stop the bullet comes out to be 1 cm.

Satellite (5)
The centripetal force required to maintain circular motion is provided by the gravitational force. When a satellite moves around in an orbit its velocity constantly changes. As soon as the force disappears, there is nothing that can alter the instantaneous velocity and so the object continues to move with that velocity. [Newton’s WHAT law? Quick. I give you 5 seconds.]

Ammeter/Voltmeter (6)
Remember that an ammeter has a very low resistance [Why?]. When we use it in place of a voltmeter, the meter will move because the readings are based on deflection. The problem is, for being a nearly ideal voltmeter [one that doesn’t alter the circuit being measured] we need the voltmeter to have a very high resistance [Why?].

Done, we connect the ammeter in parallel to the points of the circuit whose potential difference is to be measured. But in order that the parallel circuit [containing the ammeter] behaves as a voltmeter, we need it to have a high resistance. Kya karein? Just connect a high resistance to the ammeter in series. As simple as that!

So, what was all that special about the question? Very special! This is a type of question which most people will attempt and get it wrong. Remember that we are converting an ammeter into a voltmeter. Generally, we convert a galvanometer into an ammeter and a galvanometer into voltmeter. Expect most study-by-rots to tick the procedure corresponding to any one of this without giving it much of a thought.

Ratio of magnetic fields (7)
$$\frac{\frac{\mu_o\left(i\right)}{2\pi\left(2R\right)}}{\frac{\mu_o\left(2i\right)}{ 2\pi\left(R\right)}} = 1$$ Of course, the number of turns of both the coil is one. Even that may have been mentioned to make this a probable candidate for negative marks.

Just keep one thing in mind, whenever solving such questions draw one big line and write down the fraction. Then evaluate it with ease. Otherwise, there are chances of messing up. Its just a personal advice though!

Plane mirrors (8)
360/n - 1 = 5. This is a pure formulae based problem. Not in your syllabi? Think again! We studied it way back around sixth or seventh standard.

Power dissipation (9)
Go by the same procedure. We will calculate the power dissipation using $$P = \frac{V^2}{R}$$. Then, since we have to calculate the ratio, we write down the ratio of the two quantities and watch everything get cancelled off. I have provided this solution specifically to point out how lengthy questions can be shortened. It will take some time in realizing what I have written, but it is only to enable you to solve the question by yourself! $$\left( \frac{1}{2}\frac{V^2\left(\frac{l}{2}\right)}{\rho A} \right) \frac{1}{\frac{\rho A}{V^2 l}} = 4$$


 * When the pieces were cut into two, the resistances of the wire changed to half of the original one. We calculated this using PL/A.
 * When the pieces were connected in parallel, their equivalent resistance was half of their individual resistance. [How?]

Note how swiftly we solved this problem! A typical suicidal variation would be to find the current through each circuit, then square it and multiply it with resistance in both cases. Even worse, you might waste your time in calculating the power dissipation across individual resistances and then summing them up in the second case.

Hydrogen atom (10)
One fourth of the total energy of the first orbit, i.e. -3.4 ev is the energy of the second orbit. If we give the electron 3.4 ev of energy, it’ll leave the atom, ionizing it.

The 13.6 ev KE data is needed to know that the energy of the first orbit is -13.6 ev. [Why? What’s the relation between TE and KE?]

Fundamental Frequencies (11)
Direct question on waves, $$\frac{\frac{v}{2l}}{\frac{v}{4l}} = 2$$

For practice: - What are the fundamental wavelengths of the open organ pipe and closed organ pipe? What happens to the wavelengths as the frequency of the organ pipes increase? What series do they form?

Also, this is a question that every deserving candidate will solve. Please, please! Get it right. Also, if you solve the question and get the answer to be 1/2 DO NOT consolidate yourself! You messed up something very obvious!

Beats (12)
The tuning fork produces 4 beats/sec with another fork of frequency of 288 cps. The possible values of the frequencies of the tuning fork are 284 and 292 cps.

When the tuning fork is loaded with wax, its frequency decreases and so does the difference between the frequencies of the two forks. This eliminates 284 cps and we conclude that the answer is 292 cps.

By the way, beats/sec is in the unit of Hertz. Is Cycles/sec in the same unit? Ofcourse it is, but convince yourself that there is no need of a conversion factor of 2π in it!

Wave equation (13)
To satisfy the given conditions,
 * The wave must be traveling in the opposite direction. (Negative X Axis) That is, the wave equation must be a function of ωt + kx.
 * At x=0 the sum of the wave equations must amount to zero for all values of time.

Potential (14)
$$Work = \Delta V$$

20x = 2J which gives the potential difference between the two points to be 0.1 V. Just recall the definition of potential difference! We do not need anything else.

Circular motion in a magnetic field (15)
Magnetic force [qvb here] is the source of centripetal force [what’s the equation?]. Equating them, we get the expression for the radius of curvature. [So that you do not need to rot the expression.]

$$r = \frac{mv}{qb}$$ This equation is enough to deduce that the radius of curvature of both the particles will be same. [Note the numerator.]

By the way, because the options mention it, the particle with the greater radius will have a less curved path. [Draw two circles of different radii and check by yourself.]

Gravitational Potential (16)
The gravitational potential difference between the two points is GM/6R [positive, why?] [What is the potential energy difference?] The work done then is nothing but m times the potential difference. [Opposite in sign to the change in potential energy, why?]

Spring Constant (17)
T is inversely proportional to the square root of k. [Go back to SHM $$T = 2\pi\sqrt\left(\frac{m}{k}\right)$$] K in the second case will be n times that of the first. [I’ll skip the explanation, but recall F/A = Y (∆l/L). From here, we can say that YA/L is the spring constant, since length has decreased by n times the spring constant has increased by n times.]

Using this, we say that the time period of an individual piece would be T/√n.

Flux (18)
The face BCFG is symmetric with respect to the two charges. Use your existing knowledge of vectors to deduce that the electric field at any point on the face would be perpendicular to the local area vector. [Do not forget the various derivations that use this step, including the deduction that electric field near an infinite planar sheet is perpendicular to it. All of this is elementary physics and discussed one too many times in all textbooks.]

...and a word of caution (based on personal experiences! Go ahead, take it! Its meant to be a friendly advice!), if you cannot satisfactorily explain such trivial points in words and diagrams then either work harder or give up the hope of scoring 95 percent in your CBSE exams. They require a harmony in your studies, fragmented knowledge will NOT help in the 95% goal!

That’s it than, the integration of E.ds will be zero over the face. Note that we did not use Gauss’ Law to calculate the flux. It cannot be used anyways, becaue the face is not a closed surface!

Power dissipation, again? (19)
$$\frac{15^2}{\frac{2R}{2+R}} = 150 $$ which gives the value of R to be 6 ohms fairly easily. Infact, if you are in haste – you might skip solving the equation once you make it. Just check among the options for the one that satisfies it [Only if you’re quick at mental calculation!]

Resolving power (20)
I am not sure if this is still a part of the standard curriculum, but here’s the solution. The resolving power would be inversely proportional to the wavelength of the light used.

Center of mass (21)
Considering the velocity of the first particle positive... $$v_\left(cm\right) = \frac{m\left(2v\right) - m\left(v\right)}{m+m} = v/2$$

i.e. V/2 in the direction of the first particle. Anyways, it is obvious that the motion will be along the direction of the faster moving particle out of the two IDENTICAL ones.

Current and spring (22)
When a current is passed through the spring, its turns will act as parallel wires. These parallel wires will attract each other [Why?] and hence the spring will be compressed!

Heat capacity (23)
Based on definition.
 * Heat capacity is nothing but the constant of proportionality in the relation Q is proportional to ΔT. Heat capacity in turn depends upon the mass (or more generally, the amount of a substance).
 * Specific heat is then the constant of proportionality between heat capacity and mass (Specific heat capacity for mass) or between heat capacity and moles (molar specific heat capacity.

Silicon at absolute zero (24)
It is factual; however there is an explanation for it in our course! It is related to the energy band gap and thermal energy of electrons.

Nature of EM waves (25)
Standard textbook stuff – the student is obliged to know it.

Optical fibers (26)
Again an obligatory question! Basically, what such easy questions do is to filter out insincere students. Otherwise, questions based on logical understanding cannot distinguish between students who have studied whatever they are obliged to (although they might not have gone further) and those who have been reckless.

Escape velocity (27)
Search for the expression of escape velocity in your textbook. It depends solely on the mass on the planet and not on the mass of the body which wants to escape. [Derive the expression and write two convincing lines why the mass of the body doesn’t feature in it.]

EM Waves (28)
Beta rays are streams of energetic electrons, so even if you do not know what cosmic rays are – it will not matter! Again we encounter an obligatory question. By the way, cosmic rays are an interesting topic for many. Do consider reading about them online if you are inclined towards such knowledge. And anyways, we have atleast heard of them once in our normal course – in carbon dating. [What exactly is their part over there?]

Seebeck Effect (30)
This topic is now out of the CBSE syllabi, but the relation they have asked is standard stuff – something which is asked in typical Board exam objective questions.

Anyways, here’s a little explanation. The graph of the thermo-emf and temperature of the hot junction is a downward parabola (what is it’s equation?) with two roots. First is when the two junctions are at the same temperature (obvious!!) and the other is at the inversion temperature. From the geometry of a parabola we know that there will be a maxima in between the two zeroes (! – neutral temperature) and from symmetry we know that the point where an extremum is obtained (in all standard parabolas with a displaced vertex – we are not talking of parabolas which have been inclined to the axis) is the middle point of the two zeroes. (Why? Convert the equation of the parabola in the sum of a perfect square and constant and gawk at it, you’ll know.)

Half life (32)
Three half lives have been spent and in each half-life the initial amount will be reduced to half (irrespective of its initial concentration.) Again, it is obligatory for a good student (good as in a mix of sincerity and aptitude, not excellent or hardworking!) to solve this question.

Specific Resistance (33)
We are talking of specific resistance (and not resistance of the sample). The question is again straight forward! Instead of testing your application or understanding, it just checks if you know this fact or not.

A little note - In examination conditions we find it difficult to recall a trivial point. Once we encounter this – we spend the next few minutes thinking about it and mentally juggling between two options. Maximum amount of time the student will mark the correct option and then change it to the incorrect one. To avoid this kind of a situation, all we can do is to study intelligently. For example, in this particular case, the reason why this happens has been explained in almost all board textbooks. If incase we forget this point as a fact, we can arrive at these conclusions by mentally reconstructing the logic we have read in our course books. [Semiconductors - something related to thermal energy and band gap, conductors - ??? I’d leave it on the student to convince themselves by putting the above hints into a proper format.]

Capacitors in parallel (34)
$$Energy = \frac{1}{2}C_\left(equivalent\right)V^2$$

Reversible cycles (36)
The answer can of course be found in your textbook. The only reason why you might not mark option A would be (I suppose) a confusion between some of the other options. You can check them in your textbook one by one (that is because this question, again, is obligatory! I’m emphasizing this again and again so that it can be seen that if you do not get it right, you’ll drag yourself behind others by a big margin! When you encounter such questions, give them due respect and try not to mark them wrongly!).

Here’s a little question – why do irreversible cycles have a lesser efficiency than reversible cycles? [Relate it with the amount of work done and the like...]

Checkpoint
If you notice, all questions since the early 20s are easy and every student who scores decent at school is expected to solve them! I say this to emphasize that whatever is asked in competitive exams is basically the same thing that we learn in school! Many people just do not realize this and labor unintelligently with stupid books, [The IIT-JEE is an exception only because it requires specific preparation. Otherwise the syllabus is much less than a typical board exam!]

Do you study by rote?
Also, if you are struggling with these questions even though you score decently, think once. Either what you think is decent is not decent enough or you have been scoring by rote instead of learning. It is common knowledge that in many places (like Gujarat) the students are made to write down the solution of even math problems a number of times so that they can memorize them! The GujCET is again notorious for lifting questions directly from the textbook (keeping the numerical values, statements and even the order of alternatives intact!). No wonder even teachers advise to memorize the answers to the objective questions. If you have been using this loophole to score well, don’t be surprised if you fail here. Extremely examination oriented atmosphere or incompetent teachers might be the cause, though I pay my due respect to the teachers who work hard but never get competent students!

Do you ever think while studying?
Finally, if you do not belong to any of the above two categories, are doing well and still struggled with the past few Qs – sit down and think. Have you ever practiced writing down on paper the descriptive questions? Have you ever avoided doing a problem or giving a verbal explanation asked in the textbook? Have you ever excused yourself from practicing by saying that if it is asked in the examination you’ll do it just because it is simple? Be wise. If you do something for the first time directly in an examination, either your answer will not be technically sound, or you’ll make some silly mistake. Worse still – you’ll waste some time and not attempt it at all.

Give it some thought. What was the use of reading something and giving it time when you couldn’t answer something as simple as 2+2=4 in the examination? Come out of your comfort zone my friends! Solving problems and writing practice is a must for every student aiming a decent score. More so if you dream of scoring 95% in your board exams. Its relatively easy and requires nothing but common sense and sincerity. So, if you really aim 95% or even 90+ please be sincere enough. Practicing and getting out of your comfort zone is more important than laboring. You don’t need to be a genius for that. I say this based on my observation with me and my friends. If you still don’t catch the warning, God bless you!

Standing wave (37)
Either use the expression for frequency (Must!) or Think! (optional). The two rigid ends will be nodes. The distance between two nodes is half the wavelength. The more the number of nodes in between, the lesser will be the wavelength. (Common sense, check and see if you can draw a diagram to convince yourself! Remember the warning that preceded!). The wavelength will be maximum when there are no nodes between the two ends. In that case, the length of the string is half the wavelength. (ends are nodes!) In other words, the maximum wavelength will be double the length of the string.

Power factor (38)
You know tanφ, (- Xl/R) and you have to find cosφ. What is the big deal? Cosφ = (secφ)-1 = (1 + tan2φ)-0.5. This last manipulation is just a brain teaser, nothing that you haven’t learned.

Escape energy (40)
This paper is turning out to be quite simple! There are two logics: -
 * The kinetic energy required would be equal to the potential energy difference at the surface and at infinity, GMMe/R or mgR.
 * The kinetic energy required would correspond to the energy of a mass moving with escape speed.

Kindly note that the first logic was used to derive the expression for the escape speed. Using the expression of the escape speed to find energy would be clumsy in a textbook. Also, GMe/R = gR (Does anyone need an explanation?)

Temperature and gas (41)
It’s a good question. The point here is that the speed of molecules is related to their temperatures. When the gas container moves, the gas particles acquire extra speed. This is the source of the confusion; does it increase the temperature of the gas?

Go back to the construction used to derive the relation between the kinetic energy and the temperature. The movement of the container walls was not mentioned at all! The change in momentum (and hence force) was with respect to the container. This resolves the confusion. There is no change in the temperature of the gas. (I’m not sure if everyone will even have this confusion – let alone resolving it!)

The above conclusion will seem more obvious when you think about it. If it were not true, wouldn’t the temperature of gas would be different in the frame of the earth and the moon even though we measure them the same time? Think about two parallel running buses and a stationary observer when a gas cooker is placed in one of the buses.

Mass energy equivalence (43)
When water is cooled to form ice, energy is given out by the system. Does it mean that the released energy was due to the disintegration of mass into energy according to E=MC2? Certainly not! It’s simply an enthalpy change! The pity is, some solutions that I have referred to say that this release of energy was due to loss of mass of water. Some even say that since water loses its energy, its mass will have to increase to maintain the equivalence! Please use your understanding and stay immune to such stupidity!

SHM (46)
At the mean position, the acceleration is zero and the velocity is maximum, i.e. KE - max, PE - min. (Qualitative nature of SHM). However, do not be surprised if you mark KE min and PE max. Such confusions happen! :)

Angular momentum (47)
This is again a direct question based on conservation of angular momentum. We do not expect people who have been studying in pieces to solve a numerical problem, but if you have worked hard enough for a Board Exam (specially the GSEB) get it right to maintain your dignity! We should not falter on direct questions! Here is the equation, solve ahead yourself!

$$\frac{MR^2\omega_1}{2} = \left(\frac{MR^2}{2}+mR^2+mR^2\right)\omega_2$$

Skidding (48)
When the car moves around the track, it needs a centripetal force to allow the circular path. $$\frac{mV^2}{R} = f_s$$, the RHS being the force of friction. Since the frictional force can have a range of values from 0 to maximum µN, the speed of the car too has a range. After all, the idea of driving at a fixed speed around a track doesn’t seem observable in nature! Now, when the frictional force reaches its maximum value, we cannot increase the speed further. If we do, the centripetal force will not be enough to maintain circular motion. At that critical speed, $$\frac{mV^2_\left(max\right)}{R} = \mu mg$$, giving Vmax = 30 m/s.

At this stage you might want to revise the case of a car moving on a banked road.

Velocity of efflux (49)
$$V =\sqrt 2gh= 20 m/s$$. As a practice exercise, derive this expression using Bernoulli’s theorem and equation of continuity.

Spring Potential Energy (50)
This is a good question. It’s a seen question, but nevertheless a good one. The spring potential energy is 1/2kx2. Here x is not the change in extension/compression! It is the absolute value of change in length from normal length!

The work done would be the extra potential energy stored. What would it be? 1/2k(15-5)2 or 1/2 k(152 – 52)? C’mon! Atleast make so much of a decision!

Or be spoon fed. Conserve energy. Initial energy + Work Done = Final Energy. $$\frac{k5^2}{2} + W = \frac{k15^2}{2}$$

Pendulum (51)
This is the case of a physical pendulum, which is a pendulum with a finite size. In the expression $$T = 2\pi\sqrt\frac{l}{g}$$ l is the distance of the center of gravity (which is practically the center of mass in this case, but not always!) from the point of suspension. (How did we arrive at this simplification?) When the child stands up, the center of gravity rises up (why? Obvious, but don’t take it factually.) and so the distance from point of suspension decreases.

The time period decreases. If the angular amplitude remains same, this will mean that the speed of swinging has increased. You can observe this yourself. Sit on a swing, swing and raise your hands! (You don’t need to stand up - your aim is simply to raise your center of mass!)

Food for thought: - If center of mass and gravity rise and the angular displacement remains same, doesn’t it mean the maximum potential energy has increased? Let us put it this way, the speed of the system at mean position has increased. Where does the extra kinetic energy/potential energy come from? Hint – stand on a swing, sit down and repeat this a hundred times. Do you feel hungry?

Lift (52)
A man standing on the ground will observe the acceleration to be g. Why did the motion of the lift have no effect on the acceleration? Well, the motion of the lift has got nothing to do with the net force on the ball. What about the man inside the lift? Either add the two accelerations according to vector rules (relative motion) or look at it this way. Since the man is inside a non inertial frame, we apply a pseudo force on the ball in direction opposite to motion - that is the upward direction. Acceleration of the ball, considering downward positive will be (mg - ma)/m = g-a downward. I leave the vector addition pathway for practice, but do not be intimidated by the word vectors! Why! You could even find the answer using normal visualization!

RMS Velocities (54)
$$\sqrt \frac{3RT_H}{M_H} = \sqrt \frac{3RT_O}{M_O}$$

Here the subscripts H and O represent hydrogen and oxygen gases respectively. Fix the values and get the answer.

There’s one problem if instead of hydrogen you were asked about helium. The molecular mass of helium is 0.004 kg/mol. That of oxygen is 0.032 kg/mol, not 0.016! In the case of hydrogen/oxygen, you might commit the same mistake of substituting atomic mass instead of molar mass twice and benefit, but not here! Also note the SI values of molar mass! Also, we need to be careful; the values of temperature to be substituted are not in Celsius but in Kelvin! To wrap it up I’d say that after a short break, we encounter one more must solve question!

Time period in magnetic field (55)
A popular question and in fact some people treat it as a fact! Here’s the math, explanation of radius can be found in the solution to Q 15.$$T = \frac{2\pi R}{V} = \frac{2\pi\left(\frac{mV}{qB}\right)}{V}$$ The time period is then independent of the speed. This fact is the working principle of the cyclotron.

Rigid bodies down a plane (56)
Note that the bodies are not rolling because there is absolutely no friction. As if that hint wasn’t enough the paper setter explicitly mentioned this! Their acceleration is then simply component of g along the plane (gsinθ). Incase of rolling, we’d require the moment of inertias and the radii of the bodies to compare. As revision, search for one such problem and solve!

Carnot engine (58)
Efficiency = 1-T2/T1. For efficiency to be 1, T2 has to be zero, which is not possible practically. Even if we achieve that temperature, it might not be feasible to operate a carnot engine. For practice, convince yourself why the other three alternatives are wrong.

Just on a side note, the NCERT textbook mentions that the pronunciation of <I>carnot</I> isn’t <I>Kar Not</I>. (Well, that is the place where I read it first, not that its the only place!) You might want to look it up for the actual pronunciation. In fact, even <I>de Broglie</I> isn’t pronounced as <I>Dee Brog Lee</I>. Watch out for it too!

Ring and its MI (59)
Note that the moment of inertia that has been asked is about a diameter, not about the axis perpendicular to its plane! This is the first mistake that people might make.

Use the perpendicular axis theorem. Let Ix and Iy be the moment of inertias about two perpendicular diameters, both (why both?) equal to X (that’s what we have to find). The equation is then, $$I_x + I_y = I_z$$ i.e. $$2x = MR^2$$.

Force and acceleration (60)
A little analysis is required. If the particle stays in equilibrium under the three forces, doesn’t it mean that F1 is opposite in direction but equal in magnitude to the resultant of F2 and F3? When the force F1 is removed, the resultant force on the particle is the resultant of F2 and F3, which we have seen is equal in magnitude to F1! For practice, arrive at this conclusion yourself by drawing rough vector diagrams. Most people have a phobia for vectors. Just keep in mind that vectors seemed tough only because they were introduced at a transition phase - tenth to eleventh!

Force vectors (61)
I’ll put forward two solutions.

$$A + B = 18$$ - (1), $$12 = \sqrt\left(A^2 + B^2 + 2AB\cos\theta\right)$$ - (2), $$\tan \frac{\pi}{2} = \frac{B\sin\theta}{A+B\Cos\theta}$$ - (3)

Solve these three equations. Here B is the magnitude of the smaller vector. Seems difficult, but its simple. To ease your work in objective exams, just eliminate θ and then plug in values of A and B from the alternatives to see which one is correct.

Alternatively, if the resultant is perpendicular to the smaller vector, it means that the component of A along B is antiparallel to B and equal in magnitude. Acosθ = B. The resultant is then simply the component of A perpendicular to B. Asinθ = 12. Eliminate θ from the two equations by $$\left(\frac{B}{A}\right)^2 + \left(\frac{12}{A}\right)^2 = 1$$ Finally, use A + B = 18

It’s a seen question nevertheless, so I don’t think it requires specific preparation for you to solve this Q in AIEEE except revision! Of course people choose such preparatory classes because of incompetent teachers in school or teachers teaching only at their private classes. But choosing slogging for two years when you can enjoy studying in a good school (if it is available) along with all the pleasures of the school life would be a decision worth contemplating. (Don’t forget that the pleasures include everything from diving under the bench to drown laughter to gearing up for school events to freedom to bunking English classes to gossiping about idiots planning a marriage life with one of your classmates to infinity!) Hostel based coaching classes have their own advantages – but they materialize only for people who are very hardworking or have a formidable level of intellect. However, ultimately this remains my own personal opinion!

Identical Cars (62)
Identical cars imply that the two cars have the same acceleration when braked. To relate speed, displacement and acceleration use V2 = u2 + 2as with a and s having opposite signs. $$\frac{u^2}{2s_1} = \frac{\left(4u\right)^2}{2s_2}$$<BR> Now find the ratio.

Adiabatic Exponent (63)
The adiabatic exponent of the mixture will be defined as the ratio of Cp and Cv for the mixture. How are the two heat capacities defined? The heat required to raise the temperature of one mole of the sample by one degree at constant volume would be this quantity. $$C_v = \frac{\left(1\right) \left(1\right)\left(\frac{3R}{2}\right) + \left(1\right) \left(1\right) \left(\frac{5R}{2}\right)}{2}$$ <BR> Similarly,<BR> $$C_p = \frac{\left(1\right) \left(1\right)\left(\frac{5R}{2}\right) + \left(1\right) \left(1\right) \left(\frac{7R}{2}\right)}{2}$$

What we have done in the two cases is to write the total heat required to raise the temperature of the two gases by using their heat capacities differently and then divide it by total number of moles so that AVERAGE HEAT CAPACITY PER MOLE of the mixture is obtained.

Charges in equilibrium (64)
This question is a simplified version of a very popular question. Now, we know that the central charge q would be in equilibrium for any value of q. [Electric field is zero in that position, check and see.) The problem here is, we haven’t fixed the other two charges. The force on both of them should be zero for the whole system to be in equilibrium. There’s nothing but the coulomb’s law to be used. Diagram here is not needed.

The charge q should be opposite in sign to Q. This is because the two charges at the end repel each other. The duty of the charge q is to balance this force to zero by attracting both the Qs. Our equation here is to balance the force on any one of the two charges. $$\frac{kQq}{\left(\frac{d}{2}\right)^2} = \frac{kQ^2}{d^2}$$ which gives q = Q/4 (in magnitude). Also, as we discussed before, the sign is opposite to Q. Alternatively, if you do not want to decide what the sign of the charge is, then just add the two forces as vectors and equate the resultant to zero. The answer will have the sign incorporated.

Pulleys and masses (66)
This happens to be an elementary problem. Here is the equation, using standard notations as in our textbooks. $$T – m_1g = m_1a$$ (m1 going down)<BR> $$m_2g – T = m_2a$$ (m2 going up)<BR> Add the two equations so T that is eliminated. $$a = \frac{m_2 – m_1}{m_1 + m_2}g = \frac{g}{8}$$ When you divide both numerator and denominator with m2, the equation will turn into a linear equation in m1/m2.

What are the needs of the following assumptions?
 * Strings are inextensible
 * Pulley is frictionless
 * Pulley is massless (Hint – this assumption is not needed here. Why?)

Stefan’s law (67)
This question again is direct and does not require any explanation. The power radiated (Energy per second) is given by E = σeAT4. Now evaluate the fraction as we always do. In our case since the bodies have the same material the value of e is same for both.

$$\frac{E_1}{E_2} = \frac{\sigmaeA_1T^4_1}{\sigmaeA_2T^4_2} = \frac{\left(r_1\right)^2T^4_1}{\left(r_2\right)^2T^4_2} = \frac{1}{16}16 = 1$$

Three blocks carried away (68)
Use Newton’s second law. The tension in the string would be equal to (mass of block C) x (acceleration of mass C) (we have used net force T = ma). What is the acceleration? Zoom out and see the whole diagram. The acceleration of the block c would be equal to the combined acceleration of the three masses, which would be $$\frac{m_a + m_b + m_c}{F}$$. Remember that when we consider the three blocks and the two strings as a system, the tension of the other two strings becomes internal force and are not counted in the equation of Newton’s law!

Climbing up a rope (69)
The weight of the man is 60g = 600 N approx! For being able to lift him, the rope should be able to exert a force more than 600 N while in the question it is mentioned that the rope cannot withstand a force greater than 360 Newton. This means that the rope cannot even support his weight, let alone the possibility of climbing up! None of the options match. Since this question paper is reconstructed based on the memory of students who have answered it (AIEEE papers were not given back to students after they answered the exams until last couple of years!) we can safely assume that the reproduction of this question was imperfect.

Angular momentum (70)
Note that the angular momentum of a particle over another point is RxL (Cross Product), where R is the position vector of that particle with respect to the point and L is the linear momentum vector (absolute, not relative to the point!). The magnitude would be then |R||L|sinθ = m|R||V|sinθ. |R|sinθ is the perpendicular distance of the point from the velocity vector. Refer the diagram. |R|sinθ is then simply l. The answer is hence mvl. Many people with incomplete knowledge of mechanics would mark mvr here, because in standard notation the perpendicular distance or radius is taken as r! AIEEE frequently does this so that people learning the formulae by heart (without knowing their exact meanings) would score a negative here!

Radioactive emission (72)
Any particle that has a charge will be deflected in a magnetic field, just go according to that.

Work function (73)
The question confuses people as to what wavelength they are talking about. It must be the threshold wavelength. When the striking photons are of threshold wavelength, their energy is just equal to the work function of the metal. $$W_f = \frac{hc}{\lambda_t}$$

So, $$\frac{\lambda_\left(Na\right)}{ \lambda_\left(Cu\right)} = \frac{\frac{hc}{W_fNa}}{\frac{hc}{W_fCu}} = \frac{4.5}{2.3}$$ which is nearly equal to 2.

Covalent Compounds (74)
Covalent compounds are formed when two halffilled orbitals (or one filled and one empty in case of co-ordinate bonds) overlap. (refer Chemical Bonding) Recall that orbitals are nothing but wave functions of the electron cloud. The overlap is then an example of the wave nature of electrons.

Induced EMF (75)
In the loop, Sides AB and DC are moving along their lengths and hence there is no EMF induced in them. Side BC is not moving in a magnetic field so there is no emf across it as well. Only the rod AD has motional EMF induced across it which is equal to vBL.


 * A is at a higher potential on the side AD. What does this say about the charge separation inside the rod? Where are the positive charges and what end has the negative charges?
 * Suppose the loop is moving as it is but is completely immersed in the magnetic field. We say that the change in flux through it is zero and hence there is no EMF induced in it. The question is, both sides AD and BC have a motional EMF, then why is the net EMF zero? [Hint - induced EMF has a positive and negative side just like the voltage of a battery.]