Solved Question Papers - IIT JEE/PhysicsSol1996

Objectives

Cylindrical capacitor

We might not have studied the structure of a cylindrical capacitor - that does not mean we can not solve this question! So lets get started Draw two concentric cylinders. For the electric field, we know we will use Gauss' law. The gaussian surface we choose is a cylinder of radius r (r = distance from the cente at which the field has to be found.)

Intg. E.dS = Qenc/Eo. We will not explain the simplifications on the LHS. However, simplifying it will tell you that the electric field is inversely proportional to r and not r^2.

Field lines through a conductor

There are two things you need to know while solving this: -

Electric field inside a conductor is zero. So field lines (direction of electric field) are not defined inside a conductor.
Electric field near a conductor is perpendicular to its surface, because a conducting surface is equipotential.

This immediately gives us the answer to be D.

Now we put forward some descriptive questions whose answers form a part of a normal High School Physics course.

Why is the electric field inside a conductor always zero? (Answer relates to charge separation till equilibrium is attained between two forces. This is just a hint, the reader has to supply a CBSE styled i.e. explanatory answer.)

Why is the electric field near an equipotent surface perpendicular to it? (Use the relation between voltage, electric field vector and displacement vector)

The conductor was placed in an uniform electric field. Why is then the electric field not uniform now? What has caused the disturbance in it? Field lines terminate at the left surface and restart from the right surface. What does this tell you about the charge distribution on the Neutral Sphere?

RMS Speed

Purely formulae based. We dont need to provide a solution! The answer is B.

Virtual Images

Note first of all that the question must be speaking of only real objects. (u < 0 in standard convention)

Anyone who has drawn ray diagrams properly knows for sure that the answer is Concave lens and a Convex Mirror.

Infact, this is factual for most students. Hence, you might want to prove this using calculations. Go ahead! Do not hesitate here! Remember that whenever something is excessively imple, just pick the pen up and solve it. if you do not do it - you might not be able to reproduce those results when required.

For convex mirror - use the mirror equation to get V (image) in terms of u. Substitute V=(-X) and F=(+F) so that every term on the RHS becomes positive and can be analyzed. The magnification will always be positive u < 0 meaning that the image is virtual!

Similarly for a concave lens.

Answer is B,C

Holes

Factual - p-type semiconductors

On projectile motion

Introduction

Collision in mid air? Right. Have you ever encountered questions which, upon reading once, make you believe they are pretty difficult, but whose solution makes you slap your forehead? This is one of those typical problems. When you get down to it, there is absolutely no analysis invovled! I emphasize through this example how simple questions can be made to look difficult by their presentation.

Solution

Shot A - The one which was launched at an angle. Shot B - The one which was launched horizontally.

This is the first point which you need to think about. Which one of the two was launched first? Think about it, they have to collide. A has to cover the same horizontal displacement as B, but with a lesser horizontal velocity (why lesser horizontal velocity?). Its but obvious that A was launched first.

If they collide, it just means that they are at the same position at the same instant of time. The two projectiles start from the same point, which means, in order to collide - their displacements must be same at a particular time instant.

For horizontal displacement, we have [Hor. velocity of Particle 1] x [Time till collision] = [Hor. Vel. of particle 2] x [Time till collision]

This is the best part about projectile motion! The horizontal motion of projectiles is uniform.

All we have to do now is to substitute the values. Note that the time till collision for the two particles is not the same! If for the second particle it is T then for the first particle, it is T+t where t is the time interval between the firings.

Sx = xVA x (T+t) = xVB x (T)

This equation gives us T=t, after substituting the two velocities. Sy = yVA x (T+t) - (0.5)(g)(T+t)^2 = yVB x (T) - (0.5)(g)(T)^2

Substituting T=t from the first equation and solving, we get t=1 second interval between the firings.

For the second part, we require horizontal and vertical displacements. Go back to the above equations, plug in t=1/T=1. We get Sx = 5(3)^0.5 and Sy = -5 m

The displacemtent is the final-initial coordinate, so X = 5root3 m, Y = -5 + 10 = 5 m. [Launched from (0,10), remember?]

On adiabatic process

Introduction

I just mentioned excessively simple problems in the projectile motion question above. Bless the heavens! This question on adiabatic expansion is a direct one! And direct means DIRECT! Pocket the 5 marks here - you'll have to work hard for them elsewhere.

Final Temperature

Use T1(V1)^(@-1) = T2(V2)^(@-1) where @ is the adiabatic exponent (gamma) of the gas, 5/3 in our case.

T2 = 300 K x (1/2)^(2/3) = 189 K

Change in internal energy

nCv(T2-T1) = -2767J. We will not tell you what n and cv are. Please forgive us.

Work done by the gas

We wont even tell you the answer over here! Use the first law to relate Q, Work done (IntegPdV) and DelE.

Note

Please note that P(V)^@ = Constant is valid only for a reversible adiabatic process. The question paper does not mention specifically that the process carried out is reversible adiabatic. If the process were irreversible with expansion carried out against constant external pressure = final pressure, use

PV = NRT over initial and final stages
P(V2-V1) [Integ.PdV] + nCv(T2-T1) = 0 [Q=0. If we are dealing with adiabatic process, how can we forget to use this equation?]

Also note that this is NOT a trick question! If the student had to consider irreversible processes as well, the JEE would have given MAJOR hints pointing towards that possibility. e.g., "Answer all possible cases" and such.

Question 4 - Fraunhofer Diffraction Pattern

DIRECT Question. Two marks not to be missed!

The angular width is 2Lamda/b where b is slit width (constant).

If by changing the wavelength we decrease the width by 30%, it is evident from the equation that the wavelength was also reduced by 30%. Hence the answer to this part is 6000 x (70/100) = 4200 Angstorms.

That is, we used light of wavelength 4200 Angstorms the second time.

Now, even if the apparatus is immersed in some liquid (A hypothetical question), the answer will be judged by the change in wavelength of the 6000 A light in that medium.

It is clear that the light of wavelength 6000 A in air behaves as light of 4200 A wavelength.

4200 = (6000)/(Mu) giving Mu = 1.429

Question 4 - The Prism

Introduction

This is the first question that we expect prepared students to mess up at the JEE. However, bear in mind that I'm saying this just because the question involves more calculation than the others. Howver, it is as simple as the others.

Critical incidence at the hypotenuse

We are told to find that angle of incidence on AB for which the ray hits the cemented interface at the critical angle. Now, some students may find this a bit intriguing, but frankly speaking - I do not have anything to chase away that vagueness.

Recall the geometrical construction you made for a prism to find R1 + R2.

Here, R2 = Sin-1 (n1/n). If you realize that the normals at the two points and the two faces form a cyclic quadrilateral, it won't take you much time to realize that R1 = 45 - R2

Now relate R1 and the angle of incidence using Snell's law.

Finally we get Sin-1(sinI/n) = 45 - Sin-1 (n1/n).

i.e. (Sini) = n/root2[root(1-(n1/n)^2) - (n1/n)]

Tha angle i obtained from here is the answer. That is, sin-1 [n/root2[root(1-(n1/n)^2) - (n1/n)]]</red> ===Angle of incidence for no deviatio

the hypotenuse, either the two refractive indices need to be same or the ray has to be incident normally on the hypotenuse.

In that case - R2 = 0. Giving R1 = 45. the angle of incidence can then be found using Snell's law.

Sini = (1.352)/(root 2), giving i = sin-1 0.956

Wrapping up the first four questions

Considering the calculations involved in the last Question 4, we do not expect everyone to do it in examination conditions without any silly mistake. Then, we can expect that majority of the prepared students would fetch 18-21 marks in this section, however scoring full 25 in the first four questions is not difficult as well. Its just that the proportion of such students would be less.