IIT-JEE 2008 Matrix-Match type Question----

After the change of pattern with effect from the 2007 exam, the degree of difficulty of IIT-JEE has been considerably reduced, as you can judge from the following Matrix-Match Type question which appeared in the 2008 question paper:

Column I gives a list of possible set of parameters measured in some experiments The variations of the parameters in the form of graphs are shown in Column II. Match the set of parameters given in column I with the graphs given in column II. Indicate your answer by darkening the appropriate bubbles of the 4×4 matrix given in the ORS.

--------------------------Column I-----------------------------------------Column II----
(A) Potential energy of a simple pendulum (y-axis) as a function of displacement (x-axis)

(B) Displacement (y-axis) as a function of time (x-axis) for a one dimensional motion at zero or constant acceleration when the body is moving along the positive x-direction

(C) Range of a projectile (y-axis) as a function of its velocity (x-axis) when projected at a fixed angle

(D) The square of the time period (y-axis) of a simple pendulum as a function of its length (x-axis)

(A) is to be matched with (p) since the potential energy (U) of a simple pendulum is given by

U = mgh = mgℓ(1– cosθ).

Here h is the height of the bob with respect to the mean position of the bob, ℓ is the length of the pendulum and θ is the angular displacement. U is thus related to θ non-linearly as indicated in the graph (p).

(B) is to be matched with (q) and (s). When the acceleration is zero, the velocity is constant and the displacement is directly proportional to the time as indicated in the graph (q). When the acceleration is constant, the velocity is directly proportional to the time and the displacement is non-linearly related to the time as indicated in the graph (s). [Remember s = ut + ½ at²].

(C) is to be matched with (s) since the range R of a projectile is given by

R = (u²sin2θ)/g where u is the velocity of projection and θ is the angle of projection. For fixed angle of projection, the range is directly proportional to the square of the velocity of projection as indicated in graph (s).

(D) is to be matched with (q) since the time period of a simple pendulum is given by

T = 2π√(ℓ/g) so that the square of the time period T is directly proportional to the length ℓ of the pendulum as indicated in graph (q).

You will have to darken the bubbles of the 4×4 matrix as shown

You can find more IIT-JEE questions (with solution) here and at other locations at physicsplus.blogspot.com

Simple Kinematics in One Dimension – AIEEE 2008 & other questions

The following question which appeared in AIEEE 2008 question paper is an interesting one. You will find that you can work it out without the knowledge of any standard formula. In fact sheer imagination is sufficient to answer this question and that’s why it is interesting. Here is the question:

A body is at rest at x = 0. At t = 0, it starts moving in the positive x-direction with a constant acceleration. At the same instant another body passes through x = 0 moving in the positive x-direction with a constant speed. The position of the first body is given by x₁(t) after time 't' and that of second body by x₂(t) after the same time interval. Which of the following graphs correctly describes (x₁ – x₂) as a function of time ‘t’?

(x₁ – x₂) must be negative in the initial stage since the first body starts from rest where as the second body has a constant speed through out its motion. The first body will be behind the second one and the separation between them will increase for some time. As the first body picks up speed, the separation between the bodies decreases and becomes zero at a certain instant. The separation (x₁ – x₂) will be negative up to this instant. Thereafter the first body overtakes the second one and the separation (x₁ – x₂) becomes positive and goes on increasing. The situation is indicated correctly by graph (1).

Alternatively, you can find the answer by applying the equation,

s = ut + (¹/₂)at² for uniformly accelerated one dimensional motion.

Thus x₁ = 0 + (¹/₂)at² and x₂ = vt where a is the constant acceleration of the first body and v is the constant speed of the second body.

Therefore, (x₁ – x₂) = (¹/₂)at² – vt

(x₁ – x₂) = 0 when t = 0 and when t = 2v/a.

Further, d(x₁ – x₂)/dt = at – v so that at t = v/a the quantity (x₁ – x₂) will be a maximum or minimum. It is in fact a minimum point on the curve obtained by plotting (x₁ – x₂) against t, as shown by the positive value of the second differential coefficient, d²(x₁ – x₂)/dt².

So the correct option is indeed curve (1).

Here is another question:

A particle starts from rest with a constant acceleration and moves along the positive x-direction. If its displacement is x₁ in the first two seconds and x₂ in the next two seconds, then

(a) x₂ = 2x₁

(b) x₂ = 3x₁

(c) x₂ = 4x₁

(d) x₂ = 5x₁

(e) x₂ = 6x₁

We have x₁ = (1/2) a×2² = 2a

x₂ = (1/2) a×4² – (1/2) a×2² = 6a

Therefore, x₂ = 3x₁

Now, consider the following question which is quite simple. But be careful while answering.

A train having a constant speed of 54 km/hour takes 10 seconds to move past a lamp post. How much time is required for this train to cross a 300 m long bridge?

(a) 18 s

(b) 20 s

(c) 30 s

(d) 36 s

(e) 40 s

The speed of the train in ms^–1 is 54×5/18 = 15.

Since the train takes 10 s to move past the lamp post, the length of the train is 15×10 = 150 m.

The train has to travel a total distance of 450 m to cross the bridge. (This includes the length of the train also).

Therefore, time required = 450/15 = 30 s.

The following question appeared in AIPMT 2008 question paper:

A particle moves in a straight line with a constant acceleration. It changes its velocity from 10 ms^–1 to 20 ms^–1 while passing through a distance 135 m in t second. The value of t is

(1) 1.8

(2) 12

(3) 9

(4) 10

We have v_t ² = v₀ ² + 2as with usual notations.

Therefore, 20² = 10² + 2a×135 from which a = 30/27

Substituting this value of acceleration a in the equation

v_t = v₀ + at

we have 20 = 10 + (30/27)t from which t = 9 second

You will find some useful posts on one dimensional motion at physicsplus

Kerala Engineering Architecture Medical (KEAM) 2008 Entrance Examination Questions on Digital Circuits

As usual, simple questions were asked from the section on digital circuits in KEAM 2008 Examination. Here is the question which appeared in the Engineering Entrance question paper:

The combination of the following gates produces

(a) AND gate

(b) NAND gate

(c) NOR gate

(d) OR gate

(e) NOT gate

The first gate is a NAND gate. The second gate also is a NAND gate. But it functions as a NOT gate since its inputs are shorted. So we have a NAND followed by a NOT and hence the combination functions as an AND gate.

Here is the question which appeared in the Kerala Medical Entrance (2008) question paper:

The output Y when all the three inputs are first high and then low, will respectively be

(a) 1, 0

(b) 1, 1

(c) 0, 0

(d) 0, 1

(e) 1, –1

When all the three inputs are high the NAND gate following the AND gate has both inputs at high level. The output Y of the NAND is therefore low. When all the three inputs are low the NAND gate has both inputs low and hence the output is high. The correct option is (d).

You will find AIEEE 2008 and AIPMT 2008 questions on digital circuits with solution here.

All India Pre-Medical/Pre-Dental Entrance Examination (AIPMT) 2008 Questions from Electrostatics

Here are the two questions from electrostatics which appeared in the All India Pre-Medical/Pre-Dental Entrance Examination (AIPMT) 2008 question paper:

(1) A thin conducting ring of radius R is given a charge +Q. The electric field at the centre O of the ring due to the charge on the part AKB of the ring is E. The electric field at the centre due to the charge on the part ACDB of the ring is:

(a) E along OK

(b) E along KO

(c) 3E along OK

(d) 3E along KO

Since the ring is conducting, the charge on it gets distributed uniformly along it. The resultant electric field at the centre of the ring is zero (since a unit positive test charge placed at the centre will be pushed by equal radial forces all around). The field due to the charge on the part ACDB of the ring is therefore equal and opposite to the field due to the charge on the part AKB of the ring. So the answer is E along OK [option (a)].

[It would have been better if it is mentioned that K is the mid point of the arc AKB].

(2) The energy required to charge a parallel plate condenser of plate separation d and plate area of cross-section A such that the uniform electric field between the plates is E, is

(a) ε₀ E²/Ad

(b) ε₀ E²Ad

(c) ½ ε₀E²Ad

(d) ½ ε₀E²/Ad

The energy (U) required for charging a capacitor of capacitance C to V volt is given by

U = ½ CV² where C = ε₀A/d for a parallel plate capacitor with air (or vacuum) as dielectric.

Since E = V/d, we have V = Ed

Therefore, U = (½) (ε₀A/d)(Ed)² = (½) ε₀E²Ad

You can find more questions (with solution) in this section at AP Physics Resources.

Kerala Engineering Entrance 2008 Questions on Electromagnetic Induction

Questions involving electromagnetic induction at your level are usually simple and interesting. Here are the two questions which appeared in Kerala Engineering Entrance 2008 question paper:

(1) If the self inductance of 500 turn coil is 125 mH, then the self inductance of similar coil of 800 turns is

(a) 48.8 mH

(b) 200 mH

(c) 187.5 mH

(d) 320 mH

(e) 78.1 mH

Self inductance is the total magnetic flux linked with the coil when unit current flows through the coil. Since the total flux and the magnetic field are directly proportional to the number of turns in the coil, the self inductance is directly proportional to the square of the number of turns in the coil. Therefore we have

500²/800² = 125/L where L is the self inductance of the 800 turn coil.

This yields L = 320 mH.

(2) The flux linked with a circuit is given by Φ= t³ + 3t – 7. The graph between time (x–axis) and induced emf (y–axis) will be a

(a) straight line through the origin

(b) straight line with positive intercept

(c) straight line with negative intercept

(d) parabola through the origin

(e) parabola not through the origin

The induced emf, V = – dΦ/dt = – 3t² – 3

If V is plotted against t, a parabola which does not pass through the origin will be obtained. So the correct option is (e).