## Wednesday, April 23, 2008

### IIT-JEE 2008 Questions on Direct Current Circuits

Two questions from direct current circuits appeared in the IIT-JEE 2008 question paper. Here are those questions:

Figure shows three resistor configurations R1, R2 and R3 connected to 3 V battery. If the power dissipated by the configurations R1, R2 and R3 are P1, P2 and P3 respectively, then (a) P1 > P2 > P3

(b) P1 > P3 > P2

(c) P2 > P1 > P3

(d) P3 > P2 > P1

For the configuration R1, the central 1 Ω resistor can be ignored since it is connected between equipotential points (similar to the galvanometer in a balanced Wheatstone bridge). The equivalent resistance connected across the battery is therefore 1 Ω. [2 Ω in parallel with 2 Ω].

The equivalent resistance in the configuration R2 is 0.5 Ω. [2 Ω, 2 Ω and 1 Ω in parallel]

The equivalent resistance in the configuration R3 is 2 Ω.

Since power P = V2/R we have P2 > P1 > P3 [Option (c)].

The second question which follows is Reasoning Type with 4 choices out of which only one is correct

STATEMENT-1

In a metre bridge experiment, null point null point for an unknown resistance is measured. Now, the unknown resistance is put inside an enclosure maintained at a higher temperature.. The null point can be obtained at the same point as before by decreasing the value of the standard resistance and

STATEMENT-2

Resistance of a metal increases with increase in temperature.

(A) STATEMENT-1 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

We have R/X = l1/l2

If the unknown resistance X is decreased on increasing its temperature, the null point can be obtained at the sme point as before by decreasing the value of the standard resistance. You should note that nothing is mentioned about the nature of the unknown resistance. It can be a negative temperature coefficient (of resistance) device such as a thermistor. So, statement 1 is true.

Metals have positive temperature coefficient of resistance. So, statement 2 is true. But it is not a correct explanation for statement-1.

The correct option is (B).

[If the unknown resistance is a metallic wire such as copper wire or nichrome wire, the resistance will increase on heating and the correct option will then be (D).

## Saturday, April 12, 2008

### Multiple Choice Questions (MCQ) on Elastic Collision

The figure shows three identical blocks A, B and C (each of mass m) on a smooth horizontal surface. B and C which are initially at rest, are connectd by a spring of negligible mass and force constant K. The block A is initially moving with velocity ‘v’ along the line joining B and C and hence it collides with B elastically. Now, answer the following questions related to the system:

(1) After the collision the velocity of the block A is

(a) v/3

(b) v

(c) v/3

(d) v/2

(e) zero

At the moment the block A collides with the block B, you need consider these two blocks only. In one dimensional elastic collision between two bodies of the same mass, if one body is initially at rest, the moving body comes to rest and the body initially at rest moves with the velocity of the moving body. After the collision, the velocity of A is therefore zero.

[Generally, if the moving block 1 has mass m1 and it moves with initial velocity v1i towards the stationary block 2 of mass m2, the final velocities of blocks 1 and 2 are given by

v1f = v1i(m1 m2) /(m1 + m2) and

v2f = 2 m1v1i /(m1 + m2)

The above relations can be easily obtained from momentum and kinetic energy conservation equations.

(2) When the compression of the spring is maximum, the velocity of block B is

(a) v

(b) v

(c) v/3

(d) v/2

(e) zero

On receiving the entire momentum (mv) of the block A, the block B moves towards the block C, compressing the spring. When the compression of the spring is maximum, both B and C move with the same velocity (v’). Since B and C have the same mass (m) and momentum has to be conserved, we have

mv = (m + m) v’

Therefore v’ = v/2.

(3) When the compression of the spring is maximum, the kinetic energy of the system is

(a) (½) mv2

(b) (1/4) mv2

(c) Zero

(d) mv2

(e) (1/8) mv2

The block A has come to rest after colliding with block B. When the compression of the spring is maximum, both B and C are moving with the same velocity v/2 as shown above. Therefore, the kinetic energy of the system under this condition is 2×(1/2)m(v/2)2 = (1/4) mv2. [Note that the spring has negligible kinetic energy since its mass is negligible].

(4) The maximum compression of the spring is

(a) √(mv2/2K)

(b) √(mv/2K)

(c) mv/2K

(d) mv2/2K

(e) √(2mv2/K)

The total energy of the system is equal to the initial kinetic energy (½ )mv2 of the colliding block A. Therfore we have

(½)mv2 = KE of A + KE of B + KE of C + PE of spring. The KE of A is zero after the collision. As shown above, the total KE of B and C is (1/4) mv2 when the compression of the spring is maximum. Therfore, the potential energy (PE) of the spring at maximum compression is given by the equation,

(½)mv2 = (1/4) mv2 + (1/2)Kx2 where x is the maximum compression of the spring.

This gives (1/2)Kx2 = (1/4) mv2 from which x = √(mv2/2K).

You will find similar questions (with solution) from different branches of physics at

## Sunday, April 06, 2008

### Two Questions from Electrostatics-- Capacitor Combinations

Here are two multiple choice questions on effective capacitance of capacitor networks. [The first question was found to be some what difficult for the students (from their response)]: (1) An unusual type of capacitor is made using four identical plates P1, P2, P3 and P4, each of area A arranged in air with the same separation d as shown in the figure. A thin wire outside the system of plates connects the plates P2 and P4. Wires soldered to the plates P1 and P3 serve as terminals T1 and T2 of the system. What is the effective capacitance between the terminals T1 and T2?

(a) (2ε0A)/3d

(b) (3ε0A)/2d

(c) (3ε0A)/d

(e) (ε0A)/2d

The arrangement contains 3 identlcal capacitors C1, C2 and C3 each of capacitance (ε0A)/d arranged as shown in the adjoining figure. Plate P2 is common for C1 and C2. Likewise, plates P3 is common for C2 and C3. The capacitors C2 and C3 are connected in parallel (to produce an effective value 2ε0A/d) and this parallel combination is connected in series with the capacitor C1 of value ε0A/d . The effective capacitance C between the terminals T1 and T2 is therfore given by

C = [(2ε0A/d) ×(ε0A/d)] / [(2ε0A/d) +(ε0A/d)].

Therefore, C = (2ε0A)/3d

(2) Four 2 μF capacitors and a 3 μF capacitor are connected as shown in the figure. The effective capacitance between the points A and B is

(b) 2 μF

(c) 1.5 μF

(d) 1 μF

(e) 0.5 μF

This is a very simple question once you identify the circuit to be a balanced Wheatstone’s bridge. The 3 μF capacitor is connected between equipotential points and you can ignore it. Without the diagonal branch, there are four 2 μF capacitors only. The capacitors in the upper pair are in series and produce a value of 1 μF. The capacitors in the lower pair also produce a value of 1 μF. Since they are in parallel the effective capacitance between the points A and B is 2 μF.