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).

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 v_1i towards the stationary block 2 of mass m₂, the final velocities of blocks 1 and 2 are given by

v_1f = v_1i(m₁ – m₂) /(m₁ + m₂) and

v_2f = 2 m₁v_1i /(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 apphysicsresources.blogspot.com

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 P_4, 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

(d) (ε₀A)/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
(a) 3 μF

(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.