JEE Advanced 2014 Questions on Electric Circuits

“Do not worry about your problems with mathematics, I assure you mine are far greater."

– Albert Einstein

The following questions appeared in  JEE Advanced 2014 question paper. Question No. 1 is a multiple choice question in which one or more than one options are correct. The second question also is multiple choice type but it has just one correct option.

(1) Heater of an electric kettle is made of a wire of length L and diameter d. It takes 4 minutes to raise the temperature of 0.5 kg water by 40 K. This heater is replaced by a new heater having two wires of the same material, each of length L and diameter 2d. The way these wires are connected is given in the options. How much time in minutes will it take to raise the temperature of the same amount of water by 40 K ?

(A) 4 if wires are in parallel

(B) 2 if wires are in series

(C) 1 if wires are in series

(D) 0.5 if wires are in parallel

The resistance R of the heater wire in the first kettle is given by

            R =  ρL/A where ρ is the resistivity (specific resistance) of the wire and A is the cross section area of the wire.

[Note that A = π(d/2)²]

The second kettle has two heater wires, each of the same length L but cross section area 4A.

[Since the diameter of each wire in the second kettle is twice that of the wire in the first kettle, the cross section area is 4 times].

The resistance of each wire in the second kettle is R/4.

When the wires are connected in parallel, the effective resistance is R/8.

The power of the first kettle is V²/R where as the power of the second kettle is V²/(R/8) which is equal to 8V²/R.

Since the power of the second kettle is 8 times that of the first one, the time required for heating the water is reduced by a factor of 8 and is equal to 4/8 minutes which is 0.5 minute. Therefore (D) is a correct option.

When the wires are connected in series, the effective resistance of the combination is 2×(R/4) = R/2. The power of the second kettle is then V²/(R/2) which is equal to 2V²/R.

Since the power of the second kettle in this case twice that of the first one, the time required for heating the water is half the time taken by the first kettle and is equal to 2 minutes. Therefore (B) too is a correct option. 

(2) During an experiment with a metre bridge, the galvanometer shows a null point when the jockey is pressed at 40.0 cm using a standard resistance of 90 Ω, as shown in the figure. The least count of the scale used in the metre bridge is 1 mm. The unknown resistance is

(A) 60 ± 0.15 Ω

(B) 135 ± 0.56 Ω

(C) 60 ± 0.25 Ω

(D) 135 ± 0.23 Ω

We have R/90 = 40/(100 – 40) from which R = 60 Ω.

Since the least count of the scale used in the metre bridge is 1 mm only, there is an error in the value of R calculated above. In order to take this inaccuracy into account, we write the balance condition of the metre bridge as

            R/90 = x/(100 – x) where x is the balancing length (on the side of R as shown in the figure).

Taking logarithms,

            ln R = ln 90 + ln x – ln(100 – x)

Therefore ∆R/R = ∆x/x – ∆(100 – x)/(100 – x)

Or, ∆R/R = ∆x/x + ∆x/(100 – x)

Since ∆x = 0.1 cm, x = 40 cm and R = 60 Ω, we have

            ∆R/60 = (0.1/40) + (0.1/60)

Therefore, ∆R = 60[(0.1/40) + (0.1/60)] = 0.25 Ω

The unknown resistance is R ± ∆R = 60 ± 0.25 Ω

BITSAT Questions (MCQ) on Direct Current Circuits

There are two ways to live your life. One is as though nothing is a miracle; the other is as though everything is a miracle.

– Albert Einstein

We shall discuss a few interesting multiple choice questions on direct current circuits, which appeared in BITSAT question papers.

The following question appeared in BITSAT 2009 question paper:

(1) In the adjacent shown circuit, a voltmeter of internal resistance R, when connected across B and C reads (100/3) V. Neglecting the internal resistance of the battery, the value of R is

(a) 100 KΩ

(b) 75 KΩ                                                                                  

(c) 50 KΩ

(d) 25 KΩ

Since the voltage drop axross B and C is (100/3) volt, which is 1/3 of the supply voltage, the effective resistance of the parallel combination of the voltmeter resistance R and the 50 KΩ resistor must be 25 KΩ.

[2/3 of the supply voltage is dropped across the 50 KΩ resistor in the gap AB and hence the effective resistance (in the gap BC) that drops 1/3 of the supply voltage must be 25 KΩ].

Since 50 KΩ  in parallel with 50 KΩ makes 25 KΩ, the internal resistance R of the voltmeter must be 50 KΩ [Option (c)].

[If you want to make things more clear, you may write the following mathematical steps:

The current I sent by the battery is given by

            I = 100/[50 + {(50×R)/(50+R)}]

In the above equation we have written the resistances in KΩ so that we will obtain the final answer in KΩ.

Since the voltage drop across B and C is (100/3) volt, we have

            100/3 = {(50×R)/(50+R)}× I

Or, 100/3 = {(50×R)/(50+R)}×100/[50 +{(50×R)/(50+R)}]

Rearranging, (50 + R) [50 +{(50×R)/(50+R)}] = 150 R

Or, 2500 + 50 R + 50 R = 150 R

This gives R = 50 and the answer is 50 KΩ since we have written resistances in KΩ].

The following question appeared in BITSAT 2005 question paper:

(2) Two resistances are connected in two gaps of a metre bridge. The balance point is 20 cm from the zero end. A resistance of 15 Ω is connected in series with the smaller of the two. The null point shifts to 40 cm.the value of the smaller resistance in ohms is:

(a) 3

(b) 6

(c) 9

(d)12

If P is the smaller resistance (Fig.) and Q is the larger resistance, we have

            P/Q = 20/80 = ¼ ………..   (i)

After connecting 15 Ω in series with P we have

            (P + 15)/Q = 40/60 = 2/3……….(ii)

On dividing Eq.(i) by Eq.(ii) we have

P/(P + 15) = 3/8

Therefore, 8P = 3P + 45 from which P = 9 Ω

The following question also appeared in BITSAT 2005 question paper:

(3) The current in a simple series circuit is 5 A. When an additional resistance of 2 Ω is inserted, the current drops to 4 A. The original resistance of the circuit in ohms was:

(a) 1.25

(b) 8

(c) 10

(d) 20

If the emf in the circuit is V volt and the original resistance of the circuit is R ohms we have

            V/R = 5 -----------------(i)

On inserting the additional resistance of 2 Ω we have

            V/ (R+2) = 4 -----------------(ii)

On dividing Eq.(i) by Eq.(ii) we have

            (R+2)/R = 5/4 

Or, 4R + 8 = 5R from which R = 8 Ω.

The following question also appeared in BITSAT 2008 question paper:

(4) A current of 2 A flows in an electric circuit as shown in the figure. The potential difference (V_R – V_S), in volts (V_R and V_S are potentials at R and S respectively) is

(a) – 4

(b) + 2

(c) + 4

(d) – 2

Since the two branches PRQ and PSQ contain equal resistances (10 Ω), the current gets divided equally at the junction P. The same current of 1 A flows through yhe branches. Taking Q as the reference point to measure the potentials at R and S we have

            V_R = + 7 volt and

            V_S = + 3 volt

[Note that V_R is the potential drop produced across the 7 Ω resistor connected between Q and R and V_S is the potential drop produced across the 3 Ω resistor connected between Q and S].

Therefore (V_R – V_S) = + 4 volt