The following MCQ’s appeared in IIT-JEE 2007 question paper:

**(1) A circuit is connected as shown in the figure with the switch S open. When the switch is closed, the total amount of charge that flows from Y to X is**

**(a) 0 (b) 54 μC (c) 27 μC (d) 81 μC**

When S is open, the P.D. across 3 μF and 6 μF capacitors are 6V and 3V respectively. [The charges on them are equal and the P.D. across them are inversely proportional to their capacities].

When S is closed, the P.D. across them become 3V and 6V respectively since the PD across the 3Ω and 6Ω resistors are 3V and 6V respectively. Positive charges have to flow from Y to X to achieve this condition.

Since the potential at point X is to be made 6V for this, the charge flowing to the 3 μF capacitor is 3μF×3V = 9μC. The charge flowing to the 6μF capacitor is 6μF×3V = 18μC. The total charge = 9 μC +18 μC =27 μC.

[You may have certain doubts regarding this solution. Once you note that the potentials of the *lower* potential plate of the 3 μF capacitor and the *higher* potential plate of the 6 μF capacitor are raised by 3 volts, by the charges flowing from Y to X, your doubts will be cleared].

**(2)**** A resistance of 2Ω is connected across one gap of a metre bridge (the length of the wire is 100 cm) and an unknown resistance, greater than 2Ω, is connected across the other gap. When these resistances are interchanged, the balance point shifts by 20 cm. Neglecting any corrections, the unknown resistance is**

**(a) 3Ω (b) 4Ω (c) 5Ω (d) 6Ω **

If the balancing length is measured initially on the side of the unknown resistance X, it will shift from 60 cm to 40 cm. [Remember that the balancing length is measured from the same side before and after interchanging. You might have noted that the balance point in a meter bridge shifts symmetrically with respect to the mid point of the bridge wire].

Therefore, we have X/2 = 60/40, from which X = 3Ω.

[If the balancing length is measured on the side of the known resistance, it will change from 40 cm to 60 cm. In this case, 2/X = 40/60, from which X =3Ω].

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