The resistances of the four arms P, Q, R and S in a Wheatstone's bridge are 10 ohm, 30 ohm, 30 ohm and 90 ohm, respectively. The e.m.f. and internal resistance of the cell are 7 volt and 5 ohm respectively. If the galvanometer resistance is 50 ohm, the current drawn from the cell will be 1.0 A, 0.2 A, 2.0 A, or 0.1 A?

Given circuit is a balanced Wheatstone bridge. Since it's balanced, the galvanometer resistance (50 ohm) doesn't affect the total current drawn from the cell. We can find the equivalent resistance of the bridge by considering the parallel combination of (P+Q) and (R+S).

(P+Q) = 10 ohm + 30 ohm = 40 ohm
(R+S) = 30 ohm + 90 ohm = 120 ohm

The equivalent resistance (Reff) of the parallel combination is given by:

Reff = [(P+Q) * (R+S)] / [(P+Q) + (R+S)] = (40 ohm * 120 ohm) / (40 ohm + 120 ohm) = 4800 / 160 = 30 ohm

The total resistance in the circuit is the sum of the equivalent resistance and the internal resistance of the cell:

Rtotal = Reff + internal resistance = 30 ohm + 5 ohm = 35 ohm

Using Ohm's law, the current (I) drawn from the cell is:

I = V / Rtotal = 7 V / 35 ohm = 0.2 A

Therefore, the current drawn from the cell is 0.2 A.