Two resistance $$\mathrm{X}$$ and $$\mathrm{Y}$$ are connected in the two gaps of a meterbridge and the null points is obtained at $$20 \mathrm{~cm}$$ from zero end. When the resistance of $$20 \Omega$$ is connected in series with the smaller of the two resistance $$\mathrm{X}$$ and $$\mathrm{Y}$$, the null point shifts to $$40 \mathrm{~cm}$$ from left end. The value of smaller resistance in ohm is

Resistance of a potentiometer wire is $$2 \Omega / \mathrm{m}$$. A cell of e.m.f. $$1.5 \mathrm{~V}$$ balances at $$300 \mathrm{~cm}$$. The current through the wire is

A potentiometer wire has length of $$5 \mathrm{~m}$$ and resistance of $$16 \Omega$$. The driving cell has an e.m.f. of $$5 \mathrm{~V}$$ and an internal resistance of $$4 \Omega$$. When the two cells of e.m.f.s $$1.3 \mathrm{~V}$$ and $$1.1 \mathrm{~V}$$ are connected so as to assist each other and then oppose each other, the balancing lengths are respectively

Two batteries, one of e.m.f. $$12 \mathrm{~V}$$ and internal resistance $$2 \Omega$$ and other of e.m.f. $$6 \mathrm{~V}$$ and internal resistance $$1 \Omega$$, are connected as shown in the figure. What will be the reading of the voltmeter 'V'?