A Wheatstone bridge has the resistances 10 Ω, 10 Ω, 10 Ω and 30 Ω in its four arms. What resistance joined in parallel to the 30 Ω resistance will bring it to the balanced condition ? 

• 2 Ω

• 5 Ω

• 10 Ω

• 15 Ω

A wire of resistance 4 Ω is stretched to twice its original length. The resistance of stretched wire would be

• 2 Ω

• 4 Ω

• 8 Ω

• 16 Ω

The internal resistance of a 2.1 V cell which gives a current of 0.2 A through a resistance of 10 Ω is

• 0.2 Ω

• 0.5 Ω

• 0.8 Ω

• 1.0 Ω

The resistances of the four arms P, Q, R and S in a Wheatstone's bridge are 10 Ω, 30 Ω, 30 Ω and 90 Ω, respectively. The emf and internal resistance of the cell are 7 V and 5 Ω respectively. If the galvanometer resistance is 50 Ω, the current drawn from the cell will be

• 1.0 A

• 0.2 A

• 0.1 A

• 2.0 A

A wire loop is rotated in a magnetic field. The frequency of change of direction of the induced emf is

• once per revolution

• twice per revolution

• four times per revolution

• six times per revolution

In the circuit shown assume the diode to be ideal. When V_i increases from 2V to 6 V, the change in the current is (in mA)

• zero

• 20

• 80/3

• 40

Four cells, each of emf E and internal resistance r, are connected in series across an external resistance R. By mistake one of the cells is connected in reverse. Then the current in the external circuit is

• $\frac{2\mathrm{E}}{4\mathrm{r} + \mathrm{R}}$

• $\frac{3\mathrm{E}}{4\mathrm{r} + \mathrm{R}}$

• $\frac{3\mathrm{E}}{3\mathrm{r} + \mathrm{R}}$

• $\frac{2\mathrm{E}}{3\mathrm{r} + \mathrm{R}}$

A circuit consists of three batteries of emf E₁ = 1 V, E₂ = 2 V and E₃ = 3 V and internal resistances 1 Ω, 2Ω and 1 Ω respectively which are connected in parallel as shown in the figure. The potential difference between points P and Q is

• 1.0 V

• 2.0 V

• 2.2 V

• 3.0 V

The current I is in the circuit shown is

• 1.33 A

• Zero

• 2.00 A

• 1.00 A

A metal wire of circular cross-section has a resistance R₁. The wire is now stretched without breaking, so that its length is doubled and the density is assumed to remain the same. If the resistance of the wire now becomes R₂ then R₂ : R₁ is

• 1 : 1

• 1 : 2

• 4 : 1

• 1 : 4