A galvanometer, having a resistance of 50 Ω gives a full scale deflection for a current of 0.05 A. The length in meter of a resistance wire of area of cross-section 2.97× 10^–2 cm² that can be used to convert the galvanometer into an ammeter which can read a maximum of 5 A current is (Specific resistance of the wire = 5 × 10^–7 Ωm) 

(1) 9

(2) 6

(3) 3

(4) 1.5

An ammeter reads up to 1 ampere. Its internal resistance is 0.81 ohm. To increase the range to 10 A the value of the required shunt is :

(1) 0.09 Ω

(2) 0.03 Ω

(3) 0.3 Ω

(4) 0.9 Ω

The length of a wire of a potentiometer is \(100~\text{cm}\), and the emf of its standard cell is \(E\) volt. It is employed to measure the emf of a battery whose internal resistance is \(0.5~\Omega\). If the balance point is obtained at \(l = 30~\text{cm}\) from the positive end, the emf of the battery is:
1. \(\frac{30E}{100}\)
2. \(\frac{30E}{100.5}\)
3. \(\frac{30E}{(100-0.5)}\)
4. \(\frac{30(E-0.5i)}{100}, \) where \(i\) is the current in the potentiometer

What is the reading of the voltmeter in the following figure?

(1) 3 V

(2) 2 V

(3) 5 V

(4) 4 V

The current flowing in a coil of resistance \(90~\Omega\) is to be reduced by \(90\%\). What value of resistance should be connected in parallel with it?
1. \(9~\Omega\)
2. \(90~\Omega\)
3. \(1000~\Omega\)
4. \(10~\Omega\)

In a metre bridge experiment, the null point is obtained at 20 cm from one end of the wire when resistance X is balanced against another resistance Y. If X < Y, then where will be the new position of the null point from the same end, if one decides to balance a resistance of 4X against Y 

(1) 50 cm

(2) 80 cm

(3) 40 cm

(4) 70 cm

In given figure, the potentiometer wire AB has a resistance of 5 Ω and length 10 m. The balancing length AM for the emf of 0.4 V is :

(1) 0.4 m

(2) 4 m

(3) 0.8 m

(4) 8 m

In a potentiometer experiment, the balancing with a cell is at length 240 cm. On shunting the cell with a resistance of 2 Ω, the balancing length becomes 120 cm. The internal resistance of the cell is :

1. 4 Ω

2. 2 Ω

3. 1 Ω

4. 0.5 Ω

A moving coil galvanometer of resistance 100Ω is used as an ammeter using a resistance 0.1Ω. The maximum deflection current in the galvanometer is 100 mA. Find the minimum current in the circuit so that the ammeter shows maximum deflection :

(1) 100.1 A

(2) 1000.1 mA

(3) 10.01 mA

(4) 1.01 mA

If the resistance of voltmeter is 10000Ω and resistance of ammeter is 2Ω then find R when voltmeter reads 12V and ammeter reads 0.1 A :

(1) 118 Ω

(2) 120 Ω

(3) 124 Ω

(4) 114Ω