A voltmeter has a range 0-V with a series resistance R. With a series resistance 2R, the range is 0-V'. The correct relation between V and V' is :

(1) $V^{\text{'}}=2V$

(2) $V^{\text{'}}>2V$

(3) $V^{\text{'}}>>2V$

(4) $V\text{'}<2V$

The measurement of the voltmeter in the following circuit is :

(1) 2.4 V

(2) 3.4 V

(3) 4.0 V

(4) 6.0 V

If an ammeter is to be used in place of a voltmeter then we must connect with the ammeter a :

(1) Low resistance in parallel

(2) High resistance in parallel

(3) High resistance in series

(4) Low resistance in series

The potential difference across the 100Ω resistance in the following circuit is measured by a voltmeter of 900 Ω resistance. The percentage error made in reading the potential difference is :

(1) $\frac{10}{9}$

(2) 0.1

(3) 1.0

(4) 10.0

If the ammeter in the given circuit reads 2 A, the resistance R is :

(1) 1 ohm

(2) 2 ohm

(3) 3 ohm

(4) 4 ohm

Two resistances of 400 Ω and 800 Ω are connected in series with a 6-volt battery of negligible internal resistance. A voltmeter of resistance 10,000 Ω is used to measure the potential difference across 400 Ω. The error in the measurement of potential difference in volts approximately is :

(1) 0.01

(2) 0.02

(3) 0.03

(4) 0.05

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