A voltmeter has resistance of 2000 ohms and it can measure upto 2V. If we want to increase its range to 10 V, then the required resistance in series will be 

(1) 2000 Ω

(2) 4000 Ω

(3) 6000 Ω

(4) 8000 Ω

In the given figure, battery E is balanced on 55 cm length of potentiometer wire but when a resistance of 10 Ω is connected in parallel with the battery then it balances on 50 cm length of the potentiometer wire then internal resistance r of the battery is 

(1) 1 Ω

(2) 3 Ω

(3) 10 Ω

(4) 5 Ω

In the circuit shown, a meter bridge is in its balanced state. The meter bridge wire has a resistance 0.1 ohm/cm. The value of unknown resistance X and the current drawn from the battery of negligible resistance is 

(1) 6 Ω, 5 amp

(2) 10 Ω, 0.1 amp

(3) 4 Ω, 1.0 amp

(4) 12 Ω, 0.5 amp

AB is a potentiometer wire of length 100 cm and its resistance is 10 ohms. It is connected in series with a resistance R = 40 ohms and a battery of e.m.f. 2 V and negligible internal resistance. If a source of unknown e.m.f. E is balanced by 40 cm length of the potentiometer wire, the value of E is  

(1) 0.8 V

(2) 1.6 V

(3) 0.08 V

(4) 0.16 V

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

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

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