A battery of emf 1.4 volt and internal resistance `2Omega` is connected to a resistor is of `100Omega` through an ammeter. The resistance of the ammter is `(4)/(3)Omega`. A voltmeter has also been connected to measure the potential difference across the resistor `100Omega`. If the ammeter reads 0.02A and voltmeter reading is 1.10 volt then error in the reading of voltmeter is x volt then find value of `x((100)/(25))` in nearest integer.

Explanation

To solve the problem, we need to analyze the circuit and calculate the error in the voltmeter reading. Here’s a step-by-step solution:

### Step 1: Understand the Circuit
We have a battery with an emf of 1.4 V and an internal resistance of 2 Ω. This battery is connected to a 100 Ω resistor through an ammeter with a resistance of \( \frac{4}{3} \) Ω. The ammeter reads a current of 0.02 A, and we need to find the error in the voltmeter reading across the 100 Ω resistor.

### Step 2: Calculate the Total Resistance in the Circuit
The total resistance \( R_{total} \) in the circuit can be calculated as:
\[ R_{total} = R_{internal} + R_{ammeter} + R_{load} \]
Where:
- \( R_{internal} = 2 \, \Omega \)
- \( R_{ammeter} = \frac{4}{3} \, \Omega \)
- \( R_{load} = 100 \, \Omega \)

Calculating \( R_{total} \):
\[ R_{total} = 2 + \frac{4}{3} + 100 = 2 + 1.33 + 100 = 103.33 \, \Omega \]

### Step 3: Calculate the Current Through the Circuit
The current \( I \) flowing through the circuit is given as 0.02 A. We can verify this with Ohm's law:
\[ V = I \cdot R_{total} \]
Calculating the voltage drop across the total resistance:
\[ V = 0.02 \, A \times 103.33 \, \Omega = 2.0666 \, V \]

### Step 4: Calculate the Voltage Across the 100 Ω Resistor
The voltage across the 100 Ω resistor can be calculated using Ohm's law:
\[ V_{100} = I \cdot R_{load} = 0.02 \, A \times 100 \, \Omega = 2 \, V \]

### Step 5: Calculate the Voltage Drop Across the Ammeter
The voltage drop across the ammeter can be calculated as:
\[ V_{ammeter} = I \cdot R_{ammeter} = 0.02 \, A \times \frac{4}{3} \, \Omega = 0.02667 \, V \]

### Step 6: Calculate the Effective Voltage Across the Resistor
The effective voltage across the 100 Ω resistor when considering the internal resistance of the battery is:
\[ V_{effective} = V_{battery} - V_{internal} - V_{ammeter} \]
Where \( V_{battery} = 1.4 \, V \) and \( V_{internal} = I \cdot R_{internal} = 0.02 \, A \times 2 \, \Omega = 0.04 \, V \):
\[ V_{effective} = 1.4 \, V - 0.04 \, V - 0.02667 \, V = 1.33333 \, V \]

### Step 7: Calculate the Error in the Voltmeter Reading
The voltmeter reads 1.10 V, but the effective voltage across the 100 Ω resistor is 1.33333 V. The error \( x \) in the voltmeter reading is:
\[ x = V_{effective} - V_{voltmeter} = 1.33333 \, V - 1.10 \, V = 0.23333 \, V \]

### Step 8: Calculate \( x \left( \frac{100}{25} \right) \)
Now, we need to find:
\[ x \left( \frac{100}{25} \right) = 0.23333 \times 4 = 0.93332 \]
Rounding to the nearest integer gives us:
\[ \text{Final Answer} = 1 \]