A battery of emf `E` is connected with there resistance `R , 2R and R` in series. The voltage across `2R` is measured with a voltmeter whose resistance a `10R` .What is the percentage error ?

Explanation

To solve the problem, we need to find the percentage error in the voltage measured across the resistor \(2R\) when a voltmeter with resistance \(10R\) is used. Here’s a step-by-step solution:

### Step 1: Calculate the total resistance in the circuit without the voltmeter
The resistors \(R\), \(2R\), and \(3R\) are in series. Therefore, the total resistance \(R_{total}\) is:
\[ R_{total} = R + 2R + 3R = 6R \]

### Step 2: Calculate the current in the circuit
Using Ohm's law, the current \(I\) flowing through the circuit can be calculated as:
\[ I = \frac{E}{R_{total}} = \frac{E}{6R} \]

### Step 3: Calculate the actual voltage across the resistor \(2R\)
The voltage \(V_{2R}\) across the resistor \(2R\) can be calculated using Ohm's law:
\[ V_{2R} = I \times 2R = \left(\frac{E}{6R}\right) \times 2R = \frac{E}{3} \]

### Step 4: Calculate the equivalent resistance when the voltmeter is connected
The voltmeter with resistance \(10R\) is connected in parallel with the resistor \(2R\). The equivalent resistance \(R_{eq}\) of \(2R\) and \(10R\) in parallel is given by:
\[ \frac{1}{R_{eq}} = \frac{1}{2R} + \frac{1}{10R} \]
Calculating this gives:
\[ \frac{1}{R_{eq}} = \frac{5 + 1}{10R} = \frac{6}{10R} = \frac{3}{5R} \]
Thus,
\[ R_{eq} = \frac{5R}{3} \]

### Step 5: Calculate the total resistance with the voltmeter
Now, the total resistance in the circuit becomes:
\[ R_{total}' = R + R_{eq} + 3R = R + \frac{5R}{3} + 3R \]
Finding a common denominator:
\[ R_{total}' = \frac{3R}{3} + \frac{5R}{3} + \frac{9R}{3} = \frac{17R}{3} \]

### Step 6: Calculate the new current in the circuit with the voltmeter
The new current \(I'\) can be calculated as:
\[ I' = \frac{E}{R_{total}'} = \frac{E}{\frac{17R}{3}} = \frac{3E}{17R} \]

### Step 7: Calculate the voltage across the resistor \(2R\) with the voltmeter connected
Now, we can calculate the voltage \(V_{2R}'\) across the resistor \(2R\) when the voltmeter is connected:
\[ V_{2R}' = I' \times R_{eq} = \left(\frac{3E}{17R}\right) \times \frac{5R}{3} = \frac{5E}{17} \]

### Step 8: Calculate the percentage error
The percentage error can be calculated using the formula:
\[ \text{Percentage Error} = \frac{\text{Actual Value} - \text{Measured Value}}{\text{Actual Value}} \times 100 \]
Substituting the values we found:
\[ \text{Percentage Error} = \frac{\frac{E}{3} - \frac{5E}{17}}{\frac{E}{3}} \times 100 \]
Finding a common denominator for the terms in the numerator:
\[ \text{Percentage Error} = \frac{\frac{17E - 15E}{51}}{\frac{E}{3}} \times 100 = \frac{\frac{2E}{51}}{\frac{E}{3}} \times 100 \]
Simplifying gives:
\[ \text{Percentage Error} = \frac{2}{51} \times 3 \times 100 = \frac{600}{51} \approx 11.76\% \]

### Final Answer
The percentage error in the measurement is approximately **11.76%**.