Question

An experiment is performed for comparing EMF of two cells using a potentiometer. For 1st cell, balancing length was achieved at 200 cm and for 2nd cell it was 150 cm. If least count of measurement of length of potentiometer wire is 1cm, the percentage error in the ratio of emf of two cells is :

• A. \(\frac{8}{7}\)
• B. \(\frac{7}{6}\)
• C. \(\frac{5}{6}\)
• D. \(\frac{3}{2}\)

Hint:

When dealing with errors, remember the simple rules: for addition/subtraction, absolute errors are added; for multiplication/division, fractional (or percentage) errors are added.
A larger measurement generally has a smaller percentage error, which is why longer balancing lengths are desirable in a potentiometer experiment for better accuracy.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We have a potentiometer experiment where two EMFs are compared by measuring their respective balancing lengths. We are given the lengths and the uncertainty (least count) in the measurement. We need to find the percentage error in the calculated ratio of the EMFs.
Step 2: Key Formula or Approach:
1. Potentiometer Principle: The EMF of a cell is directly proportional to its balancing length, \(\mathcal{E} \propto l\). Therefore, the ratio of two EMFs is equal to the ratio of their balancing lengths: \(y = \frac{\mathcal{E}_1}{\mathcal{E}_2} = \frac{l_1}{l_2}\).
2. Propagation of Errors: For a quantity \(y = \frac{l_1}{l_2}\), the maximum fractional error is the sum of the individual fractional errors: \(\frac{\Delta y}{y} = \frac{\Delta l_1}{l_1} + \frac{\Delta l_2}{l_2}\). The percentage error is this fractional error multiplied by 100.
Step 3: Detailed Explanation:
Part A: Identify the variables and their errors.
- Balancing length for the first cell, \(l_1 = 200\) cm.
- Balancing length for the second cell, \(l_2 = 150\) cm.
- The least count is the absolute error in each measurement, so \(\Delta l_1 = \Delta l_2 = 1\) cm.
Part B: Calculate the fractional error in the ratio.
The ratio is \(y = l_1/l_2\).
\[ \frac{\Delta y}{y} = \frac{\Delta l_1}{l_1} + \frac{\Delta l_2}{l_2} \] \[ \frac{\Delta y}{y} = \frac{1}{200} + \frac{1}{150} \] To add these fractions, we find a common denominator, which is 600.
\[ \frac{\Delta y}{y} = \frac{3}{600} + \frac{4}{600} = \frac{7}{600} \] Part C: Calculate the Percentage Error
The percentage error is given by \(\left(\frac{\Delta y}{y}\right) \times 100%\).
\[ \text{Percentage Error} = \left(\frac{7}{600}\right) \times 100 = \frac{7}{6} % \] The question asks for the numerical value of the percentage error.
Step 4: Final Answer:
The percentage error in the ratio of the EMFs is \(\frac{7}{6}\).