Problem Explanation:
We are given two cells with electromotive forces (emfs) E1E_1 and E2E_2, where E1>E2E_1 > E_2, connected as shown in the figure (not provided). The potentiometer is used to measure the voltage difference across different points.
When connected between points AA and BB, the balancing length is found to be 300 cm300 \, \text{cm}, and when connected between points AA and CC, the balancing length is 100 cm100 \, \text{cm}.
We are tasked with finding the ratio E1E2\frac{E_1}{E_2}.
Key Concepts:
1. Potentiometer Principle: The potentiometer is an instrument used to measure the potential difference (voltage) between two points. The balancing length of the potentiometer is directly proportional to the potential difference between the points.
2. Balancing Length Formula: If the potentiometer is balanced over a length LL with a potential difference VV, and the total length of the wire is LtotalL_{\text{total}}, then:
VL=VtotalLtotal\frac{V}{L} = \frac{V_{\text{total}}}{L_{\text{total}}}
Where:
o VV is the potential difference across the section of the potentiometer wire,
o LL is the corresponding length for the balancing point,
o VtotalV_{\text{total}} is the total potential difference across the entire potentiometer.
Step-by-Step Solution:
1. When the potentiometer is connected between AA and BB:
The balancing length is 300 cm300 \, \text{cm}, and the potential difference across the section of the wire is E1−E2E_1 - E_2 (because the cells are connected in series with their emfs E1E_1 and E2E_2).
Thus, the potential difference corresponding to the balancing length 300 cm300 \, \text{cm} is E1−E2E_1 - E_2.
2. When the potentiometer is connected between AA and CC:
The balancing length is 100 cm100 \, \text{cm}, and the potential difference across the section of the wire is E1E_1 (since point CC is assumed to be at the terminal of E1E_1).
So, the potential difference corresponding to the balancing length 100 cm100 \, \text{cm} is E1E_1.
3. Setting up the ratios:
The ratio of the balancing lengths can be used to set up a proportion between the potential differences.
E1−E2E1=300 cm100 cm\frac{E_1 - E_2}{E_1} = \frac{300 \, \text{cm}}{100 \, \text{cm}}
Simplifying:
E1−E2E1=3\frac{E_1 - E_2}{E_1} = 3
4. Solving for E1E2\frac{E_1}{E_2}:
E1−E2=3E1E_1 - E_2 = 3E_1 E2=E1−3E1=−2E1E_2 = E_1 - 3E_1 = -2E_1
Therefore, the ratio of E1E_1 to E2E_2 is:
E1E2=E1−2E1=3:2\frac{E_1}{E_2} = \frac{E_1}{-2E_1} = 3:2
Conclusion:
The ratio of E1E_1 to E2E_2 is 3:2, so the correct answer is:
A. 3:2\boxed{\text{A. 3:2}}
