Sure, let's solve this step-by-step.
We're given:
1. The electromotive force (e.m.f) of the first cell: [tex]\( \text{emf}_1 = 2.06 \)[/tex] volts
2. The balance point for the first cell: [tex]\( \text{BP}_1 = 72 \)[/tex] cm
3. The electromotive force (e.m.f) of the second cell: [tex]\( \text{emf}_2 = 1.54 \)[/tex] volts
We need to find the new balance point, [tex]\( \text{BP}_2 \)[/tex], corresponding to the second cell's e.m.f.
We know that the balance point is directly proportional to the e.m.f of the cell, which can be expressed as:
[tex]\[ \frac{\text{BP}_1}{\text{emf}_1} = \frac{\text{BP}_2}{\text{emf}_2} \][/tex]
Rearranging this equation to solve for [tex]\( \text{BP}_2 \)[/tex] gives:
[tex]\[ \text{BP}_2 = \left( \frac{\text{emf}_2}{\text{emf}_1} \right) \times \text{BP}_1 \][/tex]
Now, substituting the given values:
[tex]\[ \text{BP}_2 = \left( \frac{1.54 \, \text{V}}{2.06 \, \text{V}} \right) \times 72 \, \text{cm} \][/tex]
Carrying out the division and multiplication steps:
[tex]\[ \text{BP}_2 = \left( 0.747572815534 \right) \times 72 \, \text{cm} \][/tex]
[tex]\[ \text{BP}_2 \approx 53.83 \, \text{cm} \][/tex]
Therefore, the new balance point when the potentiometer is connected to the second cell with an e.m.f of 1.54 V is approximately [tex]\( 53.83 \)[/tex] cm.
