MHT CET · Physics · Current Electricity
The length of a potentiometer wire is 'L'. A cell of e.m.f. E is balanced at a length \(\frac{L}{5}\) from the positive end of the wire. If the length of the wire is increased by \(\frac{L}{2}\), the same cell will give balance point at distance ' \(x\) '. The value of ' \(x\) ' is
- A \(\frac{5 \mathrm{~L}}{12}\)
- B \(\frac{4 \mathrm{~L}}{15}\)
- C \(\frac{3 \mathrm{~L}}{10}\)
- D \(\frac{2 \mathrm{~L}}{15}\)
Answer & Solution
Correct Answer
(C) \(\frac{3 \mathrm{~L}}{10}\)
Step-by-step Solution
Detailed explanation
\(\frac{\text{balance length}}{\text{total wire length}} = \text{constant for given cell}\) \(\frac{l_1}{L_1} = \frac{l_2}{L_2}\)
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