MHT CET · Physics · Current Electricity
A potentiometer wire of length \(100 \mathrm{~cm}\) and resistance \(3 \Omega\) is connected in series
with resistance of \(8 \Omega\) and an accumulator of 4 volt whose internal resistance is
\(1 \Omega\)
A cell of e.m.f. 'E' is balanced by \(50 \mathrm{~cm}\) length of the wire. The e.m.f. of the
cell is
- A \(1 \cdot 00\) volt.
- B \(0 \cdot 75\) volt.
- C \(0 \cdot 50\) volt.
- D \(0 \cdot 25\) volt.
Answer & Solution
Correct Answer
(C) \(0 \cdot 50\) volt.
Step-by-step Solution
Detailed explanation
(C)
Total resistance \(=3+8+1=12 \Omega\) current. \(\quad I=\frac{E}{R}=\frac{4}{12}=\frac{1}{3} A\)
P.D. across the wire \(=I R_{\omega}=\frac{1}{3} \times 3=1 \mathrm{~V}\)
potential gradient \(\mathrm{K}=\frac{\mathrm{V}}{\mathrm{L}}=\frac{1}{1}=1 \mathrm{v} / \mathrm{m}\) \(\mathrm{E}^{\prime}=\mathrm{K} \ell^{\prime}=1 \times 0.5=0.5 \mathrm{~V}\)
Total resistance \(=3+8+1=12 \Omega\) current. \(\quad I=\frac{E}{R}=\frac{4}{12}=\frac{1}{3} A\)
P.D. across the wire \(=I R_{\omega}=\frac{1}{3} \times 3=1 \mathrm{~V}\)
potential gradient \(\mathrm{K}=\frac{\mathrm{V}}{\mathrm{L}}=\frac{1}{1}=1 \mathrm{v} / \mathrm{m}\) \(\mathrm{E}^{\prime}=\mathrm{K} \ell^{\prime}=1 \times 0.5=0.5 \mathrm{~V}\)
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