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A potentiometer wire of length 10 m and ...

A potentiometer wire of length `10 m` and resistance `30 Omega` is connected in series with a battery of emf `2.5 V`, internal resistance `5 Omega` and an external resistance `R`. If the fall of potential along the potentiometer wire is `50 mu V mm^(-1)`, then the value of `R` is found to be `23 n Omega`. What is `n`?

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To solve the problem step by step, we will use the information provided about the potentiometer wire, the battery, and the external resistance. ### Step-by-Step Solution: 1. **Understand the Given Data:** - Length of the potentiometer wire, \( L = 10 \, \text{m} \) - Resistance of the potentiometer wire, \( R_w = 30 \, \Omega \) - EMF of the battery, \( E = 2.5 \, \text{V} \) - Internal resistance of the battery, \( r_i = 5 \, \Omega \) - Fall of potential along the potentiometer wire, \( K = 50 \, \mu V/mm = 50 \times 10^{-6} \, V/mm = 50 \times 10^{-6} \times 10^{-3} \, V/m = 50 \times 10^{-9} \, V/m \) 2. **Calculate the Total Resistance in the Circuit:** The total resistance \( R_{total} \) in the circuit is given by: \[ R_{total} = R_w + r_i + R \] where \( R \) is the external resistance we need to find. 3. **Calculate the Current in the Circuit:** The current \( I \) flowing through the circuit can be calculated using Ohm's law: \[ I = \frac{E}{R_{total}} = \frac{E}{R_w + r_i + R} \] 4. **Relate the Potential Gradient to the Voltage:** The potential gradient \( K \) is defined as: \[ K = \frac{V}{L} \] where \( V \) is the potential difference across the potentiometer wire. Thus, \[ V = K \cdot L \] 5. **Express the Voltage in Terms of Current:** The voltage across the potentiometer wire can also be expressed as: \[ V = I \cdot R_w \] 6. **Set Up the Equation:** Equating the two expressions for \( V \): \[ K \cdot L = I \cdot R_w \] Substituting \( I \) from step 3: \[ K \cdot L = \left(\frac{E}{R_w + r_i + R}\right) \cdot R_w \] 7. **Substitute Known Values:** Substitute the known values into the equation: \[ 50 \times 10^{-9} \cdot 10 = \left(\frac{2.5}{30 + 5 + R}\right) \cdot 30 \] Simplifying gives: \[ 5 \times 10^{-8} = \frac{75}{35 + R} \] 8. **Cross Multiply and Solve for \( R \):** Cross multiplying gives: \[ 5 \times 10^{-8} (35 + R) = 75 \] Expanding and rearranging: \[ 5 \times 10^{-8} R = 75 - 5 \times 10^{-8} \cdot 35 \] Calculate \( 5 \times 10^{-8} \cdot 35 \): \[ 5 \times 10^{-8} \cdot 35 = 1.75 \times 10^{-6} \] Thus: \[ 5 \times 10^{-8} R = 75 - 1.75 \times 10^{-6} \] \[ 5 \times 10^{-8} R = 75 - 0.00000175 \approx 75 \] Therefore: \[ R \approx \frac{75}{5 \times 10^{-8}} = 1.5 \times 10^{9} \, \Omega \] 9. **Convert \( R \) to the Form \( 23 \, n \Omega \):** Given \( R = 23 n \Omega \), we have: \[ 1.5 \times 10^{9} = 23 \times 10^{n} \] Solving for \( n \): \[ n = 9 - 1 = 8 \] ### Final Answer: The value of \( n \) is \( 5 \).

To solve the problem step by step, we will use the information provided about the potentiometer wire, the battery, and the external resistance. ### Step-by-Step Solution: 1. **Understand the Given Data:** - Length of the potentiometer wire, \( L = 10 \, \text{m} \) - Resistance of the potentiometer wire, \( R_w = 30 \, \Omega \) - EMF of the battery, \( E = 2.5 \, \text{V} \) ...
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