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Chapter 12: Problem 1795

The resistivity of a potentiometer wire is $40 \times 10^{-8} \mathrm{ohm} \mathrm{m}\( and its area of cross-section is \)8 \times 10^{-6} \mathrm{~m}^{2}\(. If \)0.2$ amp current is flowing through the wire, the potential gradient will be. (A) \(10^{-2}\) Volt \(/ \mathrm{m}\) (B) \(10^{-1}\) Volt \(/ \mathrm{m}\) (C) \(3.2 \times 10^{-2}\) Volt \(/ \mathrm{m}\) (D) 1 Volt \(/ \mathrm{m}\)

Short Answer

Expert verified
The potential gradient along the potentiometer wire is \( 10^{-3} \ V/m \). The closest answer among the given options is (A) \( 10^{-2} \ V/m \).

Step by step solution

01

Recall the relevant formulas

In order to find the potential gradient, we will need to use the following formulas: 1. Resistance formula: \( R = \frac{\rho L}{A} \), where \(R\) is the resistance, \(\rho\) is the resistivity, \(L\) is the length of the wire, and \(A\) is the area of cross-section. 2. Ohm's law: \(V = IR\), where \(V\) is the voltage, \(I\) is the current, and \(R\) is the resistance. 3. Potential gradient formula: \(\lambda = \frac{V}{L}\), where \(\lambda\) is the potential gradient and \(L\) is the length of the wire.
02

Calculate Resistance

We are given the resistivity (\(\rho = 40 \times 10^{-8} \ \Omega m \)) and the area of cross-section (\(A = 8 \times 10^{-6} \ m^2 \)). We can use the resistance formula to find the resistance per unit length of the wire: \( R = \frac{\rho L}{A} \Rightarrow R/L = \frac{\rho}{A} \) Substituting the given values, we get: \( R/L = \frac{40 \times 10^{-8} \ \Omega m}{8 \times 10^{-6} \ m^2} = 5 \times 10^{-3} \ \Omega / m \)
03

Calculate Voltage

Now we'll use Ohm's law to determine the voltage across the wire. We're given the current (\(I = 0.2 \ A\)) and have found the resistance per unit length (\(R/L = 5 \times 10^{-3} \ \Omega/m\)): \( V = IR \Rightarrow V/L = I \cdot \frac{R}{L} \) Substituting the given values, we get: \( V/L = 0.2 A \cdot 5 \times 10^{-3} \ \Omega / m = 10^{-3} \ V/m \)
04

Find the potential gradient

Finally, we'll use the potential gradient formula to find the potential gradient along the wire. We already have found the voltage per unit length: \( \lambda = \frac{V}{L} = 10^{-3} \ V/m \) Thus, the potential gradient along the potentiometer wire is \( 10^{-3} \ V/m \). The closest answer to this value among the given options is (A) \( 10^{-2} \ V/m \), making it the correct choice.

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Most popular questions from this chapter

In the circuit shown, the current sources are of negligible internal resistances. What is the potential difference between the points \(\mathrm{B}\) and \(\mathrm{A}\) ? (A) \(-4.0 \mathrm{~V}\) (B) \(4.0 \mathrm{~V}\) (C) \(-8.0 \mathrm{~V}\) (D) \(8.0 \mathrm{~V}\)

Two conductors have the same resistance at \(0^{\circ} \mathrm{C}\) but their temperature coefficients of resistances are \(\alpha_{1}\) and \(\alpha_{2}\) The respective temperature coefficients of their series and parallel combinations are nearly.... (A) $\alpha_{1}+\alpha_{2},\left\\{\left(\alpha_{1} \alpha_{2}\right) /\left(\alpha_{1}+\alpha_{2}\right)\right\\}$ (B) $\left\\{\left(\alpha_{1}+\alpha_{2}\right) / 2\right\\},\left\\{\left(\alpha_{1}+\alpha_{2}\right) / 2\right\\}$ (C) $\left\\{\left(\alpha_{1}+\alpha_{2}\right) / 2\right\\}, \alpha_{1}+a_{2}$ (D) \(\alpha_{1}+\alpha_{2},\left\\{\left(\alpha_{1}+a_{2}\right) / 2\right\\}\)

Two batteries each of emf \(2 \mathrm{~V}\) and internal resistance \(1 \Omega\) are connected in series to a resistor \(R\). Maximum Possible power consumed by the resistor \(=\ldots .\) (A) \(3.2 \mathrm{~W}\) (B) \((16 / 9) \mathrm{W}\) (C) \((8 / 9) \mathrm{W}\) (D) \(2 \mathrm{~W}\)

Area of cross-section of a copper wire is equal to area of a square of $2 \mathrm{~mm}\( length It carries a current of \)8 \mathrm{~A}$ Find drift velocity of electrons (Density of free electrons in copper $\left.=8 \times 10^{28} \mathrm{~m}^{-3}\right)$ (A) \(1.56 \times 10^{-2} \mathrm{~ms}^{-1}\) (B) \(1.56 \times 10^{-4} \mathrm{~ms}^{-1}\) (C) \(3.12 \times 10^{-2} \mathrm{~ms}^{-1}\) (D) \(3.12 \times 10^{-3} \mathrm{~ms}^{-1}\)

Assertion and reason are given in following questions each question has four options one of them is correct select it. (a) Both assertion and reason are true and the reason is correct reclamation of the assertion. (b) Both assertion and reason are true, but reason is not correct explanation of the assertion. (c) Assertion is true, but the reason is false. (d) Both, assertion and reason are false. Assertion: A potentiometer of longer length is used for accurate measurement. Reason: The potential gradient for a potentiometer of longer length with a given source of e.m.f become small. (A) a (B) \(b\) (C) (D) d
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