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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

The potential difference between the terminals of a battery is $10 \mathrm{~V}\( and internal resistance \)1 \Omega\( drops to \)8 \mathrm{~V}$ when connected across an external resistor find the resistance of the external resistor. (A) \(40 \Omega\) (B) \(0.4 \Omega\) (C) \(4 \mathrm{M} \Omega\) (D) \(4 \Omega\)

The network is made of uniform wire. The resistance of portion EL is $2 \Omega\(. Find the resistance of star between points \)F \& C .$ (A) \(0.985 \Omega\) (B) \(1.25 \Omega\) (C) \(1.946 \Omega\) (D) \(1.485 \Omega\)

Match the physical quantities given in column I with their dimensional formulae given in column II - I stands for the dimension of current. \(\begin{array}{ll}\text { Column I } & \text { Column II }\end{array}\) (a) Electromotive force (emf) (p) \(\mathrm{ML}^{2} \mathrm{~T}^{-3} \mathrm{~A}^{-2}\) (b) Resistance (q) \(\mathrm{ML}^{3} \mathrm{~T}^{-3} \mathrm{~A}^{-2}\) (c) Resistivity (r) \(\mathrm{M}^{-1} \mathrm{~L}^{-3} \mathrm{~T}^{3} \mathrm{~A}^{2}\) (d) Conductivity (s) \(\mathrm{ML}^{2} \mathrm{~T}^{-3} \mathrm{~A}^{-1}\) (A) \(a-s, b-p, c-q, d-r\) (B) \(a-p, b-s, c-r, d-p\) (C) \(a-p, b-s, c-r, d-q\) (D) \(a-r, b-p, c-q, d-s\)

Potentiometer wire of length \(1 \mathrm{~m}\) is connected in series with $490 \Omega\( resistance and \)2 \mathrm{v}\( battery If \)0.2 \mathrm{mv} / \mathrm{cm}$ is the potential gradient, then resistance of the potentiometer wire is. (A) \(4.9 \Omega\) (B) \(7.9 \Omega\) (C) \(5.9 \Omega\) (D) \(6.9 \Omega\)

Resistors \(P\) and \(Q\) connected in the gaps of the meter bridge. the balancing point is obtained \(1 / 3 \mathrm{~m}\) from the zero end. If a \(6 \Omega\) resistance is connected in series with \(\mathrm{p}\) the balance point shifts to \(2 / 3 \mathrm{~m}\) form same end. \(\mathrm{P}\) and \(\mathrm{Q}\) are. (A) 4,2 (B) 2,4 (C) both (a) and (b) (D) neither (a) nor (b)
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