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Chapter 12: Problem 1685

Length of a wire of resistance \(R \Omega\) is increased to 10 times, so its resistance becomes \(1000 \Omega\), therefore \(R=\ldots .\) (The volume of the wire remains same during increase in length) (A) \(0.01 \Omega\) (B) \(0.1 \Omega\) (C) \(1 \Omega\) (D) \(10 \Omega\)

Short Answer

Expert verified
The initial resistance of the wire is \(10 \ \Omega\), which matches the answer choice (D).

Step by step solution

01

Understand the resistance formula

The resistance (R) of a wire is given by the formula: \[R = \rho \frac{L}{A}\] Where: - \(R\) = resistance (in ohms, Ω) - \(\rho\) = resistivity of the wire material (a constant) - L = length of the wire (in meters, m) - A = cross-sectional area of the wire (in square meters, m²) Since the volume remains constant, we will also use the relationship between length, cross-sectional area, and volume to solve the problem.
02

Relate volume and cross-sectional area

The volume of a wire is given by the formula: \[V = A \times L\] Since the problem states that the volume remains constant, we can represent the initial volume (V₁) and the final volume (V₂) as follows: \[V_1 = A_1 \times L_1\] and \[V_2 = A_2 \times L_2\] The lengths of the wires are related by the given condition, that is \(L_2 = 10 \times L_1\). Thus, we can write: \[V_1 = A_1 \times L_1\] \[V_2 = A_2 \times (10 \times L_1)\]
03

Use the resistance formula for both cases and create an equation

Now, we can apply the resistance formula for the initial and final cases: \[R_1 = \rho \frac{L_1}{A_1}\] \[1000 = \rho \frac{10 \times L_1}{A_2}\] We are looking for the initial resistance R₁. Since the resistivity \(\rho\) is constant, we can create an equation involving the areas and lengths of the wire by dividing the second equation by the first equation: \[\frac{1000}{R_1} = \frac{10 \times L_1}{L_1} \times \frac{A_1}{A_2}\]
04

Use the volume relationship to solve for R₁

Since we know that the volume remains constant: \[V_1 = A_1 \times L_1 = A_2 \times (10 \times L_1)\] \[\frac{A_1}{A_2} = \frac{10 \times L_1}{L_1}\] Now we substitute this relationship into the equation: \[\frac{1000}{R_1} = 10 \times \frac{10 \times L_1}{L_1}\] \[\frac{1000}{R_1} = 100\]
05

Solve for R₁ and match with given answers

We can now solve for R₁ by dividing both sides by 100: \[R_1 = \frac{1000}{100}\] \[R_1 = 10 \ \Omega\] Thus, the initial resistance of the wire is 10 Ω, which matches the answer choice (D).

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Most popular questions from this chapter

A parallel combination of three resistors takes a current of \(7.5 \mathrm{~A}\) form a \(30 \mathrm{~V}\) supply, It the two resistors are \(10 \Omega\) and $12 \Omega$ find which is the third one? (A) \(4 \Omega\) (B) \(15 \Omega\) (C) \(12 \Omega\) (D) \(22 \Omega\)

Find the equivalent resistance a cross \(\mathrm{AB}\) (A) \(1 \Omega\) (B) \(2 \Omega\) (C) \(3 \Omega\) (D) \(4 \Omega\)

Figure, shows a network of seven resistors number 1 to 7, each equal to $1 \Omega\( connection to a \)4 \mathrm{~V}$ battery of negligible internal resistance The current I in the circuit is.... (A) \(0.5 \mathrm{~A}\) (B) \(1.5 \mathrm{~A}\) (C) \(2.0 \mathrm{~A}\) (D) \(3.5 \mathrm{~A}\)

On applying an electric field of \(5 \times 10^{-8} \mathrm{Vm}^{-1}\) across a conductor, current density through it is \(2.5 \mathrm{Am}^{-2}\) The resistivity of the conductor is \(\ldots .\) (A) \(1 \times 10^{-8} \Omega \mathrm{m}\) (B) \(2 \times 10^{-8} \Omega \mathrm{m}\) (C) \(0.5 \times 10^{-8} \Omega \mathrm{m}\) (D) \(12.5 \times 10^{-8} \Omega \mathrm{m}\)

A wire in a circular shape has \(10 \Omega\) resistance. The resistance per one meter is \(1 \Omega\) The resultant between \(A \& B\) is equal to \(2.4 \Omega\), then the length of the chord \(\mathrm{AB}\) will be equal to (A) \(2.4\) (B) 4 (C) \(4.8\) (D) 6
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