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Chapter 12: Problem 1703

A wire of length \(L\) is drawn such that its diameter is reduced to half of its original diameter. If the resistance of the wire were \(10 \Omega\), its new resistance would be. (A) \(40 \Omega\) (B) \(60 \Omega\) (C) \(120 \Omega\) (D) \(160 \Omega\)

Short Answer

Expert verified
The new resistance of the wire after its diameter is reduced to half of its original diameter would be \(160 \Omega\), which corresponds to option (D).

Step by step solution

01

Understanding the Problem

Initially the wire has a diameter \( D \) and length \( L \) giving it a resistance \( R = \rho \frac{L}{A} \). This resistance is given as \( 10 \Omega \).
02

Changes in the Wire

Consider the wire is drawn out such that its diameter is reduced to half. The new diameter \( D' = D/2 \), and this results in a quarter of the original area \( A' = A/4 \) and increased length \( L' \), because volume of the wire remains constant during this process.
03

Applying the Formula with the New Parameters

Plug the new values into the resistance formula \( R' = \rho \frac{L'}{A'} \).
04

Determine the Relationship Between Original and New Lengths

Since volume of wire remains same before and after, it follows that: \( A \cdot L = A' \cdot L' \). Plugging the new values of \( A \) and \( A' \) gives us \( L' = 4L \).
05

Calculate the New Resistance

Replacing \( L' \) and \( A' \) in the new resistance formula, we get \( R' = \rho \frac{4L}{A/4} \). Simplifying this we find \( R' = 16R \).
06

Determine the New Resistance Using the Given Original Resistance

Substituting the original resistance \( R = 10 \Omega \) into the expression we derived for new resistance \( R' \), we have \( R' = 16 \cdot 10 \Omega \).
07

Concluding the exercise

We find that the new resistance \( R' = 160 \Omega \) which corresponds to option (D) in the exercise given.

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Most popular questions from this chapter

Two resistances \(\mathrm{R}_{1}\) and \(\mathrm{R}_{2}\) have effective resistance \(\mathrm{R}_{\mathrm{s}}\) when connected in sires combination and \(R_{p}\) when connected in parallel combination if $\mathrm{R}_{8} \mathrm{R}_{\mathrm{p}}=16\( and \)\left(\mathrm{R}_{1} / \mathrm{R}_{2}\right)=4\( the values of \)\mathrm{R}_{1}\( and \)\mathrm{R}_{2}$ are (A) \(2 \Omega\) and \(0.5 \Omega\) (B) \(1 \Omega\) and \(0.25 \Omega\) (C) \(8 \Omega\) and \(2 \Omega\) (D) \(4 \Omega\) and \(1 \Omega\)

There are n resistors having equal value of resistance \(\mathrm{r}\). First they are connected in such a way that the possible minimum value of resistance is obtained. Then they are connected in such a way that possible maximum value of resistance is obtained the ratio of minimum and maximum values of resistances obtained in these way is.... (A) \((1 / n)\) (B) \(\mathrm{n}\) (C) \(\mathrm{n}^{2}\) (D) \(\left(1 / \mathrm{n}^{2}\right)\)

At what temperature will the resistance of a copper wire be three times its value at \(0^{\circ} \mathrm{C}\) ? (Given: temperature coefficient of resistance for copper \(=4 \times 10^{-3}{ }^{\circ} \mathrm{C}^{-1}\) ) (A) \(400^{\circ} \mathrm{C}\) (B) \(450^{\circ} \mathrm{C}\) (C) \(500^{\circ} \mathrm{C}\) (D) \(550^{\circ} \mathrm{C}\)

The temperature co-efficient of resistance of a wire is $0.00125^{\circ} \mathrm{k}^{-1}\(. Its resistance is \)1 \Omega\( at \)300 \mathrm{~K}$. Its resistance will be \(2 \Omega\) at. (A) \(1400 \mathrm{~K}\) (B) \(1200 \mathrm{~K}\) (C) \(1000 \mathrm{~K}\) (D) \(800 \mathrm{~K}\)

What maximum power can be obtained from a battery of \(\mathrm{emf} \mathrm{E}\) and internal resistance \(\mathrm{r}\) connected with an external resistance \(\mathrm{R}\) ? (A) \(\left(E^{2} / 4 r\right)\) (B) \(\left(E^{2} / 3 r\right)\) (C) \(\left(E^{2} / 2 r\right)\) (D) \(\left(\mathrm{E}^{2} / \mathrm{r}\right)\)
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