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Chapter 18: Problem 11

A wire of length \(l\) is drawn such that its diameter is reduced to half of its original diameter. If initial resistance of the wire were \(10 \Omega\), its resistance would be (a) \(160 \Omega\) (b) \(120 \Omega\) (c) \(140 \Omega\) (d) \(100 \Omega\)

Short Answer

Expert verified
The new resistance of the wire is 160 惟.

Step by step solution

01

Understanding Initial Conditions

The initial resistance of the wire is given as \(10 \Omega\). The original diameter of the wire is \(D\), and this will be reduced to \(\frac{D}{2}\) upon drawing the wire. The length of the wire is initially \(l\).
02

Finding the Volume Constraint

The wire is made of the same material throughout, so its volume remains constant. The volume of a cylindrical wire can be expressed as \(V = \pi \left(\frac{D}{2}\right)^2 l\). Since the diameter is reduced to half, the new diameter is \(\frac{D}{2}\), and the new length should compensate to maintain the same volume.
03

Calculating the New Length

As the diameter is halved, the cross-sectional area becomes \(\frac{1}{4}\)th of the original. Thus, the new length \(l'\) must be four times the original \(l\) to keep the volume constant, i.e., \(l' = 4l\).
04

Relating Resistance to Geometry

The resistance \(R\) of a wire is given by \(R = \rho \frac{l}{A}\), where \(\rho\) is the resistivity, \(l\) is the length, and \(A\) is the cross-sectional area. The initial cross-sectional area \(A_1\) is \(\pi \left(\frac{D}{2}\right)^2\), and for the halved diameter it becomes \(\frac{1}{4}A_1\).
05

Computing the New Resistance

Substitute the new geometry into the resistance formula: the new resistance \(R' = \rho \frac{4l}{\left(\frac{1}{4}A_1\right)} = 16 \rho \frac{l}{A_1}\), which is 16 times the original resistance. Thus, \(R' = 16 \times 10 \Omega = 160 \Omega\).
06

Verify and Conclude

The new resistance is calculated to be \(160 \Omega\), corresponding to option (a) \(160 \Omega\). Therefore, after drawing, the resistance of the wire will be \(160 \Omega\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Resistance Calculation
In any electrical circuit, resistance determines how much current flows for a given voltage. Knowing how to calculate resistance is key, especially when physical dimensions change. The resistance of a cylindrical wire is given by the formula \( R = \rho \frac{l}{A} \), where:
  • \( R \) is the resistance,
  • \( \rho \) is the resistivity of the material,
  • \( l \) is the length of the wire,
  • \( A \) is the cross-sectional area of the wire.
Resistance is directly proportional to the length \( l \) and inversely proportional to the cross-sectional area \( A \). If the length of the wire increases, the resistance increases. Conversely, if the area increases, resistance decreases. When a wire is stretched and its dimensions altered, it's essential to reassess these parameters to accurately calculate the new resistance.
Geometric Properties and Resistance
Understanding the geometric properties of the wire helps in determining how resistance changes. Consider a wire with an initial diameter \( D \) that is reduced to half, or \( \frac{D}{2} \). This change directly impacts the cross-sectional area \( A \). Originally, the area is \( A_1 = \pi \left( \frac{D}{2} \right)^2 \).When the diameter is halved, the new cross-sectional area becomes \( A_2 = \pi \left( \frac{D/2}{2} \right)^2 = \frac{1}{4}A_1 \). Because the diameter reduction significantly decreases the area, the same volume constraint changes the wire's length.These geometric alterations mean the length must adjust to maintain volume, resulting in a notable increase in resistance. For a direct relationship perspective, as geometry dictates area and length, they in turn influence the wire's resistance based on our understanding from resistance calculations.
Cylinder Volume Conservation
In this process, the principle of volume conservation plays a crucial role. For a cylinder, volume \( V \) remains constant if the wire is stretched, given by the formula \( V = \pi \left( \frac{D}{2} \right)^2 l \), where:
  • \( D \) is the original diameter,
  • \( l \) is the original length.
When the diameter is halved to \( \frac{D}{2} \), the cross-sectional area reduces by a factor of four. To conserve volume, the length \( l' \) of the cylinder increases to compensate for the reduced area.Mathematically, the new length \( l' \) becomes \( 4l \), ensuring that the product of the reduced area and increased length equals the original volume. This highlights the interplay between geometric properties, as conserving volume despite changes affects both the wire鈥檚 physical structure and its resistance. Understanding this interdependence is key in predicting how resistance changes with geometric alteration.

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Most popular questions from this chapter

A fuse wire of circuit cross-section and having diameter of \(0.4 \mathrm{~mm}\), allows \(3 \mathrm{~A}\) of current to pass through it. But if another fuse wire of same material and circular cross-section and having diameter of \(0.6 \mathrm{~mm}\) is taken, then the amount of current passed through the fuse is (a) \(3 \mathrm{~A}\) (b) \(3 \times \sqrt{\frac{3}{2}} \mathrm{~A}\) (c) \(3 \times\left(\frac{3}{2}\right)^{\not 2} \mathrm{~A}\) (d) \(3 \times\left(\frac{3}{2}\right) \mathrm{A}\)

Two heaters designed for the same voltage \(V\) have different power ratings. When connected individually across a source of voltage \(V\), they produce \(H\) amount of heat each in times \(t_{1}\) and \(t_{2}\) respectively. When used together across the same source, they produce \(H\) amount of heat in time \(t\). (a) If they are in series, then \(t=t_{1}+t_{2}\) (b) If they are in series, then \(t=2\left(t_{1}+t_{2}\right)\) (c) If they are in parallel, then \(t=\frac{t_{1} \times t_{2}}{\left(t_{1}+t_{2}\right)}\) (d) If they are in parallel, then \(t=\frac{t_{1} t_{2}}{2\left(t_{1}+t_{2}\right)}\)

In a thermocouple, one junction which is at \(0^{\circ} \mathrm{C}\) and the other at \(t^{\circ} \mathrm{C}\), the emf is given by \(E=a t^{2}-b t^{3}\). The neutral temperature is given by (a) \(a / b\) (b) \(2 a / 3 b\) (c) \(3 a / 2 b\) (d) \(b / 2 a\)

What must be the efficiency of an electric kettle marked \(500 \mathrm{~W}, 230 \mathrm{~V}\), if it was found to bring \(1 \mathrm{~kg}\) of water at \(15^{\circ} \mathrm{C}\) to boiling point in 15 min? (Given specific heat capacity of water \(=4200 \mathrm{~J} / \mathrm{kg}^{\circ} \mathrm{C}\) ) (a) \(79 \%\) (b) \(81 \%\) (c) \(72 \%\) (d) \(69 \%\)

In cosmic rays \(0.15\) protons \(\mathrm{cm}^{-2} \mathrm{sec}^{-1}\) are entering the earth's atmosphere. If the radius of the earth is \(6400 \mathrm{~km}\), the current received by the earth in the form of cosmic rays is nearly. (a) \(0.12 \mathrm{~A}\) (b) \(1.2 \mathrm{~A}\) (c) \(12 \mathrm{~A}\) (d) \(120 \mathrm{~A}\)
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