Question

When more than \(20 \mathrm{~kg}\) mass is tied to the end of wire it breaks what is maximum mass that can be tied to the end of a wire of same material with half the radius? (A) \(20 \mathrm{~kg}\) (B) \(5 \mathrm{~kg}\) (C) \(80 \mathrm{~kg}\) (D) \(160 \mathrm{~kg}\)

Short answer

The maximum mass that can be tied to the end of a wire of the same material with half the radius is \(5\mathrm{~kg}\) (B).

Step-by-step solution

Analyze the given information

We are given that a wire breaks when more than 20 kg mass is tied to its end. We need to find the maximum mass that can be tied to the end of a wire of the same material with half the radius.

Calculate the cross-sectional areas of the two wires

The cross-sectional area of a wire is given by the formula: \[A = \pi R^2\] where A is the area and R is the radius. Let the radius of the first wire be R1 and that of the second wire be R2. Then, the cross-sectional areas of the two wires are: \[A_1 = \pi R_1^2 \]and \[A_2 = \pi R_2^2\] Since R2 is half of R1, we have R2 = 0.5*R1. Substituting this into the equation for A2, we get: \[A_2 = \pi (0.5R_1)^2\] which simplifies to: \[A_2 = 0.25\pi R_1^2\]

Find the ratio of the cross-sectional areas

To find the ratio of the cross-sectional areas of the two wires, divide A2 by A1. We have: \[\frac{A_2}{A_1} = \frac{0.25\pi R_1^2}{\pi R_1^2}\] The \(\pi R_1^2\) factors cancel out, leaving: \[\frac{A_2}{A_1} = 0.25\] This means that the square of the radius of the second wire is 0.25 times that of the first wire.

Calculate the maximum mass that can be tied to the end of the second wire

Since breaking strength is proportional to the cross-sectional area of a wire, and the mass tied to the wire is directly related to the breaking strength, we can find the maximum mass that can be tied to the second wire using the ratio of the cross-sectional areas: \[m_2 = m_1 \frac{A_2}{A_1}\] where \(m_1\) = 20 kg and \(m_2\) is the mass we want to find. Substituting the values, we get: \[m_2 = 20\mathrm{~kg} \cdot 0.25\] Solving for \(m_2, we get: \[m_2 = 5\mathrm{~kg}\] Therefore, the maximum mass that can be tied to the end of a wire of the same material with half the radius is 5 kg, which corresponds to answer (B).

Key concepts

Cross-Sectional Area

Understanding the cross-sectional area is crucial when working with wires and their breaking strength. The cross-sectional area of a wire is the surface area of its circular cross-section. Imagine slicing the wire perpendicular to its length – the area of this circle is what we refer to as the cross-sectional area. It's given by the formula:

• \(A = \pi R^2\)

where \(A\) is the area and \(R\) is the radius of the wire. This area is directly linked to the wire's mechanical properties. Essentially, a larger cross-sectional area indicates that the wire can handle more stress and weight before breaking. Therefore, if we have a wire with a higher cross-sectional area, it will typically have a greater breaking strength than one with a smaller area. For our original problem, reducing the wire radius by half decreases the cross-sectional area by a factor of four due to its squared dependence on the radius.

Proportionality in Physics

Proportionality in physics refers to the relationship between two quantities where a change in one causes a direct and consistent change in the other. This is often expressed in the form of simple ratios or equations. In the context of our exercise, the breaking strength of a wire is proportional to its cross-sectional area. This means if the cross-sectional area decreases, the breaking strength will decrease in a consistent manner. When more than 20 kg causes the original wire to break, its breaking strength is equivalent to holding precisely that weight. Now, since the cross-sectional area of the second wire is only 0.25 times that of the original (as shown in the solution steps), the maximum weight it can hold will also be reduced proportionally by the same factor. Thus, understanding proportionality allows us to quantitatively predict how changes in physical dimensions affect breaking strength, predicting the reduced capacity to hold only 5 kg for the second wire.

Radius and Strength Relationship

The radius of a wire is vital in determining its strength. Since the cross-sectional area \(A\) of a wire can be calculated via \(A = \pi R^2\), any change in the radius impacts the area significantly due to the squared relationship. For example, if the radius is halved, the area does not just halve; it becomes one-fourth because

• \(A_2 = \pi (0.5R_1)^2 = 0.25\pi R_1^2\).

This reduction drastically affects the wire's strength.Understanding how the radius affects breaking strength is crucial because it tells us how much weight a wire of specific dimensions can sustain. In the given problem, halving the radius means the wire is weaker by a factor of 4, as shown through the decreased cross-sectional area factor of 0.25. As a result, the maximum mass it can handle shrinks proportionally, down to 5 kg from the original 20 kg, capturing the essence of the relationship between radius, area, and breaking strength.