Problem 857 - JEE Physics

Question

Two wires of equal lengths are made of the same material wire A has a diameter that is twice as that of wire \(B\). If identical weights are suspended from the ends of these wires the increase in length is......... (A) Four times for wire \(\mathrm{A}\) as for wire \(\mathrm{B}\). (B) Twice for wire \(\mathrm{A}\) as for wire \(\mathrm{B}\). (C) Half for wire \(\mathrm{A}\) as for wire \(\mathrm{B}\). (D) One-fourth for wire \(\mathrm{A}\) as for wire \(\mathrm{B}\).

Step-by-Step Solution

Answer

The increase in length for wire A is one-fourth that of wire B, which corresponds to option (D).

Step 1: Identify the Young's modulus formula

Young's modulus (Y) is a measure of the stiffness of a material, and it is defined as the ratio of tensile stress to tensile strain. The formula for Young's modulus is given by: \[Y = \frac{F/A}{\Delta L / L}\] Where: - Y is Young's modulus - F is the applied force (weight) - A is the cross-sectional area - ΔL is the change in length - L is the original length

Step 2: Relate the diameter and cross-sectional area of the two wires

First, let's find the relationship between the cross-sectional areas of both wires. The cross-sectional area of a wire is given as: \[A = \pi (\frac{d}{2})^{2}\] Where d is the diameter of the wire. Since wire A has a diameter twice the diameter of wire B: \[d_{A} = 2 d_{B}\] Now, we find the ratio of the cross-sectional areas of both wires. \[\frac{A_{A}}{A_{B}} = \frac{\pi (\frac{d_{A}}{2})^{2}}{\pi (\frac{d_{B}}{2})^{2}}\]

Step 3: Simplify the ratio of cross-sectional areas

Now, simplify the equation by substituting the relationship between the diameters: \[\frac{A_{A}}{A_{B}} = \frac{\pi (\frac{2 d_{B}}{2})^{2}}{\pi (\frac{d_{B}}{2})^{2}}\] \[\frac{A_{A}}{A_{B}} = \frac{d_{A}^2}{d_{B}^2} = \frac{(2d_{B})^2}{d_{B}^2} = 4\] The cross-sectional area of wire A is 4 times larger than that of wire B.

Step 4: Use the Young's modulus formula to compare the increase in length

We will use the Young's modulus formula to find the relationship between the increase in lengths for both wires. The wires are made of the same material, so their Young's modulus is the same. Since Y, F, and L are constant for both wires, we have: \[\frac{A_{A}}{\Delta L_{A}} = \frac{A_{B}}{\Delta L_{B}}\] Now, substitute the relationship of the cross-sectional areas and solve for the ratio of the increase in lengths: \[\frac{4A_{B}}{\Delta L_{A}} = \frac{A_{B}}{\Delta L_{B}}\] \[\frac{\Delta L_{A}}{\Delta L_{B}} = \frac{1}{4}\] So, when identical weights are suspended from the ends of both wires, the increase in length for wire A is one-fourth that of wire B, which corresponds to option (D).

Key Concepts

Tensile Stress

Imagine pulling on a rope tied to a heavy object. The force you apply is what creates tensile stress in the rope. Tensile stress is an important concept when studying the strength and behavior of materials under force. It refers to the force per unit area exerted on an object when it is stretched or pulled. For instance, if you have a string and you pull it with a force, the tensile stress is defined by the amount of force you are applying divided by the cross-sectional area of the string. The formula for tensile stress (σ) is:\[σ = \frac{F}{A}\]Where:

• \(σ\): Tensile stress (in Pascals)
• \(F\): Force applied (in Newtons)
• \(A\): Cross-sectional area (in square meters)

So, in the case of the exercise, if the diameter of wire A is larger, its cross-sectional area is greater and therefore will have less tensile stress compared to wire B when the same force is applied. This difference impacts how much each wire stretches under the force.

Cross-Sectional Area

The cross-sectional area of a wire gives us an insight into how large the wire is when we look "inside" it, across its cut surface. For a wire with a circular cross-section, this area is calculated using the formula for the area of a circle:\[A = \pi \left(\frac{d}{2}\right)^2\]Where:

• \(A\): Cross-sectional area (in square meters)
• \(d\): Diameter of the wire (in meters)

In the exercise, wire A has a diameter twice that of wire B. When plugged into the formula, because of the square of the radius, wire A's cross-sectional area ends up being four times larger than that of wire B. This is an important factor in the problem because a larger cross-sectional area means that the wire can bear more force before stretching, thus affecting how much the wire elongates under tension. This concept helps explain why wire A stretches less compared to wire B despite having the same applied force.

Tensile Strain

Tensile strain gives us a measure of how much a material stretches when a force is applied. It is the ratio of the change in length (ΔL) of a material to its original length (L). This is a dimensionless quantity and can be thought of as the relative elongation of the material.The formula for tensile strain (ε) is:\[ε = \frac{ΔL}{L}\]Where:

• \(ε\): Tensile strain (unitless)
• \(ΔL\): Change in length (in meters)
• \(L\): Original length (in meters)

In terms of Young's modulus, tensile strain is used alongside tensile stress to help determine the stretching behavior of materials. Since both wires in the problem have the same material, they share the same Young's modulus. The larger initial cross-sectional area of wire A results in a lower tensile strain because although the original lengths are the same, the change in length for wire A is less compared to wire B. Therefore, despite applying the same force, wire B stretches more because its smaller cross-sectional area results in higher tensile stress and strain.