Problem 851 - JEE Physics

Question

A steel wire of cross-sectional area \(3 \times 10^{-6} \mathrm{~m}^{2}\) can with stand a maximum strain of \(10^{-3}\) Young's modulus of steel is \(2 \times 10^{11}\left(\mathrm{~N} / \mathrm{m}^{2}\right)\). The maximum mass the wire can hold is..... \(\left[\mathrm{g}=10\left(\mathrm{~m} / \mathrm{s}^{2}\right)\right]\) (A) \(40 \mathrm{~kg}\) (B) \(60 \mathrm{~kg}\) (C) \(80 \mathrm{~kg}\) (D) \(100 \mathrm{~kg}\)

Step-by-Step Solution

Answer

The maximum mass the wire can hold is 60 kg, which corresponds to option (B).

Step 1: Calculate Stress

Calculate the stress (σ) using the Young's Modulus (Y) and the maximum strain (strain_max): \[σ = Y × strain\_max\]

Step 2: Find the maximum force

Use the calculated stress (σ) and the cross-sectional area (A) to find the maximum force (F_max) the wire can withstand: \[F\_max = σ × A\]

Step 3: Calculate the maximum mass

Now, use the formula for the force due to gravity to find the maximum mass (m_max) the wire can hold: \[F\_max = m\_max × g\] Solving for m_max, we get: \[m\_max = \frac{F\_max}{g}\] Now, let's plug in the given values and find the maximum mass.

Step 4: Plug in values

Use the given values from the problem: A = \(3 × 10^{-6} m^2\) strain_max = \(10^{-3}\) Y = \(2 × 10^{11}(N/m^2)\) g = 10 m/s²

Step 5: Calculate σ and F_max

Calculate the stress (σ) and the maximum force (F_max): σ = \((2 × 10^{11})(10^{-3}) = 2 × 10^8 N/m^2\) F_max = \((2 × 10^8)(3 × 10^{-6}) = 600 N\)

Step 6: Compute m_max

Finally, calculate the maximum mass (m_max): m_max = \(\frac{600}{10} = 60 kg\) Thus, the maximum mass the wire can hold is 60 kg, which corresponds to option (B).

Key Concepts

Stress and Strain

In physics and engineering, understanding the relationship between stress and strain is crucial when analyzing materials under force. Stress is defined as the force applied per unit area of the material. This can be thought of as how much force is trying to stretch or compress the material.
For instance, if you pull on a wire, the force distributed over the wire's area creates stress. Mathematically, it's expressed as:

• Stress ( σ ) = Force (F) / Area (A).

Strain, on the other hand, measures how much the material deforms in response to stress. It's the ratio of the change in length to the original length of the material. It tells us the fractional change experienced by the material, expressed as:

• Strain = Change in Length / Original Length.

In our example, the steel wire can handle a maximum strain of 0.001, meaning it can stretch by a tiny fraction of its original length before failing. Combining the two, Young's Modulus ( Y ) describes a material's ability to withstand changes in length under tension or compression. It's calculated as:

• Y = Stress / Strain.

Using this relationship, we can determine how the steel wire will perform under different forces.

Cross-Sectional Area

The cross-sectional area is a crucial element to consider when examining the physical properties of materials under force, like our steel wire. It refers to the area of the material's slice or section, perpendicular to the applied force.
Imagine cutting through a piece of steel wire and observing the surface that is exposed; this is its cross-sectional area. Essential for calculating stress, this area helps determine the material's capacity to withstand force.
In our exercise, the steel wire has a tiny cross-sectional area of \(3 \times 10^{-6} \text{ m}^2\). Such a small area indicates that the wire is thin. The calculation of stress (σ) relies heavily on this measurement because stress is proportionate to how the force spreads over this small section:

• Stress (σ) = Applied Force / Cross-Sectional Area (A).

This implies that even a large force can cause a huge stress if the cross-sectional area is small. Thus, when you apply force to a wire, its durability largely depends on that cross-sectional area.

Force Due to Gravity

Force due to gravity is a fundamental concept that describes how objects with mass experience attraction towards the Earth. In our day-to-day lives, we often call this weight. It's the pull that the planet exerts on every object within its vicinity, calculated using an object's mass and the acceleration due to gravity, which is approximately \(9.81 \text{ m/s}^2\). For ease, it is sometimes approximated to \(10 \text{ m/s}^2\) in exercises.
The formula for the force due to gravity (or weight) is:

• F = m \times g,

where:\

• m is the mass of the object,
• g is the acceleration due to gravity.

In solving the maximum mass a wire can hold, we rearrange this formula to solve for mass (\(m_{\text{max}}\)), using the maximum force the wire can withstand:

• \[\text{Maximum Mass } (m_{\text{max}}) = \frac{\text{Maximum Force } (F_{\text{max}})}{g}.\]

By calculating this, we conclude that the steel wire can support a maximum mass of \(60 \text { kg}\), as shown in our solution. This highlights the importance of understanding gravitational force when determining how structures and materials interact with mass.