Problem 909 - JEE Physics

Question

A brass rod of cross sectional area \(1 \mathrm{~cm}^{2}\) and length \(0.2 \mathrm{~m}\) is compressed length wise by a weight of \(5 \mathrm{~kg}\). If young's modulus of elasticity of brass is \(1 \times 10^{11}\left(\mathrm{~N} / \mathrm{m}^{2}\right)\) and \(\mathrm{g}=10\left(\mathrm{~m} / \mathrm{s}^{2}\right)\) Then what will be increase in the energy of rod ? (A) \(10^{-5} \mathrm{~J}\) (B) \(2.5 \times 10^{-5} \mathrm{~J}\) (C) \(5 \times 10^{-5} \mathrm{~J}\) (D) \(2.5 \times 10^{-4} \mathrm{~J}\)

Step-by-Step Solution

Answer

The increase in energy of the rod is \(2.5 \times 10^{-5}\ \mathrm{J}\).

Step 1: Calculate the stress acting on the rod

To calculate the stress acting on the rod, first, we need to find the force applied to the rod due to the weight. The force is given by \(F = m \times g\), where \(m\) is the mass of the weight and \(g\) is the acceleration due to gravity. Given, \(m = 5\ \mathrm{kg}\) and \(g = 10\ \mathrm{m/s^2}\), thus: \[F = 5\ \mathrm{kg} \times 10 \ \mathrm{m/s^2} = 50\ \mathrm{N} \] Next, calculate the stress acting on the rod with formula \(\sigma = \frac{F}{A}\), where \(\sigma\) represents stress, \(F\) is force, and \(A\) is the cross-sectional area. Here, we are given \(A = 1\ \mathrm{cm^2} = 10^{-4}\ \mathrm{m^2}\). So: \[\sigma = \frac{50\ \mathrm{N}}{10^{-4}\ \mathrm{m^2}} = 5 \times 10^5\ \mathrm{N/m^2} \]

Step 2: Calculate the strain on the rod

Next, we will calculate the strain on the rod using Young's modulus formula: \[Y = \frac{\sigma}{\epsilon} \] where \(Y\) is Young's modulus, \(\sigma\) is stress, and \(\epsilon\) is the strain. Rearranging for \(\epsilon\): \[\epsilon = \frac{\sigma}{Y} \] Given, \(Y = 1 \times 10^{11}\ \mathrm{N/m^2}\) and \(\sigma = 5 \times 10^5\ \mathrm{N/m^2}\). Thus: \[\epsilon = \frac{5 \times 10^5\ \mathrm{N/m^2}}{1 \times 10^{11}\ \mathrm{N/m^2}} = 5 \times 10^{-6} \]

Step 3: Calculate increase in the energy of the rod

Now, we need to calculate the increase in the energy stored in the rod due to compression: Using the formula for elastic potential energy, which states that the energy stored in a rod is given by: \[U = \frac{1}{2} V \times \sigma \times \epsilon\] Let's first calculate the volume, \(V\) of the brass rod using the formula \(V = A \times l\), where \(l\) is the length of the rod. Given, \(l = 0.2\ \mathrm{m}\) and \(A = 10^{-4}\ \mathrm{m^2}\): \[V = 10^{-4}\ \mathrm{m^2} \times 0.2\ \mathrm{m} = 2 \times 10^{-5}\ \mathrm{m^3} \] Now, plugging the values of \(\sigma\), \(\epsilon\), and \(V\) into the potential energy formula: \[U = \frac{1}{2} \times 2 \times 10^{-5}\ \mathrm{m^3} \times 5 \times 10^5\ \mathrm{N/m^2} \times 5 \times 10^{-6} \] \[U = 2.5 \times 10^{-5}\ \mathrm{J} \] The increase in energy of the rod is \(2.5 \times 10^{-5}\ \mathrm{J}\). Therefore, the correct option is (B).

Key Concepts

Stress and Strain

Stress and strain are fundamental concepts when studying elasticity and materials under load. **Stress** is defined as the force applied per unit area. In our exercise, the stress is caused by the 5 kg weight compressing the brass rod. Stress can be calculated using the formula:

• \(\sigma = \frac{F}{A}\)

where \(\sigma\) is the stress, \(F\) is the force, and \(A\) is the cross-sectional area.

**Strain** is the measure of deformation representing the displacement between particles in the material body. It is a dimensionless quantity, expressed as:

• \(\epsilon = \frac{{\Delta L}}{L}\)

where \(\Delta L\) is the change in length, and \(L\) is the original length.

In this context, strain is directly proportional to stress according to the Young's modulus of elasticity, making it essential for predicting how materials deform under various loads.

Elastic Potential Energy

Elastic potential energy is the energy stored in an object when it is deformed. For a material like our brass rod, this energy is stored as it is compressed by the applied weight. The formula to calculate the elastic potential energy in a rod is:

• \(U = \frac{1}{2} V \times \sigma \times \epsilon\)

where:

• \(U\) is the elastic potential energy.
• \(V\) is the volume of the rod.
• \(\sigma\) is the stress.
• \(\epsilon\) is the strain.

To find \(V\), use the formula \(V = A \times L\) where \(A\) is the cross-sectional area and \(L\) is the length of the rod. The step-by-step calculation shows how \(U\) was determined to be \(2.5 \times 10^{-5}\) J, emphasizing the role of Young's modulus in calculating how much energy is stored due to the rod's compression.

Compression in Rods

Compression in rods involves reducing the length of the rod by applying a force along its length. This happens commonly in structural materials where they are subjected to axial compression forces. In this exercise, the brass rod underwent compression due to the applied weight of 5 kg.

The interaction of stress and strain during compression is described by Young's modulus, a measure of the ability of a material to withstand changes in length when under lengthwise tension or compression. It is crucial for understanding how materials behave under various load conditions:

• A higher Young's modulus indicates a stiffer material.
• A compressional force will cause a decrease in length, which results in a strain \(\epsilon\) in the rod.

Applying these principles, we can predict and calculate the length changes in the rod and the energy stored during the compression process, as shown in the solution where the calculated elastic potential energy shows the efficiency of energy storage due to the material's properties.