Question

Each of three blocks \( P \), \( Q \), and \( R \) shown in the figure has a mass of 3 kg. Each of the wires \( A \) and \( B \) has a cross-sectional area of 0.005 cm\(^2\) and Young's modulus \( 2 \times 10^{11} \, \text{N} \, \text{m}^{-2} \). Neglecting friction, the longitudinal strain on wire \( B \) is \( \ldots \times 10^{-4} \). (Take \( g = 10 \, \text{m/s}^2 \))

Answer 1

Correct Answer: 2

This problem requires the calculation of the longitudinal strain in wire B within a system of interconnected blocks and wires. The provided information includes the masses of the blocks, the cross-sectional area and Young's modulus of the wires, and the acceleration due to gravity.

Underlying Principles:

The solution employs principles from mechanics and material science:

1. Newton's Second Law of Motion: To determine the tension in the wires, the acceleration of the block system must first be calculated. The acceleration (\(a\)) is determined by the formula \(a = \frac{F_{net}}{M_{total}}\), where \(F_{net}\) represents the net external force and \(M_{total}\) is the total mass undergoing acceleration.

2. Young's Modulus (Y): Young's modulus establishes a relationship between stress (\(\sigma\)) in a material and the resulting longitudinal strain (\(\epsilon\)). The governing equation is:

\[Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\epsilon}\]

In this equation, \(F\) denotes the tension in the wire and \(A\) is its cross-sectional area. Consequently, the strain can be computed as:

\[\epsilon = \frac{\text{Stress}}{Y} = \frac{F}{A \cdot Y}\]

Detailed Solution:

Step 1: System Acceleration Calculation.

The system comprises three blocks (P, Q, and R), each with a mass of 3 kg. The downward pull of the hanging block R serves as the driving force. The total mass subject to acceleration is the combined mass of all three blocks.

Net driving force, \(F_{net} = m_R \cdot g = 3 \, \text{kg} \times 10 \, \text{m/s}^2 = 30 \, \text{N}\).

Total mass, \(M_{total} = m_P + m_Q + m_R = 3 + 3 + 3 = 9 \, \text{kg}\).

The acceleration of the system is calculated as:

\[a = \frac{F_{net}}{M_{total}} = \frac{30 \, \text{N}}{9 \, \text{kg}} = \frac{10}{3} \, \text{m/s}^2\]

Step 2: Tension Calculation for Wire B (\(T_B\)).

Wire B is attached to block R. The tension can be found by analyzing the free-body diagram of block R. The forces acting on R are its weight (\(m_R g\)) directed downwards and the tension (\(T_B\)) acting upwards. The net force results in a downward acceleration \(a\).

\[m_R g - T_B = m_R a\]\[30 \, \text{N} - T_B = 3 \, \text{kg} \times \frac{10}{3} \, \text{m/s}^2\]\[30 - T_B = 10\]\[T_B = 20 \, \text{N}\]

This tension \(T_B\) represents the force exerted on wire B.

Step 3: Longitudinal Strain Calculation for Wire B.

First, convert the cross-sectional area to the standard SI unit (m²).

\[A = 0.005 \, \text{cm}^2 = 0.005 \times (10^{-2} \, \text{m})^2 = 0.005 \times 10^{-4} \, \text{m}^2 = 5 \times 10^{-7} \, \text{m}^2\]

The provided Young's modulus is \(Y = 2 \times 10^{11} \, \text{N/m}^2\).

Apply the formula for strain:

\[\text{Strain on B} = \frac{T_B}{A \cdot Y}\]\[\text{Strain on B} = \frac{20 \, \text{N}}{(5 \times 10^{-7} \, \text{m}^2) \times (2 \times 10^{11} \, \text{N/m}^2)}\]

Final Calculation and Outcome:

Simplify the strain expression.

\[\text{Strain on B} = \frac{20}{5 \times 2 \times 10^{-7} \times 10^{11}}\]\[\text{Strain on B} = \frac{20}{10 \times 10^{4}} = \frac{2}{10^{4}} = 2 \times 10^{-4}\]

The problem requires the value of x, where the strain is expressed as \(x \times 10^{-4}\).

By direct comparison, \(x = 2\).

The longitudinal strain experienced by wire B is 2 \( \times 10^{-4} \).