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

Step 1. Calculate the Total Force Acting on Block R: The total force on block R due to its weight is:

\( F = m \times g = 3 \, \text{kg} \times 10 \, \text{m/s}^2 = 30 \, \text{N} \)

Step 2. Determine the Tension T₁ in Wire B: Assuming the system is in equilibrium, the net force acting on P, Q, and R needs to balance out, with wire B supporting the tension:

\( T_1 = F - T_2 = 20 \, \text{N} \)

Step 3. Calculate Longitudinal Strain: Strain = \( \frac{\text{stress}}{Y} \) where stress = \( \frac{T_1}{A} \) and \( A = 0.005 \, \text{cm}^2 = 0.5 \times 10^{-6} \, \text{m}^2 \):

\( \text{strain} = \frac{T_1}{A \times Y} = \frac{20}{0.5 \times 10^{-6} \times 2 \times 10^{11}} = 2 \times 10^{-4} \)