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

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 _T 1_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 $A = 0.005 \, \text{cm}^2 = 0.5 \times 10^{-6} \, \text{m}^2$:

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