Let's solve the problem step by step: **Given:** - Modulus of rigidity, \( G = 77 \text{ GPa} = 77 \times 10^9 \text{ Pa} \) - Diameter of shaft, \( d = 20 \text{ mm} = 0.02 \text{ m} \) - Torque, \( T = 12 \text{ Nm} \) - Allowable angular twist, \( \theta = 0.0584 \text{ rad} \) **Formula for the angle of twist:** \[ \theta = \frac{T L}{G J} \] Where: - \( \theta \) = angle of twist (radians) - \( T \) = applied torque (Nm) - \( L \) = length of shaft (m) - \( G \) = modulus of rigidity (Pa) - \( J \) = polar moment of inertia (for a solid shaft, \( J = \frac{\pi d^4}{32} \)) **Step 1: Calculate the polar moment of inertia \( J \)** \[ J = \frac{\pi d^4}{32} \] \[ d = 0.02 \text{ m} \] \[ J = \frac{\pi}{32} \times (0.02)^4 \] \[ J = \frac{\pi}{32} \times (1.6 \times 10^{-7}) \] \[ J = 1.57 \times 10^{-8} \text{ m}^4 \] **Step 2: Rearrange the formula to solve for \( L \):** \[ \theta = \frac{T L}{G J} \implies L = \frac{\theta G J}{T} \] **Step 3: Substitute the values:** \[ L = \frac{0.0584 \times 77 \times 10^9 \times 1.57 \times 10^{-8}}{12} \] **Step 4: Calculate the numerator:** \[ 0.0584 \times 77 \times 10^9 \times 1.57 \times 10^{-8} = 0.0584 \times 77 \times 1.57 \times 10^{1} = 0.0584 \times 77 \times 1.57 \times 10 = 0.0584 \times 77 \times 15.7 = 0.0584 \times 1208.9 = 70.642 \text{ (approx)} \] **Step 5: Divide by the torque:** \[ L = \frac{70.642}{12} \approx 5.886 \text{ m} \] **Final Answer:** The maximum possible length of the shaft is **5.886 m**. So, the correct option is: - **L = 5.886 m**