Let's break down and solve the problem step by step: ### Given: - **Shear modulus (G)** = 77 GPa = \(77 \times 10^9\) Pa - **Diameter (d)** = 20 mm = 0.02 m - **Torque (T)** = 12 N·m - **Allowable angular twist (θ)** = 0.0584 rad ### Formula for angular twist: \[ \theta = \frac{T L}{G J} \] Where: - \(\theta\): Angular twist (radians) - \(T\): Torque (N·m) - \(L\): Length of shaft (m) - \(G\): Shear modulus (Pa) - \(J\): Polar moment of inertia (m\(^4\)), for a solid shaft: \[ J = \frac{\pi d^4}{32} \] ### Rearranging for \(L\): \[ L = \frac{\theta G J}{T} \] ### Calculate \(J\): \[ J = \frac{\pi}{32} \times (0.02)^4 \] \[ = \frac{3.1416}{32} \times (0.00016) \] \[ = 0.09817 \times 0.00016 \] \[ \approx 1.5708 \times 10^{-5} \text{ m}^4 \] ### Substitute all values: \[ L = \frac{0.0584 \times 77 \times 10^9 \times 1.5708 \times 10^{-5}}{12} \] Calculate step by step: 1. \(77 \times 10^9 \times 1.5708 \times 10^{-5} = 1.209516 \times 10^6\) 2. \(0.0584 \times 1.209516 \times 10^6 = 70,670.8\) 3. \(L = \frac{70,670.8}{12} = 5,889.23\) meters (This value is too high; let's check the calculation again for correct unit conversion and values.) Let's recalculate: \[ J = \frac{\pi}{32} \times (0.02)^4 = \frac{3.1416}{32} \times 1.6 \times 10^{-7} = 0.098175 \times 1.6 \times 10^{-7} = 1.5708 \times 10^{-8} \text{ m}^4 \] Now recalculate: \[ L = \frac{0.0584 \times 77 \times 10^9 \times 1.5708 \times 10^{-8}}{12} \] \[ = \frac{0.0584 \times 1,208,516}{12} = \frac{70,584.36}{12} = 5,882 \text{ m} \] But this value is still too high. Let's check units: - \(G = 77 \times 10^9\) Pa - \(J = 1.5708 \times 10^{-8}\) m\(^4\) - \(T = 12\) Nm \[ 0.0584 \times 77 \times 10^9 = 4,496,800,000 \] \[ 4,496,800,000 \times 1.5708 \times 10^{-8} = 70.61 \] \[ L = \frac{70.61}{12} = 5.88 \text{ m} \] ### Final Answer: The maximum possible length is **5.88 m**. ### Closest option: - **L = 5.886 m** (Correct answer) --- **Final selection: L = 5.886 m**