Let's solve the problem step by step. ### Given Data: - Shear modulus, \( G = 77 \) GPa = \( 77 \times 10^9 \) Pa - Diameter, \( d = 20 \) mm = \( 0.02 \) m - Torque, \( T = 12 \) N·m - Maximum allowable angular twist, \( \theta = 0.0584 \) rad ### Formula for Angular Twist: \[ \theta = \frac{T L}{J G} \] Where: - \( T \) = Applied Torque - \( L \) = Length of the shaft - \( J \) = Polar moment of inertia \( = \frac{\pi d^4}{32} \) - \( G \) = Shear modulus ### Rearranging for \( L \): \[ L = \frac{\theta J G}{T} \] ### Calculating \( J \): \[ J = \frac{\pi d^4}{32} \] \[ J = \frac{\pi (0.02)^4}{32} \] \[ J = \frac{\pi \times 1.6 \times 10^{-7}}{32} \] \[ J \approx \frac{5.0265 \times 10^{-7}}{32} \] \[ J \approx 1.57 \times 10^{-8} \text{ m}^4 \] ### Plug in the values: \[ L = \frac{0.0584 \times 1.57 \times 10^{-8} \times 77 \times 10^9}{12} \] Calculate numerator: \[ 0.0584 \times 1.57 \times 10^{-8} = 9.1688 \times 10^{-10} \] \[ 9.1688 \times 10^{-10} \times 77 \times 10^9 = 70.59876 \] Now divide by \( 12 \): \[ L = \frac{70.59876}{12} \approx 5.883 \text{ m} \] ### Closest Answer: \[ \boxed{L = 5.886 \text{ m}} \] So, the correct answer is: - **L = 5.886 m**