Let's 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 angle of twist, \( \theta = 0.0584 \) rad **Required:** Maximum possible length \( L \) of the shaft. --- ### Formula for Angle of Twist \[ \theta = \frac{T L}{G J} \] Where: - \( \theta \) = angle of twist in radians - \( T \) = applied torque - \( L \) = length of shaft - \( G \) = shear modulus - \( J \) = polar moment of inertia for a circular shaft \[ J = \frac{\pi d^4}{32} \] --- ### Substitute Values First, calculate \( J \): \[ J = \frac{\pi}{32} \times (0.02)^4 \] \[ J = \frac{\pi}{32} \times 1.6 \times 10^{-7} \] \[ J \approx 1.5708 \times 10^{-8} \, \text{m}^4 \] Now rearrange the angle of twist equation to solve for \( L \): \[ L = \frac{\theta G J}{T} \] --- ### Plug in the Values \[ L = \frac{0.0584 \times 77 \times 10^9 \times 1.5708 \times 10^{-8}}{12} \] Calculate numerator: \[ 0.0584 \times 77 \times 10^9 = 4.4968 \times 10^9 \] \[ 4.4968 \times 10^9 \times 1.5708 \times 10^{-8} \approx 70.603 \text{ (since } 10^9 \times 10^{-8} = 10^1 = 10) \] Now divide by 12: \[ L = \frac{70.603}{12} \approx 5.8836 \text{ m} \] --- ### Closest Answer The closest answer is: **L = 5.886 m** --- **So, the correct choice is:** > L = 5.886 m Let me know if you need a breakdown or more explanation!