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Soln\begin{array}{l}=\left(\frac{T L}{G J}\right)_{1}+\left(\frac{T L}{G J}\right)_{2} \\=\frac{50 \times 1.8 \mathrm{Nm} \times \mathrm{m}}{37 \times 10^{9} \mathrm{~N} / \mathrm{m}^{2} \times \frac{\pi \times\left(32 \times 10^{-3}\right)^{4}}{32} \times \mathrm{m}^{4}} \\+\frac{50 \times 0.8 \mathrm{~N} \times \operatorname{m\times m}}{26 \times 10^{9} \mathrm{~N} / \mathrm{m}^{2} \times \frac{\pi \times d_{2}^{4}}{32} \mathrm{~m}^{4}} \\\end{array}Note\theta_{C A}=0.02363+\frac{50 \times 0.8 \times 32}{26 \times 10^{9} \times \pi \times d_{2}^{4}}Given \theta_{C A}=3^{\circ}\begin{array}{l}=\frac{3 \times \pi}{180} \text { radians } \\=0.05236 .\end{array}\begin{array}{l} \therefore 0.05236=0.02363+\frac{1.57 \times 10^{-8}}{d_{2}^{4}} \\0.02873=\frac{1.57 \times 10^{-8}}{d_{2}^{4}} \\d_{2}^{4}=5.46 \times 10^{-7} d_{2}=0.02719 \mathrm{~m} \\\therefore d_{2}=27.19 \mathrm{~mm}\end{array}NOTE:as the same torque is transmitted through the whole steppedshaft, the angle rotation will get added from section to sectiontill the end i.e the support ...