Correct Answer - Option 4 : 128.6 mm

__Concept:__

**Torque transmitted by the shaft**

**\(P = \frac{{2 \times \pi \times N \times T}}{{60}}\;\)**

**Also, shear stress is given by:**

**\(\tau = \frac{{16 \times T}}{{\pi \times {d^3}}}\;\)**

P = power, N = speed, T = torque, d = diameter of shaft, τ = shear stress

__Calculation:__

__Given:__

P = 3.5 MW = 3.5 × 10^{6} W, N = 200 rpm, τ = 50 MPa = 50 × 10^{6} Pa

We have,

\(P = \frac{{2 \times \pi \times N \times T}}{{60}} \Rightarrow T = \frac{{3.5 \times {{10}^6} \times 60}}{{2 \times \pi \times 200}} = 167.112 \times {10^3}\;N/m\;\)

Also, shear stress is:

\(\tau = \frac{{16 \times T}}{{\pi \times {d^3}}}\;\)

\(50 \times {10^6} = \frac{{16 \times 167.112 \times {{10}^3}}}{{\pi \times {d^3}}}\; \Rightarrow {d^3} = \frac{{16 \times 167.112 \times {{10}^3}}}{{50 \times {{10}^6}}}\)

**d = 0.2572 m**

The radius of the shaft

\(r = \frac{{0.2572}}{2} = 0.1286\;m = 128.6\;mm\)