Correct Answer - Option 2 : 16

__Concept:__

Maximum permissible shear stress can be calculated using:

\({τ _{max}} = \frac{T}{{{Z_p}}}\)

where Z_{p} = polar moment of inertia, T = torque applied

__Given:__

T = 50 Nm, τ_{allowable} = 140 MPa

F0S = 2

\({τ _{max}} = \frac{{{τ _{allowable\;}}}}{{FOS}} = \frac{{140}}{{2}}=70~MPa\)

Polar moment of inertia for the circular road is:

\({Z_P} = \frac{{\pi {d^3}}}{{16}}\)

Now maximum shear stress is

\({τ _{max}} = \frac{T}{{{Z_p}}}\)

⇒ \({Z_p} = \frac{T}{{{τ _{max}}}}\)

\(\frac{{\pi {d^3}}}{{16}} = \frac{T}{{{τ _{max\;}}}}\)

\(d = \sqrt[3]{{\frac{{16T}}{{\pi {τ _{max}}}}}} = \sqrt[3]{{\frac{{16 \times 50}}{{\pi \times 70 \times {{10}^6}}}}}\)

d = 0.01537 m = 15.37 mm ≃ **16 mm**