Correct Answer - Option 4 : 157 N-m

__Concept:__

Using torsion equation:

\(\frac{T}{{{I_P}}} = \frac{{{{\rm{τ }}_{max}}}}{R} = \frac{{G\theta }}{L}\)

Where,

T = applied torque, IP = Polar section modulus, τmax = Maximum shear stress in the shaft material, R = Radius of the shaft, G = Modulus of Rigidity, θ = Angle of twist, L = Length of the shaft

\({I_P} = \frac{π }{{32}}{D^4}\)

R = \(\frac{D}{2}\)

From shear stress criteria:

\(\frac{{{{\rm{τ }}_{{\rm{max}}\left( {allowable} \right)}} \times 2}}{D} = \frac{T}{{{I_P}}}\)

Substituting the values of \({I_P} \) and R we get

Torque (T) = \(\frac{π }{{16}}τ {d^3}\)

__Calculation:__

__Given:__

d = 2 m, τ = 100 N/m2

T = \(\frac{π }{{16}}τ {d^3}\)

Taking π = 3.14

T = \(\frac{3.14\;\times\;100\;\times\;2^3}{16}\) = 157 N-m