Correct Answer - Option 3 : remain same

__Concept:__

The power transmitting capacity of the shaft is given by

\(P = \frac{{2\pi NT}}{{60}}\)

P ∝ NT

Torsion equation:

\(\frac{T}{J} = \frac{{{\tau _{max}}}}{R} = \frac{{G\theta }}{l}\)

\(T = {\tau _{max}} \times \frac{J}{R}\)

\(J = \frac{\pi }{{32}} \times {D^4}\)

\(T = \frac{{{\tau _{max}}\; \times \;\pi \; \times {D^3}\;}}{{16}}\)

T ∝ D^{3}

P ∝ NT

P ∝ ND^{3}

__Calculation__:

**Given**:

P_{2} = 2P_{1}, N_{2} = 2N_{1}

\(\frac{P_2}{P_1}=\frac{N_2}{N_1}\times\frac{D_2^3}{D_1^3}\)

\(\frac{2P_1}{P_1}=\frac{2N_1}{N_1}\times\frac{D_2^3}{D_1^3}\)

∴ **D**_{1} = D_{2}