Correct Answer - Option 3 : 66.67%
Concept:
The propulsion efficiency is defined as the ratio of Propulsive power (or) thrust power to the power - output of the engine.
It is given as
\(\eta_p = \frac {2u}{C_j + u}\)
Where Cj – Exhaust gases velocity, u – Flight speed;
Propulsive efficiency is the measure of the effectiveness with which kinetic energy imparted to the fluid is transferred into useful work.
Calculation:
Given u = 1400 m/s, Cj = 2800 m/s;
From the propulsive efficiency relation,
\({\eta _p} = \;\frac{{2{u}}}{{{c_j} + {u}}} = \frac{{2 \times 1400}}{{2800 + 1400}} = 66.67\;\% \)