Correct Answer - Option 1 : 5/3 kW

__Concept: __

**Indicated Power for four-stroke engine is given by,**

\(I.P.~=~\frac{P_mLAN}{60,000× 2}\)

where P_{m} = mean effective pressure in N/m^{2}, L = stroke length in m, A = Area of cross-section of the cylinder in m^{2}, N = rpm of the engine crankshaft

\(η_m~=~\frac{B.P.}{I.P.}\)

where η_{m} = mechanical efficiency, B.P. = brake power, I.P. = Indicated power

__Calculation:__

__Given:__

N = 1000 rpm, P_{m} = 400 kPa = 400 × 10^{3} Pa, η_{m }= 0.5, V_{s} = L × A = 1000 cm^{3 }= 1000 × 10^{-6} m^{3 }

From the given data we have,

\(I.P.~=~\frac{400~×~1000~× ~1000~× ~10^{-6}~×~1000} {60,000~× ~2}~=~\frac{10}{3}~kW\)

Now, by using the value of I.P. we will get B.P. as follows,

B.P. = η_{m} × I.P.

\(B.P.~=~\frac{0.5~\times ~10}{3}~=~\frac{5}{3} ~kW\)