Correct Answer - Option 1 : 3

**Given:**

A 4-stroke, 6 – cylinder gas engine

**Volume of stroke** = 1.75 ℓ

**Power developed **= 26.25 kW,** ω** = 506 rpm.

**Mean effective pressure (MEP)** = 600 kN/m^{2}

To find no. of Misfires per minute

**Calculation:**

**Total volume swept (V**_{s}) = 1.75 ℓ = 1.75 × 10^{-3} m^{3}

Let N be the no. of fires in which engine produces power per minute

\({\rm{Power}} = \frac{{\left( N \right)}}{{60}}\;{P_{mean}} \times {V_s}\)

\({\rm{Power}} = \frac{N}{{60}}\;\left( {600} \right) \times {10^3}\left( {1.75} \right) \times {10^{ - 3}}\) -------- (1)

Also power = 26.25 × 10^{3} W ------ (2) **(Given)**

**Now, equating equation 1 & 2**

\(\therefore \left( {\frac{N}{{60}}} \right)\left( {600} \right) \times {10^3}\left( {1.75} \right) \times {10^{ - 3}} = 26.25 \times {10^3}\)

**∴ ****N = 1500 **

**Hence there should be 1500 fires.**

Now, No. of fires expected per minute for on engine speed of 506 rpm is:

\({N_{actual}} = \left( {\frac{{506}}{2}} \right) \times No.\;of\;cylinders = \left( {253} \right)6\)

N_{actual}= 1518.

**But by calculation we can see that there are 1518 fires.**

∴ Number of misfires for engine per minute = N – N_{actual} = 18.

∴ Misfires for engine per minute per cylinder is \(\frac{{18}}{6} = 3\) (since there are 6 cylinders)

**∴**** Number of misfires per minute per cylinder will be 3.**