A petrol engine of a car develops 125 Nm torque at 2700 r.p.m. The car is driven in second gear having gear ratio of 1.75. The final drive ratio is 4.11. If the overall transmission efficiency is 90%, then the torque available at the driving wheels is
A. 8.091 Nm
B. 80.91 Nm
C. 809.1 Nm
D. 8091 Nm
Answer: Option C
Solution(By Examveda Team)
Torque = gear ratio × final drive ratio × overall transmission efficiency$$\eqalign{ & {\text{T}} = 1.75 \times 4.11 \times \frac{{90}}{{100}} \times 125 \cr & {\text{T}} = 809.1{\text{ Nm}} \cr} $$
This Question Belongs to Mechanical Engineering >> Automobile Engineering
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