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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
Answer & Solution
Correct Answer:
Option
C
Torque = gear ratio × final drive ratio × overall transmission efficiency
$$\eqalign{ & {\text{T}} = 1.75 \times 4.11 \times \frac{{90}}{{100}} \times 125 \cr & {\text{T}} = 809.1{\text{ Nm}} \cr} $$
$$\eqalign{ & {\text{T}} = 1.75 \times 4.11 \times \frac{{90}}{{100}} \times 125 \cr & {\text{T}} = 809.1{\text{ Nm}} \cr} $$
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