A bus weighing 100 quintals moves on a rough road with a constant speed of 72km/h. The friction of the road is 9% of its weight and that of air is 1% of its weight. What is the power of the engine. Take g = 10m/s²

To find the power of the engine, we can break down the problem into systematic, easy-to-follow steps.

Step 1: Convert the given values to standard SI units.
The mass of the bus is given in quintals. 1 quintal is equal to 100 kg.
Mass, $m=100\text{quintals}=100\times 100\text{kg}=\mathrm{10,000}\text{kg}$.
The acceleration due to gravity is:
$g=10{\text{m/s}}^2$.
Therefore, the weight of the bus ($W$) is:
$W=mg=\mathrm{10,000}\text{kg}\times 10{\text{m/s}}^2=\mathrm{100,000}\text{N}$.

The constant speed ($v$) of the bus is given in km/h. To convert it to m/s, we multiply by $\frac{5}{18}$:
$v=72\text{km/h}=72\times \frac{5}{18}\text{m/s}=20\text{m/s}$.

Step 2: Calculate the total opposing force.
The bus experiences two resistive forces:
1. Road friction force ($F_{\text{road}}$), which is 9% of the weight of the bus:
$F_{\text{road}}=9\%\text{of}W=0.09\times \mathrm{100,000}\text{N}=\mathrm{9,000}\text{N}$.
2. Air resistance force ($F_{\text{air}}$), which is 1% of the weight of the bus:
$F_{\text{air}}=1\%\text{of}W=0.01\times \mathrm{100,000}\text{N}=\mathrm{1,000}\text{N}$.

The total resisting force ($F$) that the engine must overcome to maintain constant speed is the sum of these forces:
$F=F_{\text{road}}+F_{\text{air}}=\mathrm{9,000}\text{N}+\mathrm{1,000}\text{N}=\mathrm{10,000}\text{N}$.

Step 3: Calculate the power of the engine.
Power ($P$) is defined as the product of the force and the velocity when the speed is constant:
$P=F\times v$.
Substituting the values we have calculated:
$P=\mathrm{10,000}\text{N}\times 20\text{m/s}=\mathrm{200,000}\text{W}$.

Converting Watts to kilowatts (kW):
$P=\frac{\mathrm{200,000}}{\mathrm{1,000}}\text{kW}=200\text{kW}$.

Thus, the power of the engine is 200 kW, which corresponds to the correct option.