Two cars start from the same place at the same time at right angles to each other. Their speeds are 54 km/hr and 72 km/hr, respectively. After 20 seconds the distance between them will be:

Let the cars start at a point A

Speed of first car = 54 km/hr =  $$54\times\frac{5}{18}$$ m/s = 15 m/s

Distance travelled by first car in 20 seconds = $$15\times20$$ = 300 m

Speed of second car = 72 km/hr = $$72\times\frac{5}{18}$$ = 20 m/s

Distance travelled by second car in 20 seconds = $$20\times20$$ = 400 m

Two cars move at right angle to each other

Let the position of the cars are B and C after 20 seconds respectively as shown in figure

From the figure,

$$AC^2+AB^2=BC^2$$

$$=$$>  $$400^2+300^2=BC^2$$

$$=$$>  $$1600+900=BC^2$$

$$=$$>  $$BC^2=2500$$

$$=$$>  $$BC=500$$ m

$$\therefore\ $$Distance between the cars after 20 seconds = BC = 500 m

Hence, the correct answer is Option C