Correct Answer - a
Here, mass of the car, M=1800 kg
Distance between front and back axles = 1.8 m
Distance of gravity G behind the front axle = 1m
Let `R_(F)` and `R_(B)` be the forces exerted by the level ground on each front wheel and each back wheel.
For translation equilibrium,
`2R_(F)+2R_(B)=M_(g)`
or `R_(F) + R_(B)= (Mg)/(2)= (1800 xx 10)/(2) = 9000N`............(i)
(As there are two front wheels and two back wheels)
For rotational equilibrium about G
`(2F_(F))(1) = (2R_(B))(0.8)=8/10=4/5 rArr R_(F)=4/5R_(B)`.............(ii)
Substituting this in Eq. (i), we get
`4/5R_(B)+R_(B)= 9000` or `9/5R_(B)=9000`
`R_(B)= (9000 xx 5)/(9) = 5000 N`
`therefore F_(F)= 4/5 F_(B) = 4/5 xx 5000 N=4000N`