A racing car of mass 1000 kg moves round a banked track at a constant speed of 108 kmh -1. Assuming the total reaction at the wheels is normal to the track and the horizontal radius of the track is 100 m , the angle of inclination of the track to the horizontal is (g=10 ms -2)

Question

A racing car of mass 1000 kg moves round a banked track at a constant speed of 108 kmh -1. Assuming the total reaction at the wheels is normal to the track and the horizontal radius of the track is 100 m , the angle of inclination of the track to the horizontal is (g=10 ms -2) (A) 12° (B) 27° (C) 42° (D) 65°

Tardigrade Admin · Accepted Answer

tan θ=(v2/Rg) v=108 kmhr -1=108 × (5/18) ms -1=30 ms -1 R=100 m ⇒ tan θ=(302/100 × 10)=(9/10) θ= tan -1((9/10))=42°

Q. A racing car of mass 1000 kg moves round a banked track at a constant speed of $108 km h^{- 1}$ . Assuming the total reaction at the wheels is normal to the track and the horizontal radius of the track is $100 m$ , the angle of inclination of the track to the horizontal is $(g = 10 m s^{- 2})$

$1 2^{\circ}$

$2 7^{\circ}$

$4 2^{\circ}$

$6 5^{\circ}$

$tan θ = \frac{v ^{2}}{R g}$
$v = 108 kmh r^{- 1} = 108 \times \frac{5}{18} m s^{- 1} = 30 m s^{- 1}$
$R = 100 m$
$\Rightarrow tan θ = \frac{3 0 ^{2}}{100 \times 10} = \frac{9}{10}$
$θ = tan^{- 1} (\frac{9}{10}) = 4 2^{\circ}$

Q. A racing car of mass 1000 kg moves round a banked track at a constant speed of 108kmh−1. Assuming the total reaction at the wheels is normal to the track and the horizontal radius of the track is 100m , the angle of inclination of the track to the horizontal is (g=10ms−2)

12∘

27∘

42∘

65∘