A car moves at a speed of 20 ms-1 on a banked track and describes an arc of a circle of radius 40 √3 m. The angle of banking is (g = 10 ms-2)

Question

A car moves at a speed of 20 ms-1 on a banked track and describes an arc of a circle of radius 40 √3 m. The angle of banking is (g = 10 ms-2) (A) 25° (B) 60° (C) 45° (D) 30°

Tardigrade Admin · Accepted Answer

Given, V=20 ms -1 r=40 √3 m ⇒ g=10 ms -2 We know that, tan θ=(v2/r g) ⇒ tan θ=((20)2/40 √3 × 10) ⇒ tan θ=(1/√3) or θ= tan -1((1/√3)) ⇒ θ=30°

Q. A car moves at a speed of $20 m s^{- 1}$ on a banked track and describes an arc of a circle of radius $403$ m. The angle of banking is $(g = 10 m s^{- 2})$

$2 5^{\circ}$

$6 0^{\circ}$

$4 5^{\circ}$

$3 0^{\circ}$

$4 0^{\circ}$

Given, $V = 20 m s^{- 1}$
$r = 403 m$
$\Rightarrow g = 10 m s^{- 2}$
We know that, $tan θ = \frac{v ^{2}}{r g}$
$\Rightarrow tan θ = \frac{( 20 ) ^{2}}{40 3 \times 10}$
$\Rightarrow tan θ = \frac{1}{3}$ or $θ = tan^{- 1} (\frac{1}{3})$
$\Rightarrow θ = 3 0^{\circ}$

Q. A car moves at a speed of 20ms−1 on a banked track and describes an arc of a circle of radius 403​ m. The angle of banking is (g=10ms−2)

25∘

60∘

45∘

30∘

40∘

Given, V=20ms−1 r=403​m ⇒g=10ms−2 We know that, tanθ=rgv2​ ⇒tanθ=403​×10(20)2​ ⇒tanθ=3​1​ or θ=tan−1(3​1​) ⇒θ=30∘