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Physics
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)
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Q. A car moves at a speed of $20 \; ms^{-1}$ on a banked track and describes an arc of a circle of radius $40 \; \sqrt{3}$ m. The angle of banking is $(g = 10 \; ms^{-2})$
KEAM
KEAM 2015
Laws of Motion
A
$25^{\circ}$
4%
B
$60^{\circ}$
5%
C
$45^{\circ}$
10%
D
$30^{\circ}$
78%
E
$40^{\circ}$
78%
Solution:
Given, $V=20\, ms ^{-1}$
$r=40 \sqrt{3}\, m$
$\Rightarrow g=10\, ms ^{-2}$
We know that, $\tan \theta=\frac{v^{2}}{r g}$
$\Rightarrow \tan \theta=\frac{(20)^{2}}{40 \sqrt{3} \times 10}$
$\Rightarrow \tan \theta=\frac{1}{\sqrt{3}}$ or $\theta=\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)$
$\Rightarrow \theta=30^{\circ}$