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4-1.Newton's Laws of Motion
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A car of mass $1000\, kg$ negotiates a banked curve of radius $90\, m$ on a frictionless road. If the banking angle is $45^o $, the speed of the car is ....... $ms^{-1}$
A$20$
B$10$
C$30$
D$5$
(AIPMT-2012)
Solution
$Here,\,m = 1000\,kg.\,R = 90\,m,\,\theta = {45^ \circ }$
For$\,banking,\tan \theta = \frac{{{v^2}}}{{Rg}}$
or$\,v = \sqrt {Rg\tan \theta } = \sqrt {90 \times 10 \times \tan {{45}^ \circ }}$
$= 30\,m\,{s^{ – 1}}$
For$\,banking,\tan \theta = \frac{{{v^2}}}{{Rg}}$
or$\,v = \sqrt {Rg\tan \theta } = \sqrt {90 \times 10 \times \tan {{45}^ \circ }}$
$= 30\,m\,{s^{ – 1}}$
Standard 11
Physics
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