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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 ${45^ \circ }$ , the speed of the car is:A. $5\,m{s^{ - 1}}$B. $10\,m{s^{ - 1}}$C. $20\,m{s^{ - 1}}$D. $30\,m{s^{ - 1}}$

Last updated date: 23rd May 2024
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Hint: Here we need to find the centripetal velocity required to overcome the banked curve. The concept of uniform circular motion is applied here.

Radius of banked curve, $r = 3\,m$

Mass of car, $m = 1000\,kg$

Acceleration due to gravity, $g = 10m{s^{ - 2}}$

Banking angle of the frictionless road,
$\theta = {45^ \circ }$

Speed of the car =?
We know that,
The velocity due to centripetal action is given by
$v = \sqrt {rg} \tan \theta \\$
$v = \sqrt {90 \times 10} \times 1 = 30m{s^{ - 1}} \\$

Therefore, the speed of the car when the banking angle is ${45^ \circ }$
is $30\,m{s^{ - 1}}$
.
Hence, option D is correct.