Laws of Motion - Result Question 59
59. 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) $20 ms^{-1}$
(b) $30 ms^{-1}$
(c) $5 ms^{-1}$
(d) $10 ms^{-1}$
[2012]
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Answer:
Correct Answer: 59. (c)
Solution:
- (c) For banking, $\tan \theta=\frac{V^{2}}{R g}$
$\tan 45^{\circ}=\frac{V^{2}}{90 \times 10}=1$
$V=30 m / s$
If friction is present then maximum safe speed, on
a banked frictional road, $V=\sqrt{\frac{R g(\mu+\tan \theta)}{1-\mu \tan \theta}}$
