Q106 / 200 · NEET 2012 (Screening) · Physics
A car of mass $1000 \mathrm{~kg}$ negotiates a banked curve of radius $90 \mathrm{~m}$ on a frictionless road. If the banking angle is $45^{\circ}$, the speed of the car is
- A$20 \mathrm{~ms}^{-1}$
- Correct answerB$30 \mathrm{~ms}^{-1}$
- C$5 \mathrm{~ms}^{-1}$
- D$10 \mathrm{~ms}^{-1}$
Solution
The angle of banking
$\tan \theta=\frac{v^2}{r g}$
Given, $\theta=45^{\circ}, r=90 \mathrm{~m}$
and $g=10 \mathrm{~m} / \mathrm{s}^2$
$\begin{aligned}
\tan 45^{\circ} & =\frac{v^2}{90 \times 10} \\
v & =\sqrt{90 \times 10 \times \tan 45^{\circ}}
\end{aligned}$
Speed of $\operatorname{car} v=30 \mathrm{~m} / \mathrm{s}$
Hence, the correct answer is option: 'B'.