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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
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The correct answer is:
$30 \mathrm{~ms}^{-1}$
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 $\operatorname{car} v=30 \mathrm{~m} / \mathrm{s}$
$\tan \theta=\frac{v^2}{r g}$
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 $\operatorname{car} v=30 \mathrm{~m} / \mathrm{s}$
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