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A truck of mass $2000 \mathrm{~kg}$ is moving along a circular path having radius of curvature $10 \mathrm{~m}$. if the banking angle is $39^{\circ}$, then the maximum permissible speed of the truck is (Acceleration due to gravity $=10 \mathrm{~ms}^{-2}$, take $\tan 39^{\circ}=$ $0.81)$
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Verified Answer
The correct answer is:
$9 \mathrm{~ms}^{-1}$
Mass of truck, M=2000 kg
Radius of curvature, $\mathrm{R}=10 \mathrm{~m}$
Banking angle, $a=39^{\circ}, \tan 39^{\circ}=0.81$
Maximum speed is given as:
$$
\begin{aligned}
& v=\sqrt{\mu R g}=\sqrt{R g \tan \theta} \\
& v=\sqrt{10 \times 10 \times 0.81} \\
& v=9 \mathrm{~m} / \mathrm{s}
\end{aligned}
$$
Radius of curvature, $\mathrm{R}=10 \mathrm{~m}$
Banking angle, $a=39^{\circ}, \tan 39^{\circ}=0.81$
Maximum speed is given as:
$$
\begin{aligned}
& v=\sqrt{\mu R g}=\sqrt{R g \tan \theta} \\
& v=\sqrt{10 \times 10 \times 0.81} \\
& v=9 \mathrm{~m} / \mathrm{s}
\end{aligned}
$$
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