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A road is $10m$ wide. Its radius of curvature is $50m.$The outer edge is above the lower edge by a distance of $1.5m$. This road is most suited for the velocity:
A) $2.5m{s^{ - 1}}$
B) $4.5m{s^{ - 1}}$
C) $6.5m{s^{ - 1}}$
D) $8.5m{s^{ - 1}}$

Last updated date: 15th Jun 2024
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Hint: This question is directly formula based. Since, in the question, radius of curvature is given, so here in this case we need to use the formula of velocity profile directly. After that we can directly get the solution for the given question.

Complete step by step solution:
As, in the given question, width of the road is given, $L = 10cm$
Radius or curvature, $R = 50cm$
The distance between the above and lower edge of the road, $h = 1.5m$
If a body is moving on the road, then a force will be acting on the body and a downward force due to its weight will also act on the body. There will be components of the force which will act on the body which is moving on the with a velocity on the road, which is given as,
$N\sin \theta = \dfrac{{m{v^2}}}{R}$………………(i)
And $N\cos \theta = mg$…………………(ii)
Now, when we divide equation (i) by (ii), we get,
$\tan \theta = \dfrac{{{v^2}}}{{gR}}$
Now, for most suited velocity, we can use the formula for velocity profile,
$v = \sqrt {gR\tan \theta } $………………(iii)
Also, we know that $\tan \theta = \dfrac{h}{L}$
Now, we need to put the values in equation (iii)
So, we will get, $v = \sqrt {10 \times 50 \times \dfrac{{1.5}}{{10}}} $
$ \Rightarrow v = \sqrt {50 \times 1.5} $
$ \Rightarrow v = \sqrt {75} $
$\therefore v = 8.5m{s^{ - 1}}$

Hence, option (D), i.e. $8.5m{s^{ - 1}}$ is the correct answer for the given question.

Note: Here, in this question, the angle of inclination is not given. But the width of the road and the distance between the upper edge and the lower edge is given, so we used the relation$\tan \theta = \dfrac{h}{L}$. If the angle would be given then directly we can use the value of$\tan \theta $. Now, the equation of velocity is given by$v = \sqrt {gR\tan \theta } $.