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A car is moving with velocity $v$ at the top of a semi-circular hill of radius $40 \mathrm{~m}$ such that the normal force on it is zero. Find the velocity $(v)$ of the car. [Use, $g=10 \mathrm{~ms}^{-2}$ ]
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Verified Answer
The correct answer is:
$20 \mathrm{~ms}^{-1}$
Given that, $g=10 \mathrm{~m} / \mathrm{s}^2$
Velocity of car at top of hil $\mathrm{l}=v$
Radius of semi-circular hill, $R=40 \mathrm{~cm}$
Let mass of car $=m$
Now, the centripetal force is equal to the net force acting on car at top.
$$
\begin{array}{rlrl}
\therefore \quad F_c =m g-N \quad \text { [Given, } N=0 \text { ] } \\
\frac{m v^2}{R} =m g & \\
\Rightarrow \quad =\sqrt{g R} =\sqrt{10 \times 40} \\
= 20 \mathrm{~m} / \mathrm{s}
\end{array}
$$
Velocity of car at top of hil $\mathrm{l}=v$
Radius of semi-circular hill, $R=40 \mathrm{~cm}$
Let mass of car $=m$
Now, the centripetal force is equal to the net force acting on car at top.
$$
\begin{array}{rlrl}
\therefore \quad F_c =m g-N \quad \text { [Given, } N=0 \text { ] } \\
\frac{m v^2}{R} =m g & \\
\Rightarrow \quad =\sqrt{g R} =\sqrt{10 \times 40} \\
= 20 \mathrm{~m} / \mathrm{s}
\end{array}
$$
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