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A body of mass $M \mathrm{~kg}$ is on the top point of a smooth hemisphere of radius $5 \mathrm{~m}$. It is released to slide down the surface of the hemisphere. It leaves the surface when velocity is $5 \mathrm{~m} / \mathrm{s}$. At this instant the angle made by the radius vector of the body with the vertical is (Acceleration due to gravity $=10 \mathrm{~ms}^{-2}$ )
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Verified Answer
The correct answer is:
$60^{\circ}$
Suppose the body leaves the surface of the hemisphere at point $P$.
Let at point $P$ radius vector of the body makes an angle $\theta$.
$$
\therefore \quad m g \cos \theta=\frac{m v^2}{r}
$$
When the body leaves the surface then
$$
\begin{aligned}
R & =0 \\
\therefore \quad m g \cos \theta- & =\frac{m v^2}{r}
\end{aligned}
$$
$\begin{aligned} \cos \theta=\frac{v^2}{r g}=\frac{(5)^2}{5 \times 10} & =\frac{25}{50}=\frac{1}{2} \\ & =\cos 60 \\ \theta & =60^{\circ}\end{aligned}$
Let at point $P$ radius vector of the body makes an angle $\theta$.
$$
\therefore \quad m g \cos \theta=\frac{m v^2}{r}
$$
When the body leaves the surface then
$$
\begin{aligned}
R & =0 \\
\therefore \quad m g \cos \theta- & =\frac{m v^2}{r}
\end{aligned}
$$
$\begin{aligned} \cos \theta=\frac{v^2}{r g}=\frac{(5)^2}{5 \times 10} & =\frac{25}{50}=\frac{1}{2} \\ & =\cos 60 \\ \theta & =60^{\circ}\end{aligned}$
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