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4-1.Newton's Laws of Motion
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A person with his hand in his pocket is skating on ice at the rate of $10 m / s$ and describes a circle of radius $50 m$. What is his inclination to vertical: $\left( g =10 m / sec ^2\right)$
A$\tan ^{-1}\left(\frac{1}{2}\right)$
B$\tan ^{-1}(1 / 5)$
C$\tan ^{-1}(3 / 5)$
D$\tan ^{-1}(1 / 10)$
Solution
(b)
Since surface $(ice)$ is frictionless, so the centripetal force required for skating will be provided by inclination of boy with the vertical and that angle is given as $\tan \theta=\frac{v^2}{r g}$ where $v$ is speed of skating and $r$ is radius of circle in which he moves.
Since surface $(ice)$ is frictionless, so the centripetal force required for skating will be provided by inclination of boy with the vertical and that angle is given as $\tan \theta=\frac{v^2}{r g}$ where $v$ is speed of skating and $r$ is radius of circle in which he moves.
Standard 11
Physics
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