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4-1.Newton's Laws of Motion
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A body is slipping from an inclined plane of height $h$ and length $l$. If the angle of inclination is $\theta $, the time taken by the body to come from the top to the bottom of this inclined plane is
A$\sqrt {\frac{{2h}}{g}} $
B$\sqrt {\frac{{2l}}{g}} $
C$\frac{1}{{\sin \theta }}\sqrt {\frac{{2h}}{g}} $
D$\sin \theta \sqrt {\frac{{2h}}{g}} $
Solution
(c) Force down the plane $ = mg\sin \theta $
$\therefore $ Acceleration down the plane $ = g\sin \theta $
Since $l = 0 + \frac{1}{2}g\sin \theta {t^2}$
$\therefore {t^2} = \frac{{2l}}{{g\sin \theta }} = \frac{{2h}}{{g{{\sin }^2}\theta }} \Rightarrow t = \frac{1}{{\sin \theta }}\sqrt {\frac{{2h}}{g}} $
$\therefore $ Acceleration down the plane $ = g\sin \theta $
Since $l = 0 + \frac{1}{2}g\sin \theta {t^2}$
$\therefore {t^2} = \frac{{2l}}{{g\sin \theta }} = \frac{{2h}}{{g{{\sin }^2}\theta }} \Rightarrow t = \frac{1}{{\sin \theta }}\sqrt {\frac{{2h}}{g}} $
Standard 11
Physics
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