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4-1.Newton's Laws of Motion
hard
The horizontal acceleration that should be given to a smooth inclined plane of angle $sin^{-1}\, (1/l)$ to keep an object stationary on the plane relative to the inclined plane is
A$g/\sqrt {{l^2} - 1} $
B$g\sqrt {{l^2} - 1} $
C$\sqrt {{l^2} - 1} /g$
D$g/\sqrt {{l^2} + 1} $
Solution
$m g \sin \theta=m a c o s \theta$
or $a=g \tan \theta$
$\because \quad \sin \theta=\frac{1}{l}$
Hence, $\tan \theta=\frac{1}{\sqrt{l^{2}-1}}$
$\therefore a=\frac{\mathrm{g}}{\sqrt{l^{2}-1}}$
or $a=g \tan \theta$
$\because \quad \sin \theta=\frac{1}{l}$
Hence, $\tan \theta=\frac{1}{\sqrt{l^{2}-1}}$
$\therefore a=\frac{\mathrm{g}}{\sqrt{l^{2}-1}}$
Standard 11
Physics
