Question

An inclined plane of height $h$ and length $l$ have the angle of inclination $\theta$/ The time taken by a body to come from the top to the bottom of this inclined plane will be

• A. $\sin \theta \sqrt{\frac{2h}{g}}$
• B. $\frac{1}{\sin \theta }\sqrt{\frac{2h}{g}}$
• C. $\sqrt{\frac{2h}{g}}$
• D. $\sqrt{\frac{2l}{g}}$

Answer 1

The correct option is B $\frac{1}{\sin \theta }\sqrt{\frac{2h}{g}}$

Force down the plane $=mg\sin \theta$
$\therefore$ Acceleration down the plane $=g\sin \theta$
Using the relation,
$h=ut+\frac{1}{2}g{t}^2$
$l=0+\frac{1}{2}g\sin \theta t^2$
$t^2=\frac{2l}{g\sin \theta }=\frac{2h}{g{\sin }^2\theta }[\because \sin \theta =\frac{h}{l}]$
or $t=\frac{1}{\sin \theta }\sqrt{\frac{2h}{g}}[l=\frac{h}{\sin \theta }]$.