JEE MAIN - Physics (2005 - No. 31)
A block is kept on a frictionless inclined surface with angle of inclination $$'\,\alpha \,'.$$ The incline is given an acceleration $$a$$ to keep the block stationary. Then $$a$$ is equal to
$$g$$ $$cosec$$ $$\alpha $$
$$g/tan$$ $$\alpha $$
$$g$$ $$tan$$ $$\alpha $$
$$g$$
Explanation
Acceleration of the block is to the right. Pseudo force acting on the block to the left.
From diagram we can say,
m$$a$$cos$$\alpha $$ = m$$g$$sin$$\alpha $$
$$ \Rightarrow a = g\tan \alpha $$
From diagram we can say,
m$$a$$cos$$\alpha $$ = m$$g$$sin$$\alpha $$
$$ \Rightarrow a = g\tan \alpha $$
Comments (0)
