Laws of Motion - Result Question 39
39. A block of mass $m$ is placed on a smooth inclined wedge $A B C$ of inclination $\theta$ as shown in the figure. The wedge is given an acceleration ’ $a$ ’ towards the right. The relation between $a$ and $\theta$ for the block to remain stationary on the wedge is
[2018]
(a) $a=\frac{g}{cosec \theta}$
(b) $a=\frac{g}{\sin \theta}$
(c) $a=g \tan \theta$
(d) $a=g \cos \theta$
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Answer:
Correct Answer: 39. (c)
Solution:
- (c) Let the mass of block is $m$. It will remains stationary if forces acting on it are in equilibrium. i.e., $ma \cos \theta=mg \sin \theta \Rightarrow a=g$ $\tan \theta$
Here $ma=$ Pseudo force on block, $mg=$ weight.
