A block of mass m is placed on a smooth inclined wedge A B C of...

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 A block of mass m is placed on a smooth inclined wedge A B C of inclination θ as shown in the figure. The wedge is

A block of mass m is placed on a smooth inclined wedge A B C of inclination θ as shown in the figure. The wedge is

Options

A $a=\frac{g}{\operatorname{cosec} \theta}$
B $a=\frac{g}{\sin \theta}$
C $a=g \tan \theta$
D $a=g \cos \theta$

Correct answer

C. $a=g \tan \theta$

Step-by-step solution

Let the mass of block is m. It will remains stationary if forces acting on it are in equilibrium. i.e.,

m a \cos \theta=m g \sin \theta\Rightarrow a=g \tan \theta

Here

\mathrm{ma}=

Pseudo force on block,

\mathrm{mg}=

weight

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