Solved: null Question for NEET

NEET Hard

A block of mass m is placed on a smooth inclined wedge ABC of inclination  as shown in the figure. The wedge is given an acceleration 'a' towards the right. The relation between a and  for the block to remain stationary on the wedge is

$a = g \cos \theta$

$a = \frac{g}{\sin \theta}$

$a = \frac{g}{\text{cosec } \theta}$

$a = g \tan \theta$

Step-by-Step Solution

In non-inertial frame,

$N \sin \theta = ma \quad \dots(i)$

$N \cos \theta = mg \quad \dots(ii)$

Dividing equation (i) by (ii), we get: $\tan \theta = \frac{a}{g}$

$a = g \tan \theta$

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