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Consider the shown arrangement. Assume all surfaces to be smooth. If N represents magnitudes of normal reaction between block and wedge, then acceleration of M along horizontal is equal to

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$\frac{N \sin θ}{M}$ (along -ve x-axis)

$\frac{N \sin θ}{M}$ (along +ve x-axis)

$\frac{N \sin θ}{m + M}$ (along -ve x-axis)

$\frac{N \cos θ}{M}$ (along -ve x-axis)

answer is C.

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Detailed Solution

Wedge moves due to horizontal component of normal reaction.

Thus, $a = \frac{N_{H}}{M} = \frac{N \sin θ}{M}$ (along -ve x-axis)

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