Physics - NLM Question with Solution | TestHub

All surfaces are smooth, so the block (resting on the wedge's horizontal top face) can have no horizontal force on it: its acceleration is purely vertical - (B) correct, (A) wrong.

Let a_w be the wedge's acceleration down the incline and a_b the block's vertical acceleration. The block stays on the horizontal top face, so

$${\mathrm{a}}_{\mathrm{b}}={\mathrm{a}}_{\mathrm{w}}\sin \theta$$

Equations of motion (N = contact force between block and wedge):

$$\text{block (vertical): }\mathrm{mg}-\mathrm{N}=\mathrm{m}{\mathrm{a}}_{\mathrm{b}}$$

$$\text{wedge (along incline): }(\mathrm{Mg}+\mathrm{N})\sin \theta =\mathrm{M}{\mathrm{a}}_{\mathrm{w}}$$

$$\Rightarrow {\mathrm{a}}_{\mathrm{w}}=\frac{(\mathrm{M}+\mathrm{m})\mathrm{g}\sin \theta }{\mathrm{M}+\mathrm{m}{\sin }^2\theta }>\mathrm{g}\sin \theta \left(\because \mathrm{M}+\mathrm{m}>\mathrm{M}+\mathrm{m}{\sin }^2\theta \right)$$

So the wedge's acceleration is greater than g sin$\theta$ - (C) correct.

Normal force of the incline on the wedge (perpendicular to the plane):

$${\mathrm{N}}_{\text{plane}}=(\mathrm{Mg}+\mathrm{N})\cos \theta \text{with }\mathrm{N}<\mathrm{mg}$$

$$\Rightarrow {\mathrm{N}}_{\text{plane}}<(\mathrm{M}+\mathrm{m})\mathrm{g}\cos \theta$$

So (D) is correct. Correct options: (B), (C), (D).