Physics - NLM Question with Solution | TestHub

Resolve the weight along the contact normals in each case (trough smooth, sphere in equilibrium).

Case (i): the sphere rests against the single 60° incline only, so that surface supports the whole weight:

$${\mathrm{R}}_{\mathrm{A}}=\mathrm{W}=10\mathrm{N},{\mathrm{R}}_{\mathrm{B}}=0$$

Statement (A) is correct.

Case (ii): the sphere sits symmetrically in the 60° V, so both reactions are equal:

$$2\mathrm{R}\cos 60^{\circ }=10\Rightarrow {\mathrm{R}}_{\mathrm{A}}={\mathrm{R}}_{\mathrm{B}}=10\mathrm{N}$$

Statement (B) is correct.

Case (iii): one contact is the 60° incline (normal at 30° to the vertical) and the other a vertical wall (normal horizontal). Vertical and horizontal balance give

$${\mathrm{R}}_{\mathrm{A}}\cos 30^{\circ }=10\Rightarrow {\mathrm{R}}_{\mathrm{A}}=\frac{20}{\sqrt{3}}\mathrm{N},{\mathrm{R}}_{\mathrm{B}}={\mathrm{R}}_{\mathrm{A}}\sin 30^{\circ }=\frac{10}{\sqrt{3}}\mathrm{N}$$

Statement (C) is correct. The reactions differ across the three cases, so (D) is wrong.

Correct options: (A), (B), (C).