Physics - NLM Question with Solution | TestHub

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

Case (i): the sphere rests on a single 60° incline with no second contact, so the plane 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. Both reactions are equal and share the weight:

$$2\mathrm{R}\cos 60^{\circ }=\mathrm{W}\Rightarrow {\mathrm{R}}_{\mathrm{A}}={\mathrm{R}}_{\mathrm{B}}=\frac{10}{2\times 0.5}=10\mathrm{N}$$

So R_B is not zero in case (ii) - statement (B) is wrong.

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

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

$$\text{horizontal balance: }{\mathrm{R}}_{\mathrm{B}}={\mathrm{R}}_{\mathrm{A}}\sin 30^{\circ }=\frac{20}{\sqrt{3}}\times \frac{1}{2}=\frac{10}{\sqrt{3}}\mathrm{N}$$

Statement (C) is correct. The reactions are not the same in all three cases, so (D) is wrong.

Correct options: (A), (C).