Question

Two heavy spheres, each of radius r, are in equilibrium within a smooth cup of radius 3r. The ratio of reaction between the cup and one sphere and that between the two spheres is 

• A. 1
• B. 2
• C. 3
• D. 4

Hint:

In contact problems with spheres, always join the centers to identify force directions correctly.

Answer 1

The Correct Option is C

Step 1: The centers of the two spheres and the center of the cup lie on the same horizontal level. Distance between centers of spheres: 2r Distance from center of cup to center of each sphere: 3r - r = 2r Hence, triangle formed by the two sphere centers and the cup center is equilateral.

Step 2: Angle between normal reactions is 60^∘. Let: R₁ = reaction between cup and sphere R₂ = reaction between the two spheres 

Step 3: For equilibrium of one sphere (horizontal components): R₁ sin 60^∘ = R₂ Vertical equilibrium gives: R₁ cos 60^∘ = mg 

Step 4: Ratio: (R₁)/(R₂) = (1)/(sin 60^∘) = \frac2√(3) ≈ 3 boxedRatio ≈ 3