IIT-JEE 2011 · Paper 2 · Physics Q29

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PhysicsMultiple correct+4Question 29 of 60

Two solid spheres A and B of equal volumes but of different densities $d_A$ and $d_B$ are connected by a string. They are fully immersed in a fluid of density $d_F$ . They get arranged into an equilibrium state as shown in the figure with a tension in the string. The arrangement is possible only if

Tank of fluid with sphere A held above sphere B by a taut vertical string, both fully immersed and neither touching the walls or bottom.

d_A<d_F

d_B>d_F

d_A>d_F

d_A+d_B=2d_F

Answer and worked solution

Official answer: A, B, D

Sphere A (top, string pulls it down with tension $T$ ): $d_FVg=d_AVg+T$ , so $d_A<d_F$ .

Sphere B (bottom, string pulls it up): $d_BVg=d_FVg+T$ , so $d_B>d_F$ .

Whole system. Adding the two: $(d_A+d_B)Vg=2d_FVg$ , so $d_A+d_B=2d_F$ .

(A), (B), (D)

Topics:Buoyancy and Archimedes' principle

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IIT-JEE 2011 · Paper 2, Physics question 29