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
A
`d_A lt d_F`
B
`d_B gt d_F`
C
`d_Agtd_F`
D
`d_A+d_B = 2d_F`
Text Solution
Verified by Experts
The correct Answer is:
A, B, D
(a,b,c) Let V be the volume of shperes. For equilibrium of A: `T + vd_Ag = VD_fg` `:. T = V_g (d_f- d_A)…..(1)` `for T gt0, d_f gt d_A or d_Altd_f` (a) is the correct option For equilibrium of B: `T + Vd_fg = Vd_Bg` `:. T = V_g(d_B - d_f) ...(2)` `For Tgt 0, d_Bgt d_f` (b) is the correct option From (1) & (2) Vg `(d_f - d_A) = Vg(d_B - d_f)` `:. d_f - d_A = d_B - d_f` `:. 2d_f = d_A + d_B` (d) is the correct option.
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