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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
$(A)$ $d_Ad_F$ $(B)$ $d_B > d_F$ $(C)$ $d_A>d_F$ $(D)$ $d_A+d_B=2 d_F$
$(A,B,D)$
$(A,B,C)$
$(A,C,D)$
$(B,C,D)$
Solution
$(a, b, c)$ Let $V$ be the volume of shperes.
For equilibrium of $A$ :
$T+v d_A g=V D_f g$
$\therefore T=V_g\left(d_f-d_A\right) \cdots(1)$
$f \text { or } T > 0, d_f > d_A \text { or } d_A < d_f$
$(a)$ is the correct option
For equilibrium of $B$ :
$T+V d_f g=V d_B g$
$\therefore T=V_g\left(d_B \cdot d_f\right) \cdots$
$F \text { or } T > 0, d_B > d_f$
$(b)$ is the correct option
$\text { From (1) and (2) Vg }\left(d_f-d_A\right)=V g\left(d_B-d_f\right)$
$\therefore d_f-d_A=d_B-d_f$
$\therefore 2 d_f=d_A+d_B$
