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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
Options:
Solution:
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Verified Answer
The correct answers are:
$d_A < d_F$,
$d_B>d_F$
,
$d_A+d_B=2 d_F$
$d_A < d_F$,
$d_B>d_F$
,
$d_A+d_B=2 d_F$
Equilibrium of $A$
$$
\begin{aligned}
V d_F g & =T+W_A \\
& =T+V d_A g
\end{aligned}
$$
Equilibrium of $B$,
$$
T+V d_F g=V d_B g
$$
Adding Eqs. (i) and (ii), we get $2 d_f=d_A+d_B$
$\therefore$ Option (d) is correct. From Eq. (i), we can see that
$$
d_F>d_A \quad \text { [as } T>0 \text { ] }
$$
$\therefore$ Option (a) is correct.
From Eq. (ii) we can see that,
$$
d_B>d_F
$$
$\therefore$ Option (a) is correct.
$\therefore$ Correct options are (a), (b) and (d).
Analysis of Question
Question is moderately difficult but conceptwise it is good.
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