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JEE Advanced · Physics · 11. Properties of Fluids

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

  1. A \(d_A < d_F\)
  2. B \(d_B>d_F\)
  3. C \(d_A>d_F\)
  4. D \(d_A+d_B=2 d_F\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(d_A < d_F\)

Step-by-step Solution

Detailed explanation



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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