JEE Advanced 2015 · Paper 2 · Physics Q9

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PhysicsMultiple correct+4 / -2Question 9 of 60

Two spheres $P$ and $Q$ of equal radii have densities $\rho_1$ and $\rho_2$ , respectively. The spheres are connected by a massless string and placed in liquids $L_1$ and $L_2$ of densities $\sigma_1$ and $\sigma_2$ and viscosities $\eta_1$ and $\eta_2$ , respectively. They float in equilibrium with the sphere $P$ in $L_1$ and sphere $Q$ in $L_2$ and the string being taut (see figure). If sphere $P$ alone in $L_2$ has terminal velocity $\vec V_P$ and $Q$ alone in $L_1$ has terminal velocity $\vec V_Q$ , then

A beaker with a lighter liquid L₁ layered over a denser liquid L₂. Sphere P floats in L₁ and is tied by a taut string to sphere Q below it in L₂.

\dfrac{|\vec V_P|}{|\vec V_Q|}=\dfrac{\eta_1}{\eta_2}

\dfrac{|\vec V_P|}{|\vec V_Q|}=\dfrac{\eta_2}{\eta_1}

\vec V_P\cdot\vec V_Q>0

\vec V_P\cdot\vec V_Q<0

Answer and worked solution

Official answer: A, D

Equilibrium of the pair. With the string internal, the total weight equals the total buoyancy (equal volumes):
$\rho_1+\rho_2=\sigma_1+\sigma_2.$

Taut string. P pulls up ( $\sigma_1>\rho_1$ ) and Q pulls down ( $\rho_2>\sigma_2$ ). L₂ is the lower, denser liquid ( $\sigma_2>\sigma_1$ ).

Each sphere alone.

P in L₂: $\sigma_2>\rho_1$ , so it rises, with speed $\propto(\sigma_2-\rho_1)/\eta_2$ .
Q in L₁: $\rho_2>\sigma_1$ , so it sinks, with speed $\propto(\rho_2-\sigma_1)/\eta_1$ .

Compare. From the equilibrium condition, $\sigma_2-\rho_1=\rho_2-\sigma_1$ . So
$\frac{|V_P|}{|V_Q|}=\frac{\eta_1}{\eta_2},$
and the velocities are opposite, giving $\vec V_P\cdot\vec V_Q<0$ .

(A), (D)

Topics:Terminal velocity Buoyancy and Archimedes' principle Viscosity and Stokes' law

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JEE Advanced 2015 · Paper 2, Physics question 9