Two spheres P and Q of equal radii have densities and , respectively. The spheres are connected by a massless string and placed in liquids and of densities and and viscosities and , respectively. They float in equilibrium with the sphere P in and s

Question

Two spheres P and Q of equal radii have densities $ρ_{1}$ and $ρ_{2}$ , respectively. The spheres are connected by a massless string and placed in liquids $L_{1}$ and $L_{2}$ of densities $σ_{1}$ and $σ_{2}$ and viscosities $η_{1}$ and $η_{2}$ , 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 [2015]

Smart Achievers · Accepted Answer

(1, 4)

$Since string is taut, ρ_{1} < σ_{1} and ρ_{2} < σ_{2}$

$For floating, net weight of system = net upthrust$

$η_{1}$

$L_{1}$

$V_{P} = \frac{2 r^{2}}{9 η_{2}} (σ_{2} - ρ_{1}) g$

$Where r is radius of sphere.$

$Downward terminal velocity$

$V_{Q} = \frac{2 r^{2} (ρ_{2} - σ_{1}) g}{9 η_{1}}$

$∴$ $| \frac{V_{P}}{V_{Q}} |$ $= \frac{η_{1}}{η_{2}}$

$(∵ ρ_{1} - σ_{2} = σ_{1} - ρ_{2})$

$Again V_{P} \cdot V_{Q} < 0 i.e., negative as V_{P} and V_{Q} are opposite to each other.$

Solution

1 $\frac{| {\vec{V}}_{P} |}{| {\vec{V}}_{Q} |} = \frac{η_{1}}{η_{2}}$

2 $\frac{| {\vec{V}}_{P} |}{| {\vec{V}}_{Q} |} = \frac{η_{2}}{η_{1}}$

3 ${\vec{V}}_{P} \cdot {\vec{V}}_{Q} > 0$

4 ${\vec{V}}_{P} \cdot {\vec{V}}_{Q} < 0$

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Solution

1 |V→P||V→Q|=η1η2

2 |V→P||V→Q|=η2η1

3 V→P·V→Q>0

4 V→P·V→Q<0

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1 $\frac{| {\vec{V}}_{P} |}{| {\vec{V}}_{Q} |} = \frac{η_{1}}{η_{2}}$

2 $\frac{| {\vec{V}}_{P} |}{| {\vec{V}}_{Q} |} = \frac{η_{2}}{η_{1}}$

3 ${\vec{V}}_{P} \cdot {\vec{V}}_{Q} > 0$

4 ${\vec{V}}_{P} \cdot {\vec{V}}_{Q} < 0$