Discussion Forum



1)

 Two spheres P and Q   for equal radii have densities  ρ1 and ρ2, respectively, The spheres are connected by a massless string and placed in liquids L1 and L2 of densities  σ1   and σ 2  and viscosities η1 and η2, respectively. They float in equilibrium with the sphere P  in L1  and sphere Q  in L2   and the string being taut (see figure). If  sphere  P alone in L2  has  terminal  velocity  vp  and Q alone in L1 has terminal velocity vQ, then

1032021544_p1.PNG

 


A) $\frac{|V_{P}|}{|V_{Q}|}=\frac{\eta_{1}}{\eta_{2}}$

B) $\frac{|V_{P}|}{|V_{Q}|}=\frac{\eta_{2}}{\eta_{1}}$

C) $[V_{P}.V_{Q}>0$

D) $V_{P}.V_{Q}<0$

Answer:

Option A,D

Explanation:

 For floating , net weight of system=  net upthrust.

  $\Rightarrow  (\rho_{1}+\rho_{2})V_{g}=(\sigma_{1}+\sigma_{2})V_{g}$

           Since string is taut,   $\rho_{1}<\sigma_{1}$     and  $\rho_{2}>\sigma_{2}$

                      $v_{p}=\frac{2r^{2}g}{2\eta_{2}}(\sigma_{2}-\rho_{1})$

                                                    (upward terminal velocity)

                          $v_{Q}=\frac{2r^{2}g}{9\eta_{1}}(\rho_{2}-\sigma_{1})$

                                                 ( downward  terminal velocity)

                      $|\frac{v_{P}}{v_{Q}}|= \frac{\eta_{1}}{\eta_{2}}$

   Further ,  $\overline{v_{P}}.\overline{v_{Q}}$   will negative  as they are opposite  to each other.