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Two spheres P and Q of equal radii have ...

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 `vecV_p` and Q alone in `L_1` has terminal velocity `vecV_Q`, then

A

`(|v_P| ) /(|v_Q|) = (eta _1 )/(eta _ 2 ) `

B

`(|v_P|)/(|v_Q|) = (eta _2 )/(eta _1 )`

C

` v_P. v_Q gt 2 `

D

`v_P. v_Q lt 0`

Text Solution

Verified by Experts

The correct Answer is:
A, D
For floating, net weight of system = net upthrust
` rArr (rho _1 + rho _2 ) V g = (sigma _ 1 + sigma_2) V g " "` … (i)
Since string is taut `, rho _1 lt rho _1 and rho_2 gt rho _2 `
(` because ` Upthrust on P is greater than its weight and Upthrust on Q is less than its weight because of Tension in the string). Also ` sigma _2 gt sigma_1`
` V_P = (2 r ^2 g )/( 9 eta _ 2 ) (sigma _2 - sigma _1 )` (upward terminal velocity ) , `V_Q = (2r ^ 2 g )/(9 eta _1 ) (rho_1 - sigma _ 1 ) ` (downward terminal velocity)
` | (V_P)/(V_Q)| = (eta _ 1 )/(eta _ 2 ) ` [Using (i) ] Further ` V_P. V_Q ` will be negative as they are opposite to each other.
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