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Question: A large bottle is fitted with a capillary tube at the bottom. The ratio of times taken to empty the ...

A large bottle is fitted with a capillary tube at the bottom. The ratio of times taken to empty the bottle when it is first filled with water and next with oil of relative density 0.8 is
[ηwater=103  Pas,ηoil=2×103  Pas]\left[ {{\eta _{water}} = {{10}^{ - 3}}\;{\rm{Pa - s, }}{\eta _{oil}} = 2 \times {{10}^{ - 3}}\;{\rm{Pa - s}}} \right]
A. 5:2
B. 2:3
C. 3:4
D. 1:2

Explanation

Solution

The above problem can be resolved by knowing the mathematical formulation for the rate of flow of liquids through the vessel of desired length. Moreover, the time taken to pour out the liquid depends on the rate of flow of liquid from the vessel.

Complete step by step answer:
The ratio of the relative density is, ρ2:ρ1=0.8{\rho _2}:{\rho _1} = 0.8.
As, the ratio of time taken to empty the vessel is proportional to the discharge rates of different liquids. Then the expression is given as,
twatertoil=QwaterQoil..........................(1)\dfrac{{{t_{water}}}}{{{t_{oil}}}} = \dfrac{{{Q_{water}}}}{{{Q_{oil}}}}..........................\left( 1 \right)
And the formula for discharge rate for water is,
Qwater=πρ1(gh)r48ηwater×L.......................(2){Q_{water}} = \dfrac{{\pi {\rho _1}\left( {gh} \right){r^4}}}{{8{\eta _{water}} \times L}}.......................\left( 2 \right)

And the formula for discharge rate for oil is,
Qoil=πρ2(gh)r48ηoil×L.......................(3){Q_{oil}} = \dfrac{{\pi {\rho _2}\left( {gh} \right){r^4}}}{{8{\eta _{oil}} \times L}}.......................\left( 3 \right)
Here, r is the radius of tube, L is length of tube, h is the vertical height of tube and g is the gravitational acceleration.
Taking the ratio of equation 1 and 2 as,

\dfrac{{{Q_{water}}}}{{{Q_{oil}}}} = \dfrac{{{\rho _2}/{\eta _{water}}}}{{{\rho _1}/{\eta _{oil}}}}\\\ \dfrac{{{Q_{water}}}}{{{Q_{oil}}}} = \dfrac{{2 \times {{10}^{ - 3}}\;{\rm{Pa - s}}}}{{0.8 \times {{10}^{ - 3}}\;{\rm{Pa - s}}}}\\\ \dfrac{{{Q_{water}}}}{{{Q_{oil}}}} = \dfrac{5}{2} \end{array}$$ As the ratio of discharge is directly proportional to the time taken to empty the vessel, then the ratio of time taken is equivalent to 5:2 Therefore, the required ratio of time is 5:2 **So, the correct answer is “Option A”.** **Note:** To resolve the given conditions, one must remember the key relation for the rate of flow in terms of length of vessel, relative density of fluid filling out the vessel and some other parameters like radius of vessel. The given condition can also be resolved using the concepts under the stokes law, where each parameter has their own significance.