Question

When kerosene and coconut oil of coefficient of viscosity $0.002$ and $0.0154\;{\rm{Ns}}{{\rm{m}}^{ - 2}}$ are allowed to flow through the same pipe under the same pressure difference in the same time interval, the coconut oil collected is 1 litre in volume. The volume of kerosene that flows is:
A. 5.5 lit
B. 6.6 lit
C. 7.7 lit
D. 8.8 lit

Answer 1

Hint: The Poiseulli’s law gives the information about the volume flow rate of the liquid and the expression used for the calculation of flow rate of liquid is $V = \dfrac{{\pi {P_1}{r^2}}}{{8\eta l}}$

Complete Step by Step Answer:
Given:
The coefficient of viscosity of kerosene is ${\eta _2} = 0.002\;{\rm{Ns}}{{\rm{m}}^{ - 2}}$.
The coefficient of viscosity of coconut is ${\eta _1} = 0.0154\;{\rm{Ns}}{{\rm{m}}^{ - 2}}$.
The volume of coconut that flows is ${V_1} = 1\;{\rm{litre}}$.
From the Poiseulli’s law, the expression of the volume flow rate of the coconut is,
${V_1} = \dfrac{{\pi {{\mathop{\rm P}\nolimits} _1}r_1^2}}{{8\eta {l_1}}}$ ……….(1)
Here, ${P_1}$ is the pressure, ${r_1}$ is the radius of the pipe, ${\eta _1}$ is the coefficient of viscosity of coconut and ${l_1}$ is the length of the pipe.
From the Poiseulli’s law, the expression of the volume flow rate of the kerosene is,
${V_2} = \dfrac{{\pi {{\mathop{\rm P}\nolimits} _2}r_2^2}}{{8\eta {l_2}}}$ ………..(2)
Here, ${P_2}$ is the pressure, ${r_2}$ is the radius of the pipe, ${\eta _2}$ is the coefficient of viscosity kerosene and ${l_2}$ is the length of the pipe.
From equation (1) and (2), the volume of the kerosene that flows is,
$\dfrac{{{V_1}}}{{{V_2}}} = \dfrac{{\dfrac{{\pi {{\mathop{\rm P}\nolimits} _1}r_1^2}}{{8\eta {l_1}}}}}{{\dfrac{{\pi {{\mathop{\rm P}\nolimits} _2}r_2^2}}{{8\eta {l_2}}}}}$
All the other parameters in the in the above equation remains same except viscosity, so the above equation becomes,
$\dfrac{{{V_1}}}{{{V_2}}} = \dfrac{{{\eta _2}}}{{{\eta _1}}}$
Substitute the values in the above equation
\[\begin{array}{l}
\dfrac{{1\;{\rm{lit}}}}{{{V_2}}} = \dfrac{{0.002\;{\rm{Ns}}{{\rm{m}}^{ - 2}}}}{{0.0154\;{\rm{Ns}}{{\rm{m}}^{ - 2}}}}\\
{V_2} = 7.7\;{\rm{lit}}
\end{array}\]
Therefore, the option (c) is the correct answer that is \[7.7\;{\rm{lit}}\].

Note:For the same flowing conditions of kerosene and coconut oils, the volume collected will be in ratio to the flow rate of kerosene and coconut oil, so this ratio is useful for the calculation of kerosene volume rate.