
An oil of relative density $0.9$ and viscosity $0.12kg/ms$ flows through a $2.5cm$ diameter pipe with a pressure drop of $38.4kN/{m^2}$ in a length of $30cm$. Determine the power required to maintain the flow.
A) $2.2W$
B) $3.84W$
C) $5.6W$
D) $9.3W$
Answer
159.6k+ views
Hint: Relative density is defined as the ratio of the densities of any two materials taken. It is calculated as the ratio of the density of the substance given and a reference substance. The reference substance taken is usually water. It is used to determine the density of unknown substances.
Complete step by step solution:
Step I: According to Poiseuille’s Law, flow of liquid is related to a number of factors like pressure gradient $\Delta P$, viscosity of fluid $(\eta )$, length of the tube $L$ through which it flows and the radius. It is written by the equation
$Q = \dfrac{{\Delta P \times \pi {r^4}}}{{8\eta L}}$---(i)
Step II: Given $\Delta P = 38.4 \times {10^3}N/{m^2}$
If diameter, $D = 2.5 \times {10^{ - 2}}m$
Then radius, $r = 1.25 \times {10^{ - 2}}$
$L = 30m$
$\eta = 0.12N/{m^2}$
Step III: Substitute all the values in the given formula of equation (i)
$Q = \dfrac{{38.4 \times {{10}^3} \times 3.14 \times {{(1.25 \times {{10}^{ - 2}})}^4}}}{{8 \times 0.12 \times 30}}$
$Q = \dfrac{{38.4 \times {{10}^3} \times 3.14 \times 2.44 \times {{10}^8}}}{{28.8}}$
$Q = 1 \times {10^{ - 4}}m/s$
Step IV: Power gained by a fluid from a pump is given by the formula:
$P = \Delta P \times Q$
Where $\Delta P$is the change in pressure and
$Q$ is the Rate of flow of volume of oil in the pipe
Substituting the value in the above formula and solving for power,
$ = 38.4 \times {10^3} \times 1 \times {10^{ - 4}}$
$P = 3.84W$
Step V:
Hence the minimum power required to maintain the flow of oil in the pipe will be $3.84W.$
Therefore, Option (B) is the right answer.
Note: It is to be noted that the particles of liquid have spaces between them and their position is not fixed. Therefore due to the movement of the particles, the liquids do not have a fixed shape. They take the shape of the container in which they are placed. But the change in shape does not affect the volume of the liquids. The volume remains the same.
Complete step by step solution:
Step I: According to Poiseuille’s Law, flow of liquid is related to a number of factors like pressure gradient $\Delta P$, viscosity of fluid $(\eta )$, length of the tube $L$ through which it flows and the radius. It is written by the equation
$Q = \dfrac{{\Delta P \times \pi {r^4}}}{{8\eta L}}$---(i)
Step II: Given $\Delta P = 38.4 \times {10^3}N/{m^2}$
If diameter, $D = 2.5 \times {10^{ - 2}}m$
Then radius, $r = 1.25 \times {10^{ - 2}}$
$L = 30m$
$\eta = 0.12N/{m^2}$
Step III: Substitute all the values in the given formula of equation (i)
$Q = \dfrac{{38.4 \times {{10}^3} \times 3.14 \times {{(1.25 \times {{10}^{ - 2}})}^4}}}{{8 \times 0.12 \times 30}}$
$Q = \dfrac{{38.4 \times {{10}^3} \times 3.14 \times 2.44 \times {{10}^8}}}{{28.8}}$
$Q = 1 \times {10^{ - 4}}m/s$
Step IV: Power gained by a fluid from a pump is given by the formula:
$P = \Delta P \times Q$
Where $\Delta P$is the change in pressure and
$Q$ is the Rate of flow of volume of oil in the pipe
Substituting the value in the above formula and solving for power,
$ = 38.4 \times {10^3} \times 1 \times {10^{ - 4}}$
$P = 3.84W$
Step V:
Hence the minimum power required to maintain the flow of oil in the pipe will be $3.84W.$
Therefore, Option (B) is the right answer.
Note: It is to be noted that the particles of liquid have spaces between them and their position is not fixed. Therefore due to the movement of the particles, the liquids do not have a fixed shape. They take the shape of the container in which they are placed. But the change in shape does not affect the volume of the liquids. The volume remains the same.
Recently Updated Pages
A steel rail of length 5m and area of cross section class 11 physics JEE_Main

At which height is gravity zero class 11 physics JEE_Main

A nucleus of mass m + Delta m is at rest and decays class 11 physics JEE_MAIN

A wave is travelling along a string At an instant the class 11 physics JEE_Main

The length of a conductor is halved its conductivity class 11 physics JEE_Main

Two billiard balls of the same size and mass are in class 11 physics JEE_Main

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Displacement-Time Graph and Velocity-Time Graph for JEE

Uniform Acceleration

Degree of Dissociation and Its Formula With Solved Example for JEE

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

Units and Measurements Class 11 Notes: CBSE Physics Chapter 1

Motion in a Straight Line Class 11 Notes: CBSE Physics Chapter 2

NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements

NCERT Solutions for Class 11 Physics Chapter 2 Motion In A Straight Line
