A fluid container containing a liquid of dens | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Consider a fluid particle of mass $m$ on the free surface. In the non-inertial frame of the container,the effective forces acting on the particle

Vedclass · Accepted Answer

$	an^{-1}\left[\frac{a+g\sin\alpha}{g\cos\alpha}ight]$

Vedclass · Answer

(B) Consider a fluid particle of mass $m$ on the free surface. In the non-inertial frame of the container,the effective forces acting on the particle are gravity ($mg$ downwards),pseudo force ($ma$ directed down the incline),and the normal force from the fluid below.The free surface of the liquid must be perpendicular to the effective acceleration vector $\vec{g}_{eff}$.The effective acceleration $\vec{g}_{eff} = \vec{g} - \vec{a}$.Resolving components along and perpendicular to the incline:Component of $\vec{g}$ perpendicular to the incline = $g\cos\alpha$ (downwards).Component of $\vec{g}$ parallel to the incline = $g\sin\alpha$ (down the incline).Component of $\vec{a}$ parallel to the incline = $a$ (up the incline).Total effective acceleration parallel to the incline $a_{||} = g\sin\alpha + a$ (downwards along the incline).Total effective acceleration perpendicular to the incline $a_{\perp} = g\cos\alpha$ (downwards).The angle $	heta$ that the free surface makes with the horizontal (or the incline,depending on the diagram definition) is given by the ratio of the components of the effective acceleration.From the geometry,$	an	heta = \frac{a_{||}}{a_{\perp}} = \frac{a + g\sin\alpha}{g\cos\alpha}$.Thus,$	heta = 	an^{-1}\left[\frac{a+g\sin\alpha}{g\cos\alpha}ight]$.

$A$ fluid container containing a liquid of density $\rho$ is accelerating upward with acceleration $a$ along an inclined plane of inclination $\alpha$ as shown. Then the angle of inclination $\theta$ of the free surface is:

Similar Questions

Two cylindrical vessels of equal cross-sectional area of $2 \ m^2$ contain water up to heights of $10 \ m$ and $6 \ m$,respectively. If the vessels are connected at their bottom,then the work done by the force of gravity is (Density of water is $10^3 \ kg/m^3$ and $g = 10 \ m/s^2$):

Water coming out of a horizontal tube at a speed $v$ strikes a vertical wall normally,close to the mouth of the tube,and falls down vertically after impact. When the speed of water is increased to $2v$:

$A$ pump motor is used to deliver water at a certain rate from a given pipe. To obtain twice as much water from the same pipe in the same time, the power of the motor has to be increased to: (in $\times$)

Give the limitations and uses of turbulence.

Vedclass Test Series

Exam Paper Generator

Online Exam Module