A cylindrical vessel filled with water up to | vedclass

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

Question

(A) The force exerted by the water jet exiting the hole is given by $F = \rho A v^2$,where $A$ is the area of the hole and $v$ is the velocity of the

Vedclass · Accepted Answer

$\sqrt{\frac{2\mu M}{\pi \rho H}}$

Vedclass · Answer

(A) The force exerted by the water jet exiting the hole is given by $F = \rho A v^2$,where $A$ is the area of the hole and $v$ is the velocity of the water.Using Torricelli's Law,the velocity of the water at the bottom is $v = \sqrt{2gH}$.Thus,the force is $F = \rho A (2gH)$.For the vessel to begin moving,this force must overcome the limiting friction force,$f_s = \mu Mg$.Equating the two: $\rho A (2gH) = \mu Mg$.Substituting $A = \frac{\pi D^2}{4}$,we get $\rho \left( \frac{\pi D^2}{4} \right) (2gH) = \mu Mg$.Simplifying the equation: $\frac{\rho \pi D^2 g H}{2} = \mu Mg$.Solving for $D$: $D^2 = \frac{2\mu M}{\pi \rho H}$.Therefore,$D = \sqrt{\frac{2\mu M}{\pi \rho H}}$.

Similar Questions

There is a hole of area $A_0$ at the bottom of a cylindrical vessel of cross-sectional area $A$. Water is filled up to a height $h$ and the water flows out in time $t$. If water is filled to a height $4h$,it will flow out in time equal to:

Water flows into a cylindrical vessel of large cross-sectional area at a rate of $10^{-4} \; m^3/s$. It flows out from a hole of area $10^{-4} \; m^2$, which has been punched through the base. How high does the water rise in the vessel (in $; m$)?

$A$ cylinder of height $20\;m$ is completely filled with water. The velocity of efflux of water (in $m/s$) through a small hole on the side wall of the cylinder near its bottom is ....... $m/s$.

The bottom of a cylindrical vessel has a hole of area $A$. If water is filled up to a height $h$,it flows out in $t$ seconds. If water is filled to a height $4h$,it will flow out in time:

Vedclass Test Series

Exam Paper Generator

Online Exam Module