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Mechanical Properties of Fluids — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysicsMechanical Properties of Fluids5 Marks
Question
A cylindrical vessel filled with water upto a height of 2m stands on a horizontal plane. The side wall of the vessel has a plugged circular hole touching the bottom. Find the minimum diameter of the hole so that the vessel begin to move on the floor, if the plug is removed. The coefficient of friction between the bottom of the vessel and the plane is 0.4 and total mass of water plus vessel is 100kg.
✓

Answer

Velocity of efflux through the hole, $\text{v}=\sqrt{2\text{gh}}$$\because$ Distance moved by water in one second $\text{v}=\sqrt{2\text{gh}}$
$\therefore$ Rate of the momentum $=\Big(\rho\text{A}\sqrt{2\text{gh}}\Big)\Big(\sqrt{2\text{gh}}\Big)=2\text{ghA}\rho$
According to Newton's second law of motion, Force due to the velocity of efflux $=2\text{ghA}\rho$ Now, according to Newton's third law of motion, Force on the vessel = Rate of the momentum Force on the vessel $=\text{2ghA}\rho$ The vessel will move, if force on the vessel = force of friction$2\text{gh}\text{A}\rho=\mu\text{Mg}$
$\text{A}=\frac{\mu\text{m}}{2\text{h}\rho}=\frac{0.4\times100}{2\times2\times1000}=\frac{1}{100}$
Since, the hole is circular$\text{A}=\pi\text{r}^2=\frac{\pi\text{D}^2}{\text{4}}$
$\therefore\text{D}=\sqrt{\frac{4\text{A}}{\pi}}=\sqrt{\frac{4\times1}{100\times3.14}}=0.113\text{m}$
So, diameter of a hole D = 0.113m.

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