A liquid of density 800 ~kg / m^3 is filled in a cylindrical vessel...

A liquid of density

800 \mathrm{~kg} / \mathrm{m}^3

is filled in a cylindrical vessel upto a height of 3 m . This cylindrical vessel stands on a horizontal plane. There is a circular hole on the side of the vessel. What should be the minimum diameter of the hole to move the vessel on the floor, if plug is removed. Take the coefficient of friction between the bottom of the vessel and the plane as 0.5 and total mass of vessel plus vessel as 95 kg .

Options

A0.107 m
B0.053 m
C0.206 m
D0.535 m

Correct answer

A. 0.107 m

Step-by-step solution

Given,

\rho=800 \mathrm{ kg} / \mathrm{m}^3 ,

h=3

\mathrm{ m}

, \mu=0.5, m=95

\mathrm{ kg} , d_{\min }=?

Let area of hole be a

\therefore

Reaction force,

F=\rho^2=\rho a \cdot 2 g h \quad[\because v=\sqrt{2 g h}] and

\begin{aligned} f_{\max } & =\mu\mathrm{N}=\mu \mathrm{mg} \\ F & \geq f_{\max } \\ 2 \text{ pagh } & \geq \mu \mathrm{mg} \\ a \geq \frac{\mu m}{2 \rho h} & =\frac{0.5 \times 95}{2 \times 800 \times 3}=0.009\end{aligned}

$\begin{aligned} \pi r^2 & \geq 0.009 \\ r & \geq \sqrt{\frac{0.009}{\pi}}\\ r_{\text{min }} & =0.0535 \mathrm{ m} \\ d_{\text{min }} & =2 r_{\text{min }}=2 \times 0.0535=0.107 \mathrm{ m}\end{aligned}

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