Q139 / 180 · JIPMER 2016 · Physics
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 .
- Correct answerA0.107 m
- B0.053 m
- C0.206 m
- D0.535 m
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
$\begin{aligned} f_{\max } & =\mu \mathrm{N}=\mu \mathrm{mg} \\ F & \geq f_{\max } \\ 2 \text { pagh } & \geq \mu \mathrm{mg} \\ a \geq \frac{\mu m}{2 \rho h} & =\frac{0.5 \times 95}{2 \times 800 \times 3}=0.009\end{aligned}$
$\begin{aligned} \pi r^2 & \geq 0.009 \\ r & \geq \sqrt{\frac{0.009}{\pi}} \\ r_{\text {min }} & =0.0535 \mathrm{~m} \\ d_{\text {min }} & =2 r_{\text {min }}=2 \times 0.0535=0.107 \mathrm{~m}\end{aligned}$
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
$\begin{aligned} f_{\max } & =\mu \mathrm{N}=\mu \mathrm{mg} \\ F & \geq f_{\max } \\ 2 \text { pagh } & \geq \mu \mathrm{mg} \\ a \geq \frac{\mu m}{2 \rho h} & =\frac{0.5 \times 95}{2 \times 800 \times 3}=0.009\end{aligned}$
$\begin{aligned} \pi r^2 & \geq 0.009 \\ r & \geq \sqrt{\frac{0.009}{\pi}} \\ r_{\text {min }} & =0.0535 \mathrm{~m} \\ d_{\text {min }} & =2 r_{\text {min }}=2 \times 0.0535=0.107 \mathrm{~m}\end{aligned}$Hence, the correct answer is option: 'A'.