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A vessel having small hole in the bottom has to hold water without leakage, if water is poured to a height of 7 $\mathrm{cm}$. Then the radius of the hole is
[surface tension of water is $0.07 \mathrm{Nm}^{-1}$, angle of contact is $0^{\circ}$ and $\mathrm{g}=10 \mathrm{~ms}^{-2}$ ]
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[surface tension of water is $0.07 \mathrm{Nm}^{-1}$, angle of contact is $0^{\circ}$ and $\mathrm{g}=10 \mathrm{~ms}^{-2}$ ]
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Verified Answer
The correct answer is:
$0.2 \mathrm{~mm}$
Height of water, $\mathrm{h}=7 \mathrm{~cm}=0.07 \mathrm{~m}$
Surface tension of water, $T=0.07 \mathrm{~N} / \mathrm{m}$
$$
\mathrm{T}=\frac{\mathrm{rh} \rho \mathrm{g}}{2}
$$
The radius of hole, $r=\frac{2 \tau}{h \rho g}$
$$
\begin{aligned}
& =\frac{2 \times 0.07}{0.07 \times 1000 \times 10} \\
& =0.2 \mathrm{~mm}
\end{aligned}
$$
Surface tension of water, $T=0.07 \mathrm{~N} / \mathrm{m}$
$$
\mathrm{T}=\frac{\mathrm{rh} \rho \mathrm{g}}{2}
$$
The radius of hole, $r=\frac{2 \tau}{h \rho g}$
$$
\begin{aligned}
& =\frac{2 \times 0.07}{0.07 \times 1000 \times 10} \\
& =0.2 \mathrm{~mm}
\end{aligned}
$$
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