MHT CET · Physics · Mechanical Properties of Fluids
A glass capillary of radius 0.35 mm is inclined at \(60^{\circ}\) with the vertical in water. The height of the water column in the capillary is (surface tension of water \(=7 \times 10^{-2} \mathrm{~N} / \mathrm{m}\), acceleration due to gravity, \(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2, \cos 0^{\circ}=1, \cos 60^{\circ}=0.5\) )
- A 6 cm
- B 8 cm
- C 10 cm
- D 12 cm
Answer & Solution
Correct Answer
(B) 8 cm
Step-by-step Solution
Detailed explanation
\(\begin{aligned}
& \begin{array}{l}
\mathrm{h}=\frac{2 \mathrm{~T} \cos \theta}{\mathrm{rgg}}=\frac{2 \times\left(7 \times 10^{-2}\right) \times \cos 0^{\circ}}{\left(0.35 \times 10^{-3}\right) \times 10^3 \times 10} \\
\quad \ldots\left(\theta=0^{\circ}, \text { for glass-water }\right) \\
\mathrm{h}=0.04 \mathrm{~m} \\
l=\frac{\mathrm{h}}{\cos \phi}
\end{array}
\end{aligned}\)
where \(\phi\) is angle of capillary with vertical as shown in figure.
\(l=\frac{0.04}{\cos (60)}=0.08 \mathrm{~m}=8 \mathrm{~cm}\)
& \begin{array}{l}
\mathrm{h}=\frac{2 \mathrm{~T} \cos \theta}{\mathrm{rgg}}=\frac{2 \times\left(7 \times 10^{-2}\right) \times \cos 0^{\circ}}{\left(0.35 \times 10^{-3}\right) \times 10^3 \times 10} \\
\quad \ldots\left(\theta=0^{\circ}, \text { for glass-water }\right) \\
\mathrm{h}=0.04 \mathrm{~m} \\
l=\frac{\mathrm{h}}{\cos \phi}
\end{array}
\end{aligned}\)
where \(\phi\) is angle of capillary with vertical as shown in figure.
\(l=\frac{0.04}{\cos (60)}=0.08 \mathrm{~m}=8 \mathrm{~cm}\)
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