A capillary tube of radius 0.1 mm is partly dipped in water (surface tension $70 \mathrm{dyn} / \mathrm{cm}$…
$\left(\right.$ Take $\left.\mathrm{g}=9.8 \mathrm{~m} / \mathrm{s}^2\right)$
- $\frac{82}{5}$
- $\frac{57}{2}$
- $\frac{71}{5}$
- $\frac{68}{5}$
Solution

$\begin{aligned} & \mathrm{h}=\frac{2 \mathrm{~T} \cos \theta}{\rho \mathrm{gr}}=\frac{2 \times 70 \times 1}{1 \times 980 \times 10^{-2}} \\ & \mathrm{~h}=\frac{100}{7} \mathrm{~cm} \\ & \sin 60^{\circ}=\frac{\mathrm{h}}{\ell} \\ & \ell=\frac{\mathrm{h} \times 2}{\sqrt{3}} \\ & \ell=\frac{100}{7} \times \frac{2}{\sqrt{3}} \\ & =\frac{200}{7 \times \sqrt{3}} \\ & =16.49 \mathrm{~cm}\end{aligned}$
Asked in: JEE Main 2025 (07 Apr Shift 2)
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