JEE MAIN - Physics (2025 - 7th April Evening Shift - No. 14)
A capillary tube of radius 0.1 mm is partly dipped in water (surface tension 70 dyn/cm and glass water contact angle ≈ 0°) with 30° inclined with the vertical. The length of water risen in the capillary is _______ cm. (Take $g = 9.8 \text{ m/s}^2$)
$\frac{82}{5}$
$\frac{68}{5}$
$\frac{57}{2}$
$\frac{71}{5}$
Explanation
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$$\begin{aligned} & \mathrm{h}=\frac{2 \mathrm{~T} \cos \theta}{\rho \mathrm{gr}}=\frac{2 \times 70 \times 1}{1 \times 980 \times 10^{-2}} \\ & \mathrm{~h}=\frac{100}{7} \mathrm{~cm} \\ & \sin 60^{\circ}=\frac{\mathrm{h}}{\ell} \\ & \ell=\frac{\mathrm{h} \times 2}{\sqrt{3}} \\ & \ell=\frac{100}{7} \times \frac{2}{\sqrt{3}} \\ & =\frac{200}{7 \times \sqrt{3}} \\ & =16.49 \mathrm{~cm} \end{aligned}$$
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