A capillary tube of radius 'r' is immersed in | vedclass

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(A) The height of a liquid column in a capillary tube is given by the formula: $h = \frac{2T \cos 	heta}{rdg}$.For water: $h_1 = \frac{2T_1 \cos 	he

Vedclass · Accepted Answer

$5 	imes 10^{-3}$

Vedclass · Answer

(A) The height of a liquid column in a capillary tube is given by the formula: $h = \frac{2T \cos 	heta}{rdg}$.For water: $h_1 = \frac{2T_1 \cos 	heta_1}{r d_1 g}$. Since $	heta_1 = 0^{\circ}$, $\cos 0^{\circ} = 1$, so $h_1 = \frac{2T_1}{r d_1 g}$.The mass of the liquid is $m = \pi r^2 h d$. Substituting $h$, we get $m = \pi r^2 d \left( \frac{2T \cos 	heta}{rdg} ight) = \frac{2 \pi r T \cos 	heta}{g}$.For the first case (water): $m_1 = \frac{2 \pi r T_1}{g} = 5 	imes 10^{-3} \ kg$.For the second liquid: $T_2 = \sqrt{2} T_1$ and $	heta_2 = 45^{\circ}$.$m_2 = \frac{2 \pi r T_2 \cos 	heta_2}{g} = \frac{2 \pi r (\sqrt{2} T_1) \cos 45^{\circ}}{g}$.Since $\cos 45^{\circ} = \frac{1}{\sqrt{2}}$, we have $m_2 = \frac{2 \pi r (\sqrt{2} T_1) 	imes \frac{1}{\sqrt{2}}}{g} = \frac{2 \pi r T_1}{g}$.Comparing $m_1$ and $m_2$, we see that $m_2 = m_1 = 5 	imes 10^{-3} \ kg$.

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