Water rises to a height h in a capillary tube | vedclass

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Question

(A) The height $h$ to which a liquid rises in a capillary tube is given by the formula: $h = \frac{2T \cos 	heta}{rdg}$,where $T$ is the surface tens

Vedclass · Accepted Answer

$6h$

Vedclass · Answer

(A) The height $h$ to which a liquid rises in a capillary tube is given by the formula: $h = \frac{2T \cos 	heta}{rdg}$,where $T$ is the surface tension,$	heta$ is the angle of contact,$r$ is the radius of the capillary,$d$ is the density of the liquid,and $g$ is the acceleration due to gravity.From the formula,it is clear that $h \propto \frac{1}{g}$.Since the acceleration due to gravity on the Moon $(g_m)$ is $\frac{1}{6}$ times the acceleration due to gravity on the Earth $(g_e)$,i.e.,$g_m = \frac{g_e}{6}$.Therefore,the height on the Moon $(h_m)$ will be $h_m = h_e 	imes \frac{g_e}{g_m} = h_e 	imes 6 = 6h$.

List-$I$	List-$II$
$(A)$ $[\mathbf{a} \mathbf{b} \mathbf{c}]$	$1. \|\mathbf{a}\|\|\mathbf{b}\|\cos(\mathbf{a}, \mathbf{b})$
$(B)$ $(\mathbf{c} \times \mathbf{a}) \times \mathbf{b}$	$2. (\mathbf{a} \cdot \mathbf{c})\mathbf{b} - (\mathbf{a} \cdot \mathbf{b})\mathbf{c}$
$(C)$ $\mathbf{a} \times (\mathbf{b} \times \mathbf{c})$	$3. \mathbf{a} \cdot \mathbf{b} \times \mathbf{c}$
$(D)$ $\mathbf{a} \cdot \mathbf{b}$	$4. \|\mathbf{a}\|\|\mathbf{b}\|$
	$5. (\mathbf{b} \cdot \mathbf{c})\mathbf{a} - (\mathbf{a} \cdot \mathbf{b})\mathbf{c}$

Water rises to a height $h$ in a capillary tube at the surface of the Earth. On the surface of the Moon,the height of the water column in the same capillary tube will be:

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