$A$ long capillary tube of mass $\pi \,g$,radius $2\,mm$ and negligible thickness,is partially immersed in a liquid of surface tension $0.1\,N/m$. Take the angle of contact as zero and neglect the buoyant force of the liquid. Find the force required to hold the tube vertically. $(g = 10\,m/s^2)$

$A$ vertical glass capillary tube of radius $r$ open at both ends contains some water (surface tension $T$ and density $\rho$). If $L$ be the length of the water column,then:

When a capillary tube of radius '$r$' is immersed in water, the rise of water is up to height '$h$'. The mass of water in the capillary tube is '$m$'. When another capillary tube of radius '$xr$' is immersed in water, the mass of water that will rise in this tube is:

Water rises to a height of $2 \,cm$ in a capillary tube. If the cross-sectional area of the tube is reduced to $\frac{1}{16}^{\text{th}}$ of the initial area, then water will rise to a height of: (in $\,cm$)

Two capillary tubes of radii $R$ and $2R$ are dipped in water. The water rises to heights $h_1$ and $h_2$ respectively in the two tubes. If $m_1$ and $m_2$ be the masses of water ascended in the tubes,then $m_1: m_2=$

When a long glass capillary tube of radius $0.015 \; cm$ is dipped in a liquid,the liquid rises to a height of $15 \; cm$ within it. If the contact angle between the liquid and glass is close to $0^{\circ}$,the surface tension of the liquid,in $milliNewton \; m^{-1}$ is $.....$
$[\rho_{\text{liquid}} = 900 \; kg \; m^{-3}, g = 10 \; ms^{-2}]$ (Give answer in closest integer)