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 $m_1: m_2=$

In a capillary tube experiment, a vertical $30 \text{ cm}$ long capillary tube is dipped in water. Water rises up to a height of $10 \text{ cm}$ due to capillarity. If this experiment is conducted in a freely falling elevator, then the length of the water column becomes

When a capillary tube is dipped into two liquids having relative densities $0.8$ and $0.6$ and surface tensions $60 \, dyne/cm$ and $50 \, dyne/cm$ respectively,the ratio of the heights of the liquids in the capillary tube $\frac{h_1}{h_2}$ 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 $\,cm$)

When a capillary is dipped in water,water rises to a height $h$. If the length of the capillary is made less than $h$,then

Two narrow bores of diameter $5.0 \, mm$ and $8.0 \, mm$ are joined together to form a $U$-shaped tube open at both ends. If this $U$-tube contains water,what is the difference in the level of two limbs of the tube? [Take surface tension of water $T = 7.3 \times 10^{-2} \, Nm^{-1}$,angle of contact $= 0$,$g = 10 \, ms^{-2}$ and density of water $\rho = 1.0 \times 10^{3} \, kg \, m^{-3}$] (in $mm$)