$A$ glass rod of radius $r_1$ is inserted symmetrically into a vertical capillary tube of radius $r_2$ $(r_1 < r_2)$ such that their lower ends are at the same level. The arrangement is dipped in water. The height to which water will rise into the tube will be ($\rho =$ density of water,$T =$ surface tension of water,$g =$ acceleration due to gravity).

  • A
    $\frac{2T}{(r_2-r_1)\rho g}$
  • B
    $\frac{T}{(r_2^2-r_1^2)\rho g}$
  • C
    $\frac{T}{(r_2-r_1)\rho g}$
  • D
    $\frac{2T}{(r_2^2-r_1^2)\rho g}$

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$A$ capillary tube of radius '$r$' is immersed in water and water rises to a height of '$h$'. The mass of water in the capillary tube is $5 \times 10^{-3} \ kg$. The same capillary tube is now immersed in a liquid whose surface tension is $\sqrt{2}$ times the surface tension of water. The angle of contact between the capillary tube and this liquid is $45^{\circ}$. The mass of liquid which rises into the capillary tube now is (in $kg$):

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:

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If $M$ is the mass of water that rises in a capillary tube of radius $r,$ then the mass of water which will rise in a capillary tube of radius $2r$ is

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In a capillary tube,water rises by $1.2 \ mm$. The height of water that will rise in another capillary tube having half the radius of the first is ........ $mm$.

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