$A$ capillary tube of radius $0.2\,cm$ is dipped vertically in a beaker containing liquid. If the liquid rises to a height of $5\,cm$ for which the angle of contact is $60^o$,then the surface tension of the liquid is ...... $dynes/cm$ (given density $d = 1\,gm/cm^3$ and acceleration due to gravity $g = 980\,cm/s^2$).

  • A
    $49$
  • B
    $98$
  • C
    $490$
  • D
    $980$

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Similar Questions

$A$ capillary tube made of glass with a radius of $0.15\, mm$ is dipped vertically in a beaker filled with methylene iodide (surface tension $= 0.05\, N m^{-1}$,density $= 667\, kg m^{-3}$),which rises to a height $h$ in the tube. It is observed that the two tangents drawn from the liquid-glass interfaces (from opposite sides of the capillary) make an angle of $60^{\circ}$ with one another. Then $h$ is close to $...... m$ $(g = 10\, m s^{-2})$

The lower end of a capillary tube is dipped into water and it is observed that the water in the capillary tube rises by $7.5 \ cm$. Find the radius of the capillary tube used,if the surface tension of water is $7.5 \times 10^{-2} \ N \ m^{-1}$. The angle of contact between water and glass is $0^{\circ}$ and the acceleration due to gravity is $10 \ m \ s^{-2}$.

$A$ $20 cm$ long capillary tube is dipped vertically in water and the liquid rises up to $10 cm$. If the entire system is kept in a freely falling platform, the length of the water column in the tube will be (in $cm$)

According to Poiseuille's law,the pressure drop per unit length required to overcome viscous forces is $\Delta P = \frac{8 \eta v}{r^2}$,where $r$ is the radius of the cross-section,$v$ is the fluid velocity,and $\eta$ is the coefficient of viscosity. $A$ capillary tube of radius $a$ is dipped in a liquid of density $\rho$,surface tension $T$,and coefficient of viscosity $\eta$. The liquid starts rising in it so that its height $h(t)$ is a function of time $t$. The resulting rate of change of the momentum of the liquid column in the capillary (taking vertically up to be the positive direction and the contact angle to be close to $0^{\circ}$) is $-\pi a^2 \rho gh + F$. Then $F$ is ($g$ is the acceleration due to gravity):

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