1.

The rise in the water level in a capillary tube of radius 0.07 cm when dipped veryically in a beaker containing water of surface tension `0.07 N m^(-1)` is (g = `10 m s^(-2)`)A. 2 cmB. 4 cmC. 1.5 cmD. 3 cm

Answer» Correct Answer - A
Rise of a liquid in a capillary tube,
`h=(2Scostheta)/(r rhog)`
Here, r=0.07cm `= 0.07xx10^(-2)m`
For water , `S=0.07 N m^(-1), rho=10^(3) kg m^(-3)`
Angle of contact `theta=0^(@)`
`therefore h=(2xx(0.07 N m^(-1))xx1)/((0.07xx10^(-2)m)(10^(3) kg m^(-3))(10 m s^(-2)))`
`=2xx10^(-2)m=2` cm


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