InterviewSolution
Saved Bookmarks
| 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 |
|