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Water rises to a height of 10 cm in a capillary tube and mercury falls to a depth of 3.42 cm in the same capillary tube. If the contact angle for mercury and surface tension of water and mercury is |
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Answer» `1:0.15` or `S=(hr rho g)/(2 cos theta) implies S prop (h rho)/(cos theta)` `:. S_(w)/S_(Hg)=h_(1)/h_(2)XX(cos theta_(2))/(cos theta_(1))xxrho_(1)/rho_(2)` `=10/((-3.42))xx(cos 135^(@))/(cos 0^(@))xx1/13.6` `S_(w)/S_(Hg)=10/3.42xx0.707/13.6=1/6.5` |
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