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The limbs of a manometer consist of uniform capillary tubes of radii are `10^(-3)&7.2xx10^(-4)`m find out the correct pressure difference if the level of the liquid in narrower tube stands 0.2 m above that in the broader tube. (density of liquid `=10^(3)kg//m^(3)`, surface tension `=72xx10^(-3)N//m`) |
Answer» If `P_(1)` and `P_(2)` are the pressures in the broader and narrower tubes of radii `r_(1)` and `r_(2)` respectively, the pressure just below the meniscus in the respective tubes will be `P_(1)-(2T)/(r_(1)) and P_(2)-(2T)/(r_(2))` So that `[P_(1)-(2T)/(r_(1))]-[P_(2)-(2T)/(r_(2))]=hrhog or P_(1)-P_(2)=hrhog-2T[(1)/(r_(2))-(1)/(r_(1))]` `P_(1)-P_(2)=0.2xx10^(3)xx9.8-2xx72xx10^(-3)[(1)/(7.2xx10^(-4))-(1)/(14xx10^(-4))]=1960-97=1863Pa` |
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