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## Homework Statement

a manometer consisting of tube that is 1.25 cm inner diameter . on one side , the manometer leg contains mercury,10cc of an oil(S.G.=1.4)and 3 cc of air as a bubble in oil . the other leg contains only mercury . both legs are open to atmosphere and static . An accident occurs in which 3cc of oil and air bubble are removed from one leg . how much do mercury height levels change?

## Homework Equations

Δp=ρgh

## The Attempt at a Solution

i calculated the change in pressure due to 3 cc air bubble

∏/4 * 1.25*1.25*l=3

where l = length of airbubble column

l=2.445 cm ...after that cudnt proceed

please help me how to do it asap