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What happens if we replace this cylindrical liquid column with another object which is made up of a material whose density is equal to the density of liquid ? |
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Answer» Solution :We KNOW that the pressure difference in the liquid, `P_(2)-P_(2)=hrhog` `P_(2)-P_(2)=h(m)/(V)grArrP_(2)-P_(1)=h(m)/(Ah)G` `rArrP_(2)-P_(1)=(m)/(A)g` `rArr(P_(2)-P_(1))A=mg` Since F = PA and W = mg We get F = W (Values of displaced liquid) 1. Here 'F' is the force applied on the object and 'W' is the weight of the liquid. 2. So, the force applied on the object by the liquid is EQUAL to the weight of the displaced liquid. |
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