Question

A manometer has two liquids of densities ${\rho }_1$ and ${\rho }_2$. The value of $P-P_0$ is $(\text{where}{P}_0=\text{atmospheric pressure})$

• A. $(\frac{1}{2}{\rho }_2{h}_1-{\rho }_2{h}_2)g$
• B. $({\rho }_2{h}_2-{\rho }_1{h}_1)g$
• C. $({\rho }_1{h}_1-{\rho }_2{h}_2)g$
• D. none of these

Answer 1

The correct option is B $({\rho }_2{h}_2-{\rho }_1{h}_1)g$

Given: Pressure in left chamber$=P_1{P}_0=\text{atmospheric pressure}$
Pressure at same horizontal level is same




On the left side pressure in chamber is $P$ and right limb is open so it will be open to atmospheric pressure at the top
$\text{Pressure at A}=\text{Pressure at B}$
$P+{\rho }_{1}g{h}_1=P_0+{\rho }_{2}g{h}_2$
$P-P_0=g({\rho }_2{h}_2-{\rho }_1{h}_1)$

FINAL ANSWER: (a)