Question

Two immiscible liquids are poured in a $U-tube$ having densities ${\rho }_1=1.0\times {10}^{3}kg/m^3$ and ${\rho }_2=3.0\times {10}^{3}kg{m}^2$. Find the ratio of heights (of the liquids above their interface) $\frac{h_1}{h_2}$.

• A. $3$
• B. $6$
• C. $9$
• D. $12$

Answer 1

The correct option is A $3$

$\text{Since pressure at}$

$\text{Point}A\&B\text{is same}So$

${\rho }_{1}g{h}_1={\rho }_{2}g{h}_2$

$1\times {10}^3\times 10h_1=3\times {10}^3\times 10h_2$

$\frac{h_1}{h_2}=3$

$\text{Answer: - (A)}$