Two liquids of densities rho_{1} and rho_{2} | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let the width of the wall be $w$. The area of each section $MN$ and $NO$ is $A = h 	imes w = 5w$.For the upper liquid (part $MN$):The pressure at

Vedclass · Accepted Answer

$\frac{1}{4}$

Vedclass · Answer

(A) Let the width of the wall be $w$. The area of each section $MN$ and $NO$ is $A = h 	imes w = 5w$.For the upper liquid (part $MN$):The pressure at the top is $0$ and at depth $h$ is $ho_{1}gh$. The average pressure is $P_{avg1} = \frac{0 + ho_{1}gh}{2} = \frac{ho_{1}gh}{2}$.The force $F_{1} = P_{avg1} 	imes A = \left(\frac{ho_{1}gh}{2}ight) (5w) = \frac{5}{2} ho_{1}ghw$.For the lower liquid (part $NO$):The pressure at the top of this section (at depth $h$) is $P_{top} = ho_{1}gh$. The pressure at the bottom (at depth $2h$) is $P_{bottom} = ho_{1}gh + ho_{2}gh = ho_{1}gh + 2ho_{1}gh = 3ho_{1}gh$.The average pressure is $P_{avg2} = \frac{P_{top} + P_{bottom}}{2} = \frac{ho_{1}gh + 3ho_{1}gh}{2} = 2ho_{1}gh$.The force $F_{2} = P_{avg2} 	imes A = (2ho_{1}gh) (5w) = 10ho_{1}ghw$.The ratio is $\frac{F_{1}}{F_{2}} = \frac{\frac{5}{2} ho_{1}ghw}{10ho_{1}ghw} = \frac{5}{20} = \frac{1}{4}$.

Similar Questions

Straws are used to take soft drinks. Why?

Gauge pressure is defined as:

$A$ tube is bent into an $L$ shape and kept in a vertical plane. If these three liquids are kept in equilibrium by the piston of area $A$,the value of $\frac{F}{A}$ is:

In which layer of the atmosphere is water vapour present?

Vedclass Test Series

Exam Paper Generator

Online Exam Module