A vertical U-tube of uniform cross-section co | vedclass

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

Question

(B) Let the height of the glycerine column be $h = 10 \, cm$ and its relative density be $\rho_g = 1.2 \, g/cm^3$. The density of water is $\rho_w = 1

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) Let the height of the glycerine column be $h = 10 \, cm$ and its relative density be $ho_g = 1.2 \, g/cm^3$. The density of water is $ho_w = 1.0 \, g/cm^3$.Let the difference in the levels of the free surfaces of the two liquids be $x$.At the interface of the glycerine and water in the $U$-tube,the pressure at the same horizontal level must be equal.Let the pressure at the level of the bottom of the glycerine column be $P$.Pressure on the left limb at this level = $P_{atm} + ho_g \cdot g \cdot h$.Pressure on the right limb at the same level = $P_{atm} + ho_w \cdot g \cdot (h - x)$.Equating the pressures:$ho_g \cdot g \cdot h = ho_w \cdot g \cdot (h - x)$$1.2 \cdot 10 = 1.0 \cdot (10 - x)$$12 = 10 - x$Wait,let's re-evaluate the geometry. The glycerine column of height $h$ pushes the water down by some amount,and the water level rises in the other arm. The difference in height between the top of the glycerine and the top of the water is $x$.From the hydrostatic balance at the level of the glycerine-water interface:Pressure due to glycerine column = Pressure due to water column of height $(h - x)$.$1.2 	imes 10 = 1.0 	imes (10 - x)$ is incorrect based on the diagram. Looking at the diagram,the water level in the right arm is higher than the interface by $(h - x)$.Actually,the pressure balance is $ho_g \cdot g \cdot h = ho_w \cdot g \cdot h_{water}$.Here,the water column height balancing the glycerine is $h_{water} = h - x$.$1.2 	imes 10 = 1.0 	imes (10 - x) \Rightarrow 12 = 10 - x \Rightarrow x = -2$. This implies the water level is higher on the glycerine side,which is wrong.Correct approach: The pressure at the level of the glycerine-water interface (left side) is equal to the pressure at the same level in the right side.Pressure (left) = $ho_g \cdot g \cdot h$.Pressure (right) = $ho_w \cdot g \cdot (h - x)$ where $x$ is the difference in free surface levels.$1.2 	imes 10 = 1.0 	imes (10 - x) \Rightarrow 12 = 10 - x$. This suggests the water level in the right arm must be higher than the glycerine surface.Actually,$h_{water} = h 	imes (ho_g / ho_w) = 10 	imes 1.2 = 12 \, cm$.The difference in levels $x = h_{water} - h = 12 - 10 = 2 \, cm$.

$A$ vertical $U$-tube of uniform cross-section contains water in both the arms. $A$ $10 \, cm$ glycerine column $(R.D. = 1.2)$ is added to one of the limbs. The level difference between the two free surfaces in the two limbs will be ...... $cm$.

Similar Questions

$A$ siphon in use is demonstrated in the following figure. The density of the liquid flowing in the siphon is $1.5 \text{ g/cm}^3$. The pressure difference between the points $P$ and $S$ will be

Why do bags and suitcases have broad (wide) handles?

In which layer of the atmosphere is water vapour present?

The term 'fluid' is used for ............

Vedclass Test Series

Exam Paper Generator

Online Exam Module