A uniform long tube is bent into a circle of | vedclass

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

Question

(A) Let the vertical line passing through the center be the reference. The interface between the two liquids is at an angle $	heta$ from the vertical

Vedclass · Accepted Answer

$	an ^{-1}\left(\frac{ho-\delta}{ho+\delta}ight)$

Vedclass · Answer

(A) Let the vertical line passing through the center be the reference. The interface between the two liquids is at an angle $	heta$ from the vertical.Since the tube is in a vertical plane, the pressure at the lowest point must be the same from both sides.Let the density of the liquid on the left be $\delta$ and on the right be $ho$.The vertical height of the center of mass of the liquid column of density $\delta$ from the lowest point is $h_1 = R(1 - \cos 	heta)$.The vertical height of the center of mass of the liquid column of density $ho$ from the lowest point is $h_2 = R(1 - \cos 	heta)$.For equilibrium, the pressure at the lowest point of the tube must be equal from both sides.The pressure exerted by a liquid column in a circular tube is proportional to the vertical depth of its center of mass.The condition for equilibrium is given by:$\delta g R(\cos 	heta + \sin 	heta) = ho g R(\cos 	heta - \sin 	heta)$Dividing both sides by $gR$:$\delta(\cos 	heta + \sin 	heta) = ho(\cos 	heta - \sin 	heta)$$\delta \cos 	heta + \delta \sin 	heta = ho \cos 	heta - ho \sin 	heta$$\sin 	heta(ho + \delta) = \cos 	heta(ho - \delta)$$	an 	heta = \frac{ho - \delta}{ho + \delta}$$	heta = 	an ^{-1}\left(\frac{ho - \delta}{ho + \delta}ight)$

$A$ uniform long tube is bent into a circle of radius $R$ and it lies in a vertical plane. Two liquids of the same volume but densities $\rho$ and $\delta$ fill half the tube. The angle $\theta$ is

Similar Questions

Depth of a river is $100 \ m$. Magnitude of compressibility of the water is $0.5 \times 10^{-9} \ N^{-1} \ m^2$. The fractional compression in water at the bottom of the river is (Acceleration due to gravity $= 10 \ m/s^2$)

$A$ force of $F = 10 \ N$ is exerted on a surface of area $A = 0.1 \ m^{2}$ at an angle of $60^{\circ}$ with the surface. Find the pressure produced on this surface.

$A$ beaker containing a liquid is kept inside a big closed jar. If the air inside the jar is continuously pumped out,the pressure in the liquid near the bottom of the liquid will

The pressure at half the depth of a lake is equal to two-third pressure at the bottom of the lake. So the depth '$h$' of the lake is ($\rho$ = density of water in the lake, $g$ = acceleration due to gravity, $P_0$ = atmospheric pressure).

Vedclass Test Series

Exam Paper Generator

Online Exam Module