CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

A thin uniform tube is bent into a circle of radius $$r$$ in the vertical plane. Equal volumes of two immiscible liquids, whose densities are $${ \rho  }_{ 1 }$$ and $${ \rho  }_{ 2 }$$ $$({ \rho  }_{ 1 }> { \rho  }_{ 2 })$$ fill half the circle. The angle $$\theta$$ between the radius vector passing through the common interface and the vertical is:


A
θ=tan1[π2(ρ1ρ2ρ1+ρ2)]
loader
B
θ=tan1π2(ρ1ρ2ρ1+ρ2)
loader
C
θ=tan1π(ρ1ρ2)
loader
D
None of above
loader

Solution

The correct option is B None of above
Let us find the pressure at the lowest point 1. Since the liquid has density $$\rho_1$$ and height $$h_1$$ on the left hand side of point 1, we have
$$P_1 = \rho_1 g h_1$$.............(1)
Since two liquid columns of height $$h_2 \ and \ h'_2$$ and densities $$\rho_1 \ and \ \rho_2$$ are situated above point 1, on the right hand side, we have
$$P_2 = \rho_1 g h_2 + \rho_2 g h'_2$$..............(2)
Equating $$P_1 and P_2$$, we get
$$\rho_1 h_2 + \rho_2 h'_2 = \rho_1 h_1$$
Substituting
$$h'_2 = R sin\theta +R cos \theta, \ h_2 = R(1 - cos \theta) \ and \ h_1 = R (1 - sin \theta)$$
$$\rho_1 R (1 - cos \theta) + \rho_2 R (sin \theta + cos\theta) = \rho_1R (1 - sin \theta)$$ 
This gives
$$tan \theta = \dfrac{\rho_1 - \rho_2}{\rho_1 + \rho_2}$$

807762_868149_ans_4866dac222334e289fe439a3364e7d33.png

Physics

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image