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 \gt \rho _2 ), fill half the circle. The angle \theta between the radius vector passing through the common interface and the vertical is :

\theta = \tan ^ - 1( \rho _1 - \rho _2 \over \rho _1 + \rho _2 )

\theta = \tan ^ - 1\pi ( \rho _1 \over \rho _2 )

\theta = \tan ^ - 1\pi \over 2( \rho _1 \over \rho _2 )

\theta = \tan ^ - 1\pi \over 2( \rho _1 + \rho _2 \over \rho _1 - \rho _2 )

Properties of Matter PYQ — JEE Main 2018 · 15 Apr · Shift 1 · Q63 with Solution

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Correct answer: C

Set up the geometry

A thin tube is bent into a vertical circle of radius $r$ . Two immiscible liquids of densities $\rho_1$ and $\rho_2$ ( $\rho_1>\rho_2$ ) occupy half the circle in total, and their volumes are equal.

Since the tube has uniform cross-section, equal volumes mean equal lengths along the tube.

So each liquid occupies a length equal to one-fourth of the circumference:

\ell=\frac{1}{4}(2\pi r)=\frac{\pi r}{2}

Hence each liquid subtends an angle

\frac{\ell}{r}=\frac{\pi}{2}

at the center.

Locate the interfaces

The liquids together fill a semicircle. Since each occupies a quarter-circle, the two liquid-air free surfaces are separated from the common interface by angles $\pi/2$ on either side.

Let the common interface be at point $P$ , and let the radius $OP$ make angle $\theta$ with the vertical.

Then the two free surfaces are located at angular positions differing by $\pi/2$ from $P$ .

Use hydrostatic pressure balance at the common interface

Pressure at the common interface computed through each liquid must be equal.

Let the free surface of liquid $\rho_1$ be at height $h_1$ above the interface, and that of liquid $\rho_2$ be at height $h_2$ above the interface.

Then

P_{\text{interface}}=P_{atm}+\rho_1 g h_1

and also

P_{\text{interface}}=P_{atm}+\rho_2 g h_2

Therefore,

\rho_1 h_1=\rho_2 h_2

Find the vertical height differences from geometry

Take the center as origin, with upward vertical as positive $y$ .

If the interface point $P$ has radius making angle $\theta$ with the vertical, then its height is

y_P=r\cos\theta

A point $\pi/2$ away on the circle has height

y=r\cos\left(\theta\pm\frac{\pi}{2}\right)=\mp r\sin\theta

Depending on which side is occupied by which liquid, the two free surfaces have heights:

y_1=-r\sin\theta, \qquad y_2=+r\sin\theta

relative to the center.

Thus the vertical rises from the free surfaces to the interface are

h_1=y_P-y_1=r(\cos\theta+\sin\theta)

h_2=y_P-y_2=r(\cos\theta-\sin\theta)

Since the denser liquid must have the smaller vertical column for equal pressure rise, we assign

\rho_1 h_2=\rho_2 h_1

that is,

\rho_1(\cos\theta-\sin\theta)=\rho_2(\cos\theta+\sin\theta)

Solve for $\theta$

Expand:

\rho_1\cos\theta-\rho_1\sin\theta=\rho_2\cos\theta+\rho_2\sin\theta

Bring like terms together:

(\rho_1-\rho_2)\cos\theta=(\rho_1+\rho_2)\sin\theta

Hence

\tan\theta=\frac{\rho_1-\rho_2}{\rho_1+\rho_2}

So,

\boxed{\theta=\tan^{-1}\left(\frac{\rho_1-\rho_2}{\rho_1+\rho_2}\right)}

Match with options

This corresponds to:

\boxed{\text{Option C}}

Comparison with stored correct answer

Stored correct answer: C

Our derived answer: C

So the derived answer agrees with the stored correct answer.

More from Properties of Matter

View written solutionFree

Correct answer: C

Set up the geometry

A thin tube is bent into a vertical circle of radius $r$ . Two immiscible liquids of densities $\rho_1$ and $\rho_2$ ( $\rho_1>\rho_2$ ) occupy half the circle in total, and their volumes are equal.

Since the tube has uniform cross-section, equal volumes mean equal lengths along the tube.

So each liquid occupies a length equal to one-fourth of the circumference:

\ell=\frac{1}{4}(2\pi r)=\frac{\pi r}{2}

Hence each liquid subtends an angle

\frac{\ell}{r}=\frac{\pi}{2}

at the center.

Locate the interfaces

The liquids together fill a semicircle. Since each occupies a quarter-circle, the two liquid-air free surfaces are separated from the common interface by angles $\pi/2$ on either side.

Let the common interface be at point $P$ , and let the radius $OP$ make angle $\theta$ with the vertical.

Then the two free surfaces are located at angular positions differing by $\pi/2$ from $P$ .

Use hydrostatic pressure balance at the common interface

Pressure at the common interface computed through each liquid must be equal.

Let the free surface of liquid $\rho_1$ be at height $h_1$ above the interface, and that of liquid $\rho_2$ be at height $h_2$ above the interface.

Then

P_{\text{interface}}=P_{atm}+\rho_1 g h_1

and also

P_{\text{interface}}=P_{atm}+\rho_2 g h_2

Therefore,

\rho_1 h_1=\rho_2 h_2

Find the vertical height differences from geometry

Take the center as origin, with upward vertical as positive $y$ .

If the interface point $P$ has radius making angle $\theta$ with the vertical, then its height is

y_P=r\cos\theta

A point $\pi/2$ away on the circle has height

y=r\cos\left(\theta\pm\frac{\pi}{2}\right)=\mp r\sin\theta

Depending on which side is occupied by which liquid, the two free surfaces have heights:

y_1=-r\sin\theta, \qquad y_2=+r\sin\theta

relative to the center.

Thus the vertical rises from the free surfaces to the interface are

h_1=y_P-y_1=r(\cos\theta+\sin\theta)

h_2=y_P-y_2=r(\cos\theta-\sin\theta)

Since the denser liquid must have the smaller vertical column for equal pressure rise, we assign

\rho_1 h_2=\rho_2 h_1

that is,

\rho_1(\cos\theta-\sin\theta)=\rho_2(\cos\theta+\sin\theta)

Solve for $\theta$

Expand:

\rho_1\cos\theta-\rho_1\sin\theta=\rho_2\cos\theta+\rho_2\sin\theta

Bring like terms together:

(\rho_1-\rho_2)\cos\theta=(\rho_1+\rho_2)\sin\theta

Hence

\tan\theta=\frac{\rho_1-\rho_2}{\rho_1+\rho_2}

So,

\boxed{\theta=\tan^{-1}\left(\frac{\rho_1-\rho_2}{\rho_1+\rho_2}\right)}

Match with options

This corresponds to:

\boxed{\text{Option C}}

Comparison with stored correct answer

Stored correct answer: C

Our derived answer: C

So the derived answer agrees with the stored correct answer.

Properties of Matter question

Properties of Matter question

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