Two liquids of densities \rho _1 an \rho _2 ( \rho _2 = 2 \rho _1 ) are filled up behind a square wall of side 10 m as shown in figure. Each liquid has a height of 5 m. The ratio of the forces due to these liquids exerted on upper part MN to that at the lower part NO is (Assume that the liquids are not mixing)

Properties of Matter PYQ — JEE Main 2020 · 8 Jan · Shift 2 · Q49 with Solution

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Correct answer: C

Interpret the setup

A square vertical wall has side $10\,\text{m}$ , so its total height is $10\,\text{m}$ .

The wall is divided into two equal vertical parts:

Upper part $MN$ : height $5\,\text{m}$
Lower part $NO$ : height $5\,\text{m}$

Two immiscible liquids are filled behind it:

Upper liquid: density $\rho_1$ , height $5\,\text{m}$
Lower liquid: density $\rho_2 = 2\rho_1$ , height $5\,\text{m}$

Let the width of the wall be $10\,\text{m}$ .

We need:

\frac{F_{MN}}{F_{NO}}

where $F_{MN}$ is force on the upper half and $F_{NO}$ on the lower half.

Force on upper part $MN$

Take depth $y$ measured from the top surface of the upper liquid. For $0 \le y \le 5$ m, pressure is:

p(y)=\rho_1 g y

A horizontal strip of width $10$ m and thickness $dy$ has area:

dA = 10\,dy

So differential force:

dF = p(y)\,dA = \rho_1 g y \cdot 10\,dy

Hence,

F_{MN} = 10\rho_1 g \int_0^5 y\,dy

F_{MN} = 10\rho_1 g \left[\frac{y^2}{2}\right]_0^5

F_{MN} = 10\rho_1 g \cdot \frac{25}{2} = 125\rho_1 g

Force on lower part $NO$

For the lower half, depth from top lies between $y=5$ and $y=10$ .

Pressure at a point in lower liquid equals:

pressure due to upper liquid of height $5$ m, plus
pressure due to lower liquid below interface

So for $5 \le y \le 10$ ,

p(y)= \rho_1 g(5) + \rho_2 g(y-5)

Since $\rho_2 = 2\rho_1$ ,

p(y)=5\rho_1 g + 2\rho_1 g(y-5)

p(y)=\rho_1 g(2y-5)

Now,

F_{NO} = 10\int_5^{10} p(y)\,dy

F_{NO} = 10\rho_1 g \int_5^{10} (2y-5)\,dy

Evaluate the integral:

\int (2y-5)dy = y^2 - 5y

So,

F_{NO} = 10\rho_1 g \left[ y^2 - 5y \right]_5^{10}

F_{NO} = 10\rho_1 g \Big((100-50) - (25-25)\Big)

F_{NO} = 10\rho_1 g \cdot 50 = 500\rho_1 g

Take the ratio

\frac{F_{MN}}{F_{NO}} = \frac{125\rho_1 g}{500\rho_1 g} = \frac{1}{4}

Match with options

\boxed{\frac{1}{4}}

So the correct option is:

C: $\dfrac{1}{4}$

Comparison with stored answer

Stored correct answer: C

Our derived answer: C

So they agree.

More from Properties of Matter

View written solutionFree

Correct answer: C

Interpret the setup

A square vertical wall has side $10\,\text{m}$ , so its total height is $10\,\text{m}$ .

The wall is divided into two equal vertical parts:

Upper part $MN$ : height $5\,\text{m}$
Lower part $NO$ : height $5\,\text{m}$

Two immiscible liquids are filled behind it:

Upper liquid: density $\rho_1$ , height $5\,\text{m}$
Lower liquid: density $\rho_2 = 2\rho_1$ , height $5\,\text{m}$

Let the width of the wall be $10\,\text{m}$ .

We need:

\frac{F_{MN}}{F_{NO}}

where $F_{MN}$ is force on the upper half and $F_{NO}$ on the lower half.

Force on upper part $MN$

Take depth $y$ measured from the top surface of the upper liquid. For $0 \le y \le 5$ m, pressure is:

p(y)=\rho_1 g y

A horizontal strip of width $10$ m and thickness $dy$ has area:

dA = 10\,dy

So differential force:

dF = p(y)\,dA = \rho_1 g y \cdot 10\,dy

Hence,

F_{MN} = 10\rho_1 g \int_0^5 y\,dy

F_{MN} = 10\rho_1 g \left[\frac{y^2}{2}\right]_0^5

F_{MN} = 10\rho_1 g \cdot \frac{25}{2} = 125\rho_1 g

Force on lower part $NO$

For the lower half, depth from top lies between $y=5$ and $y=10$ .

Pressure at a point in lower liquid equals:

pressure due to upper liquid of height $5$ m, plus
pressure due to lower liquid below interface

So for $5 \le y \le 10$ ,

p(y)= \rho_1 g(5) + \rho_2 g(y-5)

Since $\rho_2 = 2\rho_1$ ,

p(y)=5\rho_1 g + 2\rho_1 g(y-5)

p(y)=\rho_1 g(2y-5)

Now,

F_{NO} = 10\int_5^{10} p(y)\,dy

F_{NO} = 10\rho_1 g \int_5^{10} (2y-5)\,dy

Evaluate the integral:

\int (2y-5)dy = y^2 - 5y

So,

F_{NO} = 10\rho_1 g \left[ y^2 - 5y \right]_5^{10}

F_{NO} = 10\rho_1 g \Big((100-50) - (25-25)\Big)

F_{NO} = 10\rho_1 g \cdot 50 = 500\rho_1 g

Take the ratio

\frac{F_{MN}}{F_{NO}} = \frac{125\rho_1 g}{500\rho_1 g} = \frac{1}{4}

Match with options

\boxed{\frac{1}{4}}

So the correct option is:

C: $\dfrac{1}{4}$

Comparison with stored answer

Stored correct answer: C

Our derived answer: C

So they agree.

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