## Grade 11 Physics Note

# Hydrostatics

**State Pascal’s law:**

Pascal's law of liquid pressure states that, "The pressure is equally exerted perpendicularly on all sides as pressure is applied on a liquid kept in a closed container.

**Archimedes principle**** and the laws of liquid pressure:**

It states that, "when an object is completely or partially immersed in a liquid it experiences upthrust which is equal to weight of liquid displaced by it.

I.e. up thrust = weight of liquid displaced by the object

Let’s take a container containing a liquid of density $\rho $ in which an object of area A and height h is completely immersed in it. Let x be the height of liquid column above the upper surface and P_{1 }be the pressure of liquid on it. Then, the pressure is P_{1} = $\rho {\rm{g*x}}$ ------- (i)

The downward thrust acting the upward surface is,

F_{1} = P_{1}${\rm{*A}}$

= $\rho {\rm{gx*A}}$ --------- (ii)

Similarly, the pressure acting on the lower surface of the object is,

P_{2} = $\rho {\rm{g}}\left( {{\rm{x}} + {\rm{h}}} \right)$ ------- (iii)

Then, the upthrust acting on lower surface of the object is,

F = P_{2}${\rm{*A}}$

=$\rho {\rm{g}}\left( {{\rm{x}} + {\rm{h}}} \right){\rm{A}}$ ------- (iv)

Since, F_{2} is more than F_{1} , the net thrust act is upward direction and it is the upthrust. So,

Upthrust = F_{2}$ - $ F_{1}

=${\rm{\: }}\rho {\rm{g}}\left( {{\rm{x}} + {\rm{h}}} \right){\rm{A}} - \rho {\rm{g*A}}$

= $\rho {\rm{gxA}} + \rho {\rm{ghA}} - \rho {\rm{gxA}}$

= $($A${\rm{*h}}){\rm{*}}\rho {\rm{*g}}$

= volume of the object ${\rm{*\: }}\rho {\rm{*g}}$

= Mass of liquid displaced${\rm{\: *\: g}}$

= Weight of liquid displaced

Hence proved

Also, the upthrust F_{3 }and F_{4} acting on two sides of the object are equal and opposite and they cancel in pair leaving no effect on the object.

**The laws of hydrostatic pressure and Archimedes’ Principle from them;**

The branch of mechanics that deals with the rest state of fluids is called hydrostatics. Pascal's law of liquid pressure (hydrostatic pressure) states that, "The pressure is equally exerted perpendicularly on all sides as pressure is applied on a liquid kept in a closed container.

Let’s take a container containing a liquid of density $\rho $ in which an object of area A and height h is completely immersed in it. Let x be the height of liquid column above the upper surface and P_{1 }be the pressure of liquid on it. Then, the pressure is P_{1} = $\rho {\rm{g*x}}$ ------- (i)

The downward thrust acting the upward surface is,

F_{1} = P_{1}${\rm{*A}}$

= $\rho {\rm{gx*A}}$ ---------(ii)

Similarly, the pressure acting on the lower surface of the object is,

P_{2} = $\rho {\rm{g}}\left( {{\rm{x}} + {\rm{h}}} \right)$ ------- (iii)

Then, the upthrust acting on lower surface of the object is,

F = P_{2}${\rm{*A}}$

=$\rho {\rm{g}}\left( {{\rm{x}} + {\rm{h}}} \right){\rm{A}}$ ------- (iv)

Since, F_{2} is more than F_{1} , the net thrust act is upward direction and it is the upthrust. So,

Upthrust = F_{2}$ - $ F_{1}

=${\rm{\: }}\rho {\rm{g}}\left( {{\rm{x}} + {\rm{h}}} \right){\rm{A}} - \rho {\rm{g*A}}$

= $\rho {\rm{gxA}} + \rho {\rm{ghA}} - \rho {\rm{gxA}}$

= $($A${\rm{*h}}){\rm{*}}\rho {\rm{*g}}$

= volume of the object ${\rm{*\: }}\rho {\rm{*g}}$

= Mass of liquid displaced${\rm{\: *\: g}}$

= Weight of liquid displaced

Hence proved

**The law of flotation:**

It states that, "For a floating object, the weight of object is equal to weight of liquid displaced by its immersed portion."

i.e. Weight of object = Weight of liquid displaced by its immersed portion

**An expression for the resultant force experienced by a body of mass m and density p when it is immersed in a fluid of density p:**

It states that, "when an object is completely or partially immersed in a liquid it experiences upthrust which is equal to weight of liquid displaced by it.

I.e. up thrust = weight of liquid displaced by the object

Let’s take a container containing a liquid of density $\rho $ in which an object of area A and height h is completely immersed in it. Let x be the height of liquid column above the upper surface and P_{1 }be the pressure of liquid on it. Then, the pressure is P_{1} = $\rho {\rm{g*x}}$ ------- (i)

The downward thrust acting the upward surface is,

F_{1} = P_{1}${\rm{*A}}$

= $\rho {\rm{gx*A}}$ --------- (ii)

Similarly, the pressure acting on the lower surface of the object is,

P_{2} = $\rho {\rm{g}}\left( {{\rm{x}} + {\rm{h}}} \right)$ ------- (iii)

Then, the upthrust acting on lower surface of the object is,

F = P_{2}${\rm{*A}}$

=$\rho {\rm{g}}\left( {{\rm{x}} + {\rm{h}}} \right){\rm{A}}$ ------- (iv)

Since, F_{2} is more than F_{1} , the net thrust act is upward direction and it is the upthrust. So,

Upthrust = F_{2}$ - $ F_{1}

=${\rm{\: }}\rho {\rm{g}}\left( {{\rm{x}} + {\rm{h}}} \right){\rm{A}} - \rho {\rm{g*A}}$

= $\rho {\rm{gxA}} + \rho {\rm{ghA}} - \rho {\rm{gxA}}$

= $($A${\rm{*h}}){\rm{*}}\rho {\rm{*g}}$

= volume of the object ${\rm{*\: }}\rho {\rm{*g}}$

= Mass of liquid displaced${\rm{\: *\: g}}$

= Weight of liquid displaced

Hence proved

Also, the upthrust F_{3 }and F_{4} acting on two sides of the object are equal and opposite and they cancel in pair leaving no effect on the object.

**Centre of buoyancy and metacentre:**

A body immersed in a liquid displaces a volume of liquid equal to the immersed portion of the body. The displaced liquid exerts an upward force on the body due to pressure difference at bottom and top levels. This force is called upthrust or buoyancy. The point at which a vertical line through the heeled centre of buoyancy crosses the line through the original, vertical centre of buoyancy is the metacentre. The metacentre remains directly above the centre of buoyancy by definition.The buoyant force on an object submerged in a fluid is caused by the pressure difference between the top and bottom of the object. The larger pressure at greater depth pushes upward on the object. *There* is *no* acceleration of the object upward or downward. The buoyant force on an *object *is the resultant of the pressure distribution on the exterior surface of the object.

The centre of buoyancy of a floating body is the point about which all the body parts exactly buoy each other—in other words, the effective centre of the displaced water. The metacentre remains directly above the centre of buoyancy regardless of the tilt of a floating body, such as a ship. When at rest on even keel, the vessel’s centre of buoyancy is directly below the centre of gravity as well as below the metacentre. (The centre of gravity is the point in a body about which all parts of the body balance each other.) When a vessel tilts, one side displaces more water than does the other and the centre of buoyancy moves and is no longer directly under the centre of gravity, but, regardless of the amount of the tilt, the centre of buoyancy remains directly below the metacentre.

Condition for sinking and flotation:

An object sinks or floats on a liquid depending upon the weight of the object and weight of the liquid displaced by it.

Let, 'W' be the weight of object and 'W^{1}' be the weight of liquid displaced by it in which '$\rho '$ be the density of the object and '${\rho ^1}{\rm{'\: }}$be the density of liquid.

Case 1

The object sinks in the liquid if W $ > $ W^{1}

or, ${\rm{V*}}\rho {\rm{*g}} > {\rm{V*}}{\rho _1}{\rm{*g}}$

or,$\rho > {\rho _1}$

Case 2

The object remains inside the liquid is W = W^{1}

or, ${\rm{V*}}\rho {\rm{*g}} = {\rm{V*}}{\rho _1}{\rm{*g}}$

or, $\rho = {\rho _1}$

Case 3

The object floats in the liquid if, W $ < {{\rm{W}}^1}$

or, ${\rm{V*}}\rho {\rm{*g}} < {\rm{V*}}{\rho _1}{\rm{*g}}$

or, ${\rm{g}} < {\rho _1}$

Meta - centre G $ \to $ Centre of gravity(C.G) of the object.

B $ \to $ C.G of liquid displaced

B^{1}$ \to $ C.G of displaced liquid when the hydrometer is loaded at base

In the first figure a straight line is drawn passing through G and B. In the second figure, a straight line is also drawn passing through B^{1}. Then, the point of intersection between the initial straight line passing through B^{1} is called metacentre. It is the combined center gravity of the floating object and it should lie in the lowest line as far as possible.

For floating object or moving vehicle, the meta center should lie on the lowest position for its stability otherwise, it topples. Due to this reason, the lower part of ship is loaded and lower part of boat is also loaded.

**Specific gravity: **

We define specific gravity as the ratio of weight of certain volume of a substance is to weight of same volume of water at 4$^{\circ}$

i.e. specific gravity = $\frac{{{\rm{weight\: of\: certain\: volume\: of\: a\: substace}}}}{{{\rm{weight\: of\: same\: volume\: of\: water\: at\: }}4^{\circ}}}$

=$\frac{\rho( \frac{\rm{gm}}{\rm{cc}})}{1(\frac{\rm{gm}}{\rm{cc}})}$

= $\rho $

Hence, specific gravity is the ratio of two similar quantities and it has no units and dimensions. It is a unit less and dimensionless pure number. The density and specific gravity are numerically equal but density has unit in C.G.S system whereas the specific gravity has no unit. If the density of glass is 2.5gm/cc then, the specific gravity of glass is only 2.5. The specific gravity of a substance can be found by the following ways. Specific gravity of a solid is heavier than water and insoluble in water. For this purpose, we take the weight of object in air and then we find its weight in water. Then, specific gravity is represented by,

Specific gravity = $\frac{{{\rm{weight\: of\: the\: object\: in\: air\: }}}}{{{\rm{loss\: of\: the\: weight\: in\: water}}}}{\rm{*}}\rho $_{t}

=$\frac{{\rm{W}}}{{{\rm{W}} - {\rm{W}}1}}{\rm{*}}\rho $_{t}

Where, W = Weight of the object in air

W_{1} = Weight of the object in water

$\rho $_{t} = density of specific gravity

Similarly, the specific gravity of liquid insoluble in water is given by,

Specific gravity = $\frac{{{\rm{Weight\: of\: certain\: volume\: of\: a\: liquid}}}}{{{\rm{Weight\: of\: same\: volume\: of\: water\: at\: }}4^{\circ}{\rm{\: }}}}{\rm{*\: }}\rho $_{t}

= $\frac{{{\rm{W}} - {\rm{W}}1}}{{{\rm{W}}2 - {\rm{W}}1}}{\rm{*\: }}\rho $_{t}

Where,

W = Weight of (liquid+ vessel)

W_{1} = Weight of vessel

W_{2} = Weight of (water+ vessel)

But in lab, we take a sinker and it is weighted in air and then in liquid Similarly, it is weighted in water then the specific gravity of liquid is given by,

Specific gravity = $\frac{{{\rm{Weight\: \: loss\: in\: liquid\: }}}}{{{\rm{loss\: of\: weight\: in\: water}}}}{\rm{*}}$$\rho $_{t}

**The principle of graduation of a common hydrometer:**

Hydrometer is a device used to measure specific gravity of a solid or liquid. It floats in a liquid and so works in the principle of floatation. Principle of flotation states that, "For a floating object, the weight of object is equal to weight of liquid displaced by its immersed portion." i.e.Weight of object = Weight of liquid displaced by its immersed portion. The form of *Hydrometer* first used was one with a uniformly *graduated *stem, a pear-shaped form of ball, and a weighted ball underneath. The *Hydrometer* is based on the *principle*, that the weight of a floating body is equal to the weight of the quantity of liquid which it displaces.

Fig12_1: Variation immersion hydrometer

In variable immersion type, the stem of the meter is graduated into the index, and up to that index of liquid level gives the specific gravity of that liquid.

**Railway tracks are laid on large sized wooden, iron or cement sleepers**:

Railway tracks are laid on large sized wooden, iron or cement sleepers. This is done so that the thrust due to weight of train is spread over a large area. Due to this there is less pressure of the train on the ground and railway line does not sink into the ground.

**It is difficult to walk bare footed on a sandy ground;**

The man feet has smaller area and weight of man falls on smaller area of sand the force per unit area or presser produced on sand is large due to large pressure It is difficult to walk bare footed on a sandy ground.

**Pins and nails are made to have pointed ends:**

Pins and nails are made to have pointed ends because due to sharp pin nails and the needle may put force on small area producing large pressure which is sufficient to pierce the object.