Physics Topics cover a broad range of concepts that are essential to understanding the natural world.

## Pressure Difference And Buoyant Force in a Liquid of a Uniformly Accelerated Container Filled with the Liquid

The mathematical expression used so far for the buoyant force acting on a body immersed in a liquid holds true for any liquid at rest. However, those equations are to be modified for a uniformly accelerated container filled with the liquid.

### In the Case of a Moving Lift

Case 1: When a lift with a liquid-filled container moves upwards with an acceleration a (or moves downwards with a retardation a):

Difference in Pressure: Let us imagine a liquid column of height h and cross-sectional area A inside the lift [Fig.]. If the density of the liquid is ρ, then the mass of the liquid column, m = hAρ.

### Different forces exerted on the liquid column:

- If p
_{1}is the pressure on the upper surface of the liquid column, then the thrust on that surface F_{1}= p_{1}A; this force acts vertically downwards. - If p
_{2}is the pressure on the lower surface of the liquid column, then the thrust on that surface F_{2}= p_{2}A; this force acts vertically upwards. - The weight of the liquid column mg = hAρg; this force acts vertically downwards.

Therefore, the equation of motion of the liquid column is

F_{2} – F_{1} – mg = ma

or, p_{2}A – p_{1}A = ma + mg

or, (p_{2} – p_{1})A = m(a + g) ……….. (1)

= hAρ(a + g)

or, p_{2} – p_{1} = hρ(g + a) ………. (2)

So the difference in pressure increases by hρa.

Buoyont force : The buoyant force or upthrust on the liquid column, W’ = (p_{2} – p_{1})A

From equation (1) we get, W’ = m(g + a) ………… (3)

So, the buoyant force increases by ma.

Case 2: When the lift with a liquid-filled container moves downwards with an acceleration a (or moves upwards with a retardation a):

Following the previous calculation, it can be seen that

P_{2} – p_{1} = hρ(g – a)

i.e., the difference in pressure, in this case, decreases by hρa.

Similarly, the buoyant force, W’ = m(g – a)

i.e., the buoyant force, in this case, decreases by ma.

Case 3: When the lift with a liquid-filled container moves downwards freely (i.e., a = g):

In this case, the difference in pressure, p_{2} – p_{1} = 0

and the buoyant force, W’ = 0

i.e., both the pressure at any point inside a liquid and the upthrust on any immersed object become nil.

Case 4: When the lift with a liquid-filled container is at rest or moves up or down with uniform speed:

Here, a = 0.

∴ Difference in pressure, p_{2} – p_{1} = hρg

and buoyant force, W’ = mg.

### Free Surface in a Liquid-filled Container Moving Horizontally with Uniform Acceleration a

A horizontal cylinder of length l and cross-sectional area a is imagined inside a liquid [Fig.]

If the density of the liquid is ρ, then the mass of the said liquid cylinder, m = lαρ.

Let the liquid exert pressures P_{A} and P_{B} respectively at the two ends A and B of the cylinder.

∴ The equation of motion of the liquid cylinder is

P_{A}α – P_{B}α = ma

or, α(p_{A} – p_{B}) = lαρa or, P_{A} – P_{B} = lρa ……… (1)

Pressures at two points on the same horizontal line inside a liquid-filled container moving horizontally with uniform acceleration are not the same.

Due to the absence of any vertical acceleration of the container, the pressure at any point inside the liquid is similar to that in the case of a liquid at rest. But in this case, the free surface of the liquid does not remain horizontal. Suppose the free surface makes angle θ with the horizontal.’

Now, p_{A} = h_{1}ρg and p_{B} = h_{2}ρg

∴ According to equation (1),

h_{1}ρg – h_{2}ρg or, \(\frac{h_1-h_2}{l}\) = \(\frac{a}{g}\)

or, tanθ = \(\frac{a}{g}\) or, θ = tan^{-1}\(\left(\frac{a}{g}\right)\)

### Numerical Examples

**Example 1.**

A body weighs 100 g × g and 40 g × g in air and water respectively. Calculate the weight of the body when immersed in a liquid of specific gravity 0.8.

**Solution:**

Weight of the body in air = 100 g × g, weight of the body in water =40 g × g.

∴ Weight of displaced water = (100 – 40) × g = 60 g × g.

∴ Volume of displaced water = 60 cm^{3}

So, the volume of the body = 60 cm^{3}

Specific gravity of the liquid = 0.8

∴ Density of the liquid = 0.8 g ᐧ cm^{-3}.

∴ Weight of the displaced liquid

= (60 × 0.8) × g = 48 g × g.

∴ Weight of the body in the liquid

= weight of the body in air – weight of the displaced liquid

= (100 – 48) × g = 52 g × g = 50960 dyn.

**Example 2.**

The mass of a steamer is 10 tonnes. When it enters a fresh water lake from the sea, it displaces 50 L more water than before. Calculate the density of sea water. [Mass oil L of fresh water = 1 kg]

**Solution:**

Mass of displaced sea water

= mass of the streamer

= 10 × 10^{3} kg = 10^{7} g.

∴ Volume of displaced sea water

= \(\frac{10^7}{\rho}\) cm^{3} [ρ = density of sea water]

Mass of displaced 1ae water 10^{7} g

As the density of fresh water is 1 g ᐧ cm^{-3}, the volume of displaced lake water = 10^{7} cm^{3}.

According to the problem,

**Example 3.**

The internal and external diameters of a sphere are 8cm and 10cm respectively. The sphere neither floats nor sinks In a liquid of specific gravity 1.5 g ᐧ cm^{-3}. Calculate the density of the material of the sphere.

**Solution:**

Let the density of the material of the given sphere be ρ.

According to the condition of floatation,

volume of the sphere × density of the material of the sphere

= volume of displaced liquid × density of that liquid

**Example 4.**

A body of density ρ is placed slowly on the surface of a liquid of density δ. If the depth of the liquid is d, then prove that the time taken by the body to reach the bottom of the liquid is \(\left[\frac{2 d \rho}{g(\rho-\delta)}\right]^{1 / 2}\) second.

**Solution:**

Let the mass of the body be in.

∴ Its volume = \(\frac{m}{\rho}\).

∴ Volume of 1iqiid displaced = \(\frac{m}{\rho}\)

∴ Mass of liquid displaced = \(\frac{m}{\rho}\) × 5

If the downward acceleration of the body in the liquid is a, then by the problem

mg – \(\frac{m}{\rho}\)ᐧδg = ma

or, a = (1 – \(\frac{\delta}{\rho}\))g = \(\left(\frac{\rho-\delta}{\rho}\right)\)g

If the time taken by the body to reach the bottom of the liquid is t, then

**Example 5.**

The weight of a body in air is 0.4 g × g. Its weight along with an sinker in water = 3.37 g × g. The weight of the sinker in air = 4 g × g. Find the specific gravity of the body. [Specific gravity of the sinker = 8]

**Solution:**

Let the density of the body be d g ᐧ cm^{-3}.

∴ Volume of the body = \(\frac{0.4}{d}\)cm^{3}; volume of the sinker = \(\frac{4}{8}\) = 0.5 cm^{3}

∴ Volume of water displaced by the body along with the sinker =(o.5 + \(\frac{0.4}{d}\)) cm^{3}

∴ Weight of displaced water = (0.5 + \(\frac{0.4}{d}\)) g × g

Again, weight of displaced water

= weight of the body along with sinker in air —

weight of the body along with sinker in water

= (0.4 + 4) × g – 3.37 × g =(4.4 – 3.37) × g

= 1.03 g × g

∴ 0.5 + \(\frac{0.4}{d}\) = 1.03 or, \(\frac{0.4}{d}\) = 0.53

or, d = \(\frac{0.4}{0.53}\) = 0.75

∴ Specific gravity of the body = 0.75.

**Example 6.**

A piece of wood of volume 20.5 cm^{3} is tied to a piece of lead of volume 1 cm^{3}. State whether the combination will sink or float in water. [Specific gravity of wood and lead are respectively 0.5 and 11.4]

**Solution:**

Weight of the piece of wood

= 20.5 × 0.5 × g = 10.25g × g

Weight of lead = 1 × 11.4 × g = 11.4g × g

∴ Total weight of the combination

=(10.25 + 11.4) × g = 21.65 g × g

Total volume of wood and lead =(20.5 + 1) = 21.5 cm^{3}

∴ Weight of displaced water by wood and lead = 21.5 g × g

∵ Weight of combination > weight of displaced water

So, when the piece of wood and lead are tied together, the combination will sink in water.

**Example 7.**

A piece of an alloy of gold and silver weighs 25 g × g in air and 23 g × g in water. Find out the amount of gold and silver in the piece of alloy. [Specific gravity of silver of the body. [Specific gravity of silver is 10.8 and that of gold is 19.3]

**Solution:**

Let the amount of gold in the alloy be x g.

∴ Amount of silver = (25 – x) g

Volume of gold = \(\frac{x}{19.3}\) cm^{3};

volume of silver = \(\frac{25-x}{10.8}\)cm

∴ Volume of the alloy = \(\left(\frac{x}{19.3}+\frac{25-x}{10.8}\right)\) cm^{3}

Again, weight of water displaced by the alloy

= (25 – 23) × g = 2g × g

i.e., mass of this displaced water = 2 g

∴ Volume of displaced water = 2 cm^{3}

∴ \(\frac{x}{19.3}\) + \(\frac{25-x}{10.8}\) = 2

or, 10.8 x + 19.3 × 25 – 19.3x = 2 × 19.3 × 10.8

or, x = 7.72

∴ Amount of gold = 7.72g and amount of silver = 25 – 7.72 = 17.28 g.

**Example 8.**

There is a cavity inside a piece of metal of mass 237.3 g. The apparent weight of the piece of metal immersed completely in water is 192.1 g × g. If the density of the metal is 7.91 g cm^{-3}, then find the volume of the cavity.

**Solution:**

Volume of the material of the metal piece

= \(\frac{237.3}{7.91}\) = 30 cm^{3};

mass of water displaced by the metal piece

= (237.3 – 192.1) × g = 45.2g × g

∴ Volume of displaced water = 45.2 cm^{3}

∴ Volume of the metal piece including the cavity

= 45.2 cm^{3}

∴ Volume of the cavity = 45.2 – 30 = 15.2 cm^{3}.

**Example 9.**

A piece of wood weighs 40g × g in air. A piece of brass of mass 12 g is tied to the wooden piece and the combination floats just totally immersed in water. If the specific gravity of brass is 8.5, calculate the specific gravity of wood. [Density of water = 1g ᐧ cm^{-3}]

**Solution:**

Let the density of the body be ρ g ᐧ cm^{-3}.

Total volume of the pieces of wood and brass = \(\left(\frac{40}{\rho}+\frac{12}{8.5}\right)\) cm^{3}

Volume of displaced water = \(\left(\frac{40}{\rho}+\frac{12}{8.5}\right)\) cm^{3}

By the condition,

weight of the piece of wood + weight of the piece of brass = weight of water displaced by both of them

or, 40 + 12 = \(\left(\frac{40}{\rho}+\frac{12}{8.5}\right)\) × 1 or, 52 = \(\frac{40}{\rho}+\frac{12}{8.5}\)

or, \(\frac{40}{\rho}\) = 52 – 1.41 or, ρ = \(\frac{40}{50.59}\) = 0.79

∴ Specific gravity of wood is 0.79.

**Example 10.**

The apparent weights of two bodies suspended from the two ends of a balance beam and immersed in water are the same. The mass and density of one body are 32 g and 8 g ᐧ cm^{-3} respectively. If the density of the other body is 5 g ᐧ cm^{-3}, then find its mass.

**Solution:**

Let the mass of the second body be m g.

Volume of the first body = \(\frac{32}{8}\) = 4 cm^{3}

∴ Volume of water displaced by the first body = 4 cm^{3}

∴ Volume of the second body = \(\frac{m}{5}\) cm^{3}

∴ Volume of water displaced by the second body = \(\frac{m}{5}\) cm^{3}

By the problem, apparent weight of the first body = apparent weight of the second body

∴ Real weight of the first body – weight of water displaced by the first body = real weight of the second body – weight of water displaced by the second body

or, 32 – 4 = m – \(\frac{m}{5}\) or, 28 = \(\frac{4 m}{5}\) or, m = 35

∴ Mass of the second body is 35 g.