Mechanical Properties Of Fluids

When a pebble is dropped in a fluid, a variable force called viscous force acts on it. This force increases the velocity of the pebble until it becomes constant. This constant velocity is called the critical velocity. Since, the force is variable, acceleration is variable and so v-t graph is curved and not a straight line to which critical velocity v_c is an asymptote.

When a pebble is dropped in a fluid, a variable force called viscous force acts on it. This force increases the velocity of the pebble until it becomes constant. This constant velocity is called the critical velocity. Since, the force is variable, acceleration is variable and so v-t graph is curved and not a straight line to which critical velocity v_c is an asymptote.

The following diagram does not represent a streamline flow.

A streamline can be straight or curved and tangent at which gives the direction of the flow.

As two streamlines cannot cross each other, the given diagram does not represent a streamline.

According to the law of continuity,

Along a streamline,

Av=constant.

This means that at a particular cross-section, the velocity of all fluid particles is constant.

Hence, we can say that the velocity of all fluid particles crossing a given position is constant.

Given:

Diameter of section1 = d₁ = 2.5 cm

Diameter of section2 = d₂ = 3.75 cm

According to the law of continuity,

Along a streamline,

Av=constant.

It is given that the meniscus is convex. This means the angle is obtuse. This indicates that the liquid in the trough is mercury. This can also, be understood from the fact as: -

The surface is always perpendicular to the resultant force. Since the meniscus is convex, the resultant force is inside the liquid. That means that cohesive force is greater than adhesive force like in the case of mercury.

Molecules inside a liquid experience a similar force of attraction due to liquid molecules around it from all directions. Molecules on the surface experience force of attraction by the liquid molecules below it and the air molecules above it. The force of attraction of liquid-liquid is greater than that of liquid-gas. Hence, it experiences a net downward force. As a result of this, the potential energy of a surface molecule is more than that of a molecule inside.

During calculation of pressure, only the component of force that is normal to the surface is taken into consideration. It is defined as the ratio of magnitude of force normal to the magnitude of the area of the surface.

Hence,

Hence, pressure is a scalar quantity as it is a ratio of two scalars.

When the coin is on the top of the block, it displaces volume V₁ of water whose weight is equal to the weight of the coin. Whereas, in the second case, the coin displaces the volume of water V₂ whose volume is equal to the volume of the coin.

According to case 1,

Here,

So,

This means in the second case less volume is displaced and hence h decreases.

As the coin falls into the water, total weight of the block-coin system decreases and less water is displaced than before and hence I decreases.

With increase in temperature, the rate of diffusion of gases increases and hence, the viscosity of gases increases.

With increase in temperature, the cohesive force between the liquid molecules decreases and hence the viscosity decreases.

Critical velocity is the velocity of the liquid up to which liquid can be streamlined and above which it is turbulent.

It is related to coefficient of viscosity and density as,

Hence, for a streamline flow the critical velocity should be high and hence density should be low and coefficient of viscosity should be high. High coefficient of viscosity indicates high viscosity of a liquid.

Viscosity is the measure of thickness of a liquid. It is a property of the liquid. It has magnitude but no direction. Hence, viscosity is a scalar and not a vector quantity.

Surface tension is defined as the force per unit length acting along the imaginary line drawn perpendicular to the tangent line on the surface of the liquid. Its direction is unique and defined. It only depends upon the magnitude force and the length taken into consideration. As it is a ratio of two scalars, surface tension is a scalar and not a vector quantity.

Let volume of iceberg be V₁.

We know,

Weight of iceberg = weight of displaced water

Volume of submerged iceberg = volume of displaced water = V₂.

Since,

Weight of iceberg = weight of displaced water

Therefore, 0.917 fraction of the total volume of the iceberg is submerged.

In the above given case, a vessel filled with water is kept on a weighing pan and the scale is adjusted to zero. Then, a block is suspended by a spring and submerged into the water.

The net force on the block zero as it is in equilibrium and it is given by,

, where x is the compression of the spring.

The reading of the scale as a result of this is the buoyant force on the block due to the water. This force is given by,

Since, the scale is adjusted to zero before the block is suspended into the vessel, the new reading is,

Let ρ be the density of the block, L be its height and A be its area and ρ_w be the density of water.

Let l be the length of the submerge part, then l = Lx

Here, in case1,

Weight of the block = weight of submerged water

In the case2, in the frame relative to the elevator, the effective coefficient of gravity becomes g^’ = g + a

Weight of the block = weight of submerged water

Given:

Radius = r = 2.5×10^-5 m

The surface tension of sap = T = 7.28×10^-2 Nm^-1

angle of contact = 0°

The maximum height h to which sap can rise due to capillary action is given by,

where ρ is density of water.

Since, a lot of trees have heights greater than 0.6m, surface tension does not alone account for the supply of water to the top of all trees.

Every surface aligns itself such that it is perpendicular to the net force so that the net force along the surface is zero.

When the tanker moves with acceleration a, a pseudo force acts on the oil in the opposite direction.

Here,

So, the slope of the free surface is a/g

Given:

Radius1 = r₁ = 0.1cm = 10^-3m

Radius2 = r₂ = 0.2cm = 2×10^-3m

Here, two droplets form a bigger drop.

So, volume of bigger drop = sum of the volumes of the smaller drops.

Now, energy released is equal to the difference in the surface energies.

Hence, of energy is absorbed.

Here, N droplets form a bigger drop.

So, volume of bigger drop = sum of the volumes of the smaller drops.

Now, energy released is equal to the difference in the surface energies.

Now, if is the temperature drop, m is the mass and c is the specific heat then,

is negative as R>r.

Hence, the drop-in temperature is,

Given:

Surface Tension = T =7.28×10^-2 Nm^-1

Vapour pressure = P = 2.33×10³ Pa

Radius = r

Now,

Excess pressure = 2T/r

We know, the drop will evaporate when excess pressure is equal to vapour pressure.

(a) As we go up in the atmosphere, pressure decreases.

Consider a layer of atmosphere at height h and area A.

Let the pressure at h be p + dp.

Let the pressure at h + dh be p.

This layer is in equilibrium and hence, upward and downward forces are balanced.

Pressure × Area = Force = ma =

Negative sign indicates that pressure decreases with height.

(b) Here,

Hence,

Integrating on both sides,

(c)

(d) We can assume only under isothermal conditions which is valid near the surface of the earth and not at great heights.

(a) Given:

Latent heat of vaporisation of water = L_v = 540 k Cal kg^-1 = 540 ×10³ ×4.2 J = 2268×10³J

Energy required to evaporate 1kmole of water is given by,

E

We know,

The number of molecules in 1kmole of water are N_A.

Energy required for evaporation of 1 molecule is given by,

(b) Assume molecules to be at a distance d from each other.

Volume around 1 molecule = d³

Also, we know, volume of N_A whose mass is M_A is given by,

So, volume of 1 molecule =

Equating we get,

(c) Volume of 1g of water =1601 cm³ = 1.601 × 10^-3m³

There are N_A molecules in 18kg of water. So, volume occupied by 1 molecule is given by,

If d^’ is the intermolecular distance and d^’3 is the volume of one molecule, then

(d) The work done by force in move a molecule from distance d to d^’ is equal to the energy required to evaporate one molecule.

(e) Here,

Given:

Inside temperature = T_i = 60°C = 333K

Outside temperature = T_o= 20°C = 293K

Outside pressure = P_o = 1.013×10⁵Nm^-2

Let the pressure inside the balloon be P_i, T be the surface tension and r be the radius of the balloon, then by formula for excess pressure,

Assume air to be ideal gas, then by formula for an ideal gas,

If there are n_i moles of air inside the balloon, then

where M_i is the mass of air inside the balloon and M_A is the molar mass.

Similarly,

where M_o is the mass of air inside the balloon and M_A is the molar mass.

Now, in order to calculate P_i, lets calculate excess pressure.

Now, since this is negligible, we can assume,

P_i = P_o = 1.013×10⁵Nm^-2

Assume, 21% of O₂ and 79% of N₂ in the atmosphere.

We can calculate M_A as,

We know, the mass M lifted by the balloon can be found as,

So, the balloon can carry a mass of 301.46kg.