Mechanical Properties Of Fluids - Class 11 Engineering Physics - Extra Questions

Using Bernoulli's equation,

$$\dfrac{dgH}{2} + 2dg(\dfrac{H}{2}-h) + P_o = P_o + \dfrac{\rho v^2}{2}$$

$$v = \sqrt{\dfrac{3gH-4gh}{4}}$$

Now, using kinematics,

Time to reach ground $$t=\sqrt{2h/g}$$ 

$$x = vt = \sqrt{\dfrac{3Hh-4h^2}{2}}$$

For the maximum horizontal distance x,
$$\dfrac {dx}{dh}=\Rightarrow \dfrac {d}{dh}(3Hh-4h^2)^{\dfrac {1}{2}}=0$$
or $$\dfrac {1}{2}(3Hh-4h^2)^{-\dfrac {1}{2}}\times (3H-8h)=0$$
This gives $$3H-8h=0\Rightarrow h=\dfrac {3H}{8}$$
This is the length of the liquid for which the range is maximum, i.e.,
$$h\rightarrow h_m=\dfrac {3H}{8}$$
The maximum horizontal distance is given by
$$x_m=\sqrt {(3H-4h_m)h_m}$$
$$=\sqrt {\left \{\left (3H-4\dfrac {3H}{8}\right )\dfrac {3H}{8}\right \}}=\dfrac {3}{4}H$$

Given that,

m= 45 kg = 45000g

A = 900 cm²

Ρ = 1000 kg/m³ = 1 g/cm³

Let, the difference in the height of the water level be $$h\ units$$.

Now,

According to the figure

  $$ {{P}_{a}}+h\rho g={{P}_{a}}+\dfrac{mg}{A} $$

 $$ h=\dfrac{m}{A\rho } $$

Put the values of all elements

  $$ h=\dfrac{45\times 1000}{900\times 1} $$

 $$ h=50\,cm $$

Hence, the difference in the levels of water in the two tubes is $$50\ $$cm.

The pressure of the liquid column will increase with the depth of the column.

Pressure in a liquid column varies as,

$$P = h\rho g$$

The volume surface of the droplet which spherical in shape is given as

$$V = \frac{4}{3}\pi {R^3}$$

The volume of the eight droplet is

$$V = 8 \times \frac{4}{3}\pi {\left( {0.5} \right)^3}$$

On equation the above two equation we get the radius of the combined droplet

The radius is$$1\;mm$$

The surface energy of the droplet will be

$$\Delta E = 8 \times 4\pi {\left( {\frac{1}{2} \times \frac{1}{{1000}}} \right)^2} \times 0.072 - 4\pi {\left( {\frac{1}{{1000}}} \right)^2} \times 0.072$$

The energy evolved will be $${\rm{9 \times 1}}{{\rm{0}}^{{\rm{ - 7}}}}{\rm{J}}$$

Let p₁ and p₂ are pressure inside and outside the roof and v₁ and v₂ are velocities of the wind inside and outside the roof.

According to Bernoulli’s theorem

$$ {{p}_{1}}+\dfrac{1}{2}\rho v_{1}^{2}={{p}_{2}}+\dfrac{1}{2}\rho v_{2}^{2} $$

$$ {{p}_{1}}-{{p}_{2}}=\dfrac{1}{2}\times 1.2\times ({{40}^{2}}-0) $$

$$ {{p}_{1}}-{{p}_{2}}=960\,N/{{m}^{2}} $$

Force is

$$ F=({{p}_{1}}-{{p}_{2}})A=960\times 250 $$

$$ 24\times {{10}^{4}}\,N $$

We have,

  $$ {{v}_{1}}=\sqrt{2g\times \left( \dfrac{h}{2} \right)} $$

 $$ {{v}_{1}}=\sqrt{gh} $$

Now, by using Bernoulli’s theorem

  $$ \rho gh+2\rho g\left( \dfrac{h}{2} \right)=\dfrac{1}{2}\left( 2\rho  \right)v_{2}^{2} $$

 $$ {{v}_{2}}=\sqrt{2gh} $$

Now, the value of $$\dfrac{{{v}_{1}}}{{{v}_{2}}}$$

  $$ \dfrac{{{v}_{1}}}{{{v}_{2}}}=\dfrac{\sqrt{gh}}{\sqrt{2gh}} $$

 $$ \dfrac{{{v}_{1}}}{{{v}_{2}}}=\dfrac{1}{\sqrt{2}} $$

Hence, this is the required solution

Let a liquid of density ρ rise a height h in a capillary of raduis r, then we can write

hrρg = 2Tcosθ

where T is surface tension of that liquid and θ is angle of contact.

Let Tm and Tw be the surface tension of water and mercury.

Tm/Tw = [(3.112 * 13.6)/(10 * 1)] * [cos0 / cos(135)]         

θw is the angle of contact of water 0 degrees

         = 5.98

Applications of Stoke's law are as follows:

• Rain drop do not acquire alarmingly high velocity during their free fall. If this does not happen a person moving in rain would get hurt.
• While jumping from an airplane, parachute helps us to land safely on the earth.
• It is used to determine the value of charge on an electron. (Millikan’s oil drop method)

When a small quantity of liquid is poured on solid, three interfaces, namely liquid – air, solid – air, and solid-liquid are formed. The surface tension corresponding to theses three layers is $$S_{LA}$$, $$S_{SA}$$, and $$S_{SL}$$ respectively. Let $$\theta$$ be the angle of contact between the liquid and solid.

The molecules in the region, where the three interfaces meet are in equilibrium. It means that net force acting on them is zero. For the molecule at K to be in equilibrium, we have
$$S_{SL} + S_{LA} cos \theta = S_{SA}$$ or $$cos \theta =\dfrac{S_{SA}−S_{SL}}{S_{LA}}$$ In case of mercury-glass, $$S_{SA} < S_{SL}$$, therefore cos θ is negative or $$\theta > 90^\circ$$ i.e. (obtuse). In case of water-glass $$S_{SA} > S_{SL}$$, therefore $$cos\,\theta$$ is positive or $$\theta < 90^\circ$$ (acute).

Stoke’s law states the backward dragging force (F) acting on a small spherical body of radius r, moving through a medium having a coefficient of viscosity η and velocity v is given by
$$F = 6\pi \eta rv$$
viscous drag (F) depends on

1. coefficient of viscosity of the medium η
2. the velocity of the body (v)
3. the radius of the spherical body (r)

The terminal velocity of an object is the maximum constant velocity acquired by the object while falling freely in a viscous medium.
Terminal velocity (v) of an object of radius (r) density ($$\rho$$) moving through a viscous medius of viscosity $$\eta$$ and density $$\rho_0$$ is given by
$$v=\dfrac { 2 }{ 9 } \dfrac {r^2}{ \eta  } \left( \rho -\rho_0 \right)g$$
Hence, terminal velocity depends on

1. radius of the object
2. coefficient of viscosity of the medium
3. density of the object.
4. density of the medium

Height of water, $$z_1 = 3\,m$$
Height of oil, $$z_2 = 1\,m$$
Density of oil
= Sp.gr of oil $$\times$$ Density of water
$$= 0.9 \times 1000\,kg/m^3 = 9000\,kg/m^3$$
$$=\rho_2$$
Density of water $$= \rho_1 =1000\,kg/m^3$$

1. At the interface, i.e., at A,
$$P=\rho_2 g \times z_2 = 900 \times 9.81 \times 1$$

$$= 8829\,N/m^2$$

2. At the bottom i.e., at B,
$$p =\rho_2gz_2 + \rho_1gz_1$$
$$= 900 \times 9.81 \times 1 + 1000 \times 9.81 \times 3$$

$$= 8829 + 29430 = 38259\,N/m$$

The constant velocity with which a body drops down after initial acceleration in a dense liquid or fluid is called terminal velocity. This is attained when the apparent weight is compensated by the viscous force. It is given by
$$v = \dfrac{2}{9}\dfrac{r^2g}{\eta}(\rho−\sigma)$$, where $$\sigma$$ and $$\rho$$ are densities of the body and liquid respectively, $$\eta$$ is the coefficient of viscosity of liquid and r is the radius of the spherical body.
The net force on the sphere becomes zero as the viscous force equals the apparent weight.
Consider a long column of dense liquid-like glycerine. As the ball is dropped in it, the forces experienced are;

1. weight $$= mg =\dfrac{4}{3}\pi r^3 \rho g$$, where ρ is the density of the ball.
2. upthrust $$= U =\dfrac{4}{3}\pi r^3 \rho g$$, where $$\rho_1$$ is the density of the liquid.
3. viscous force $$F_v = 6 \pi \eta \rho v$$

1. The pressure at the base of the container is due to the liquids both at the upper and lower half.
∴ $$P_{Total} = P_1 + P_2$$
$$P_{Total} = ρgh_1 + ρgh_2$$
$$P_{Total} = \rho g(h_1 + h_2)$$

2. Force exerted on the bottom of the container,
$$P = P_{Total} \times A_2$$
$$F = \rho g(h_1 + h_2) \times A_2$$.

Since there is no acceleration in horizontal direction, we have 
$$\text{S(distance travelled)} \propto \text{vt (velocity of efflux} \times \text{time taken by the jet to reach the bottom)}\\\Rightarrow \dfrac { { S }_{ 1 } }{ { S }_{ 2 } } =\dfrac { { v }_{ 1 }{ t }_{ 1 } }{ v_{ 2 }{ t }_{ 2 } } \quad \quad \quad \quad .......(I)$$
For the top hole $$v_1 =\sqrt {2g \left(\dfrac{h}{4}\right)} = \sqrt{\dfrac{gh}{2}}\quad \quad \quad \quad .......(II)$$
Time taken by the jet reach the bottom from the first hole $${ t }_{ 1 }=\sqrt { \dfrac { 2\left( h-\dfrac { h }{ 4 }  \right)  }{ g }  } =\sqrt { \dfrac { 3h }{ 2g }  } \quad \quad \quad .......(III)$$
Applying Bernoulli's theorem for the bottom hole, we have 
$$\sigma g\left( \dfrac { h }{ 2 }  \right) +2\sigma g\left( \dfrac { h }{ 4 }  \right) =\dfrac { 1 }{ 2 } (2\sigma )v_{ 2 }^{ 2 }\\ \Rightarrow v_{ 2 }=\sqrt { gh } \quad \quad \quad .......(IV)$$
Time taken by the jet reach the bottom from the second hole $${ t }_{ 2 }=\sqrt { \dfrac { 2\left( \dfrac { h }{ 4 }  \right)  }{ g }  } =\sqrt { \dfrac { h }{ 2g }  } \quad \quad \quad .......(V)$$
From $$(I)$$,  $$(II)$$, $$(III)$$,$$(IV)$$ and $$(V)$$, we have
$$\dfrac { { S }_{ 1 } }{ { S }_{ 2 } } =\dfrac { { v }_{ 1 }{ t }_{ 1 } }{ v_{ 2 }{ t }_{ 2 } } =\dfrac { \left( \sqrt { \dfrac { gh }{ 2 }  }  \right) \times \left( \sqrt { \dfrac { 3h }{ 2g }  }  \right)  }{ \left( \sqrt { gh }  \right) \times \left( \sqrt { \dfrac { h }{ 2g }  }  \right)  } =\dfrac { \sqrt { 3 }  }{ \sqrt { 2 }  } \\\Rightarrow \boxed{x=3}$$ 

The applied force is given as,

$$F = PA$$

$$F = 15 \times {10^{ - 4}}$$

$$F = 0.0015\;{\rm{N}}$$

toward x-dircetion, Angle bisect into half, as show in fgure

Inlet velocity $${{V}_{i}}$$ = outlet velocity $${{V}_{o}}$$ = $$V$$

Change in velocity in y-direction, $$\Delta {{V}_{y}}={{V}_{o}}\sin {{30}^{o}}-{{V}_{i}}\sin {{30}^{o}}=0$$

Change in velocity in x-direction, $$\Delta {{V}_{x}}={{V}_{o}}\cos {{30}^{o}}-\left( -{{V}_{o}}\cos {{30}^{o}} \right)=2{{V}_{o}}\dfrac{\sqrt{3}}{2}=\sqrt{3}{{V}_{o}}$$

Rate of fluid flow, $$\overset{\bullet }{\mathop{m}}\,=\rho AV$$

$$ Force=rate\,of\,change\,of\,momentum=\overset{\bullet }{\mathop{m}}\,\Delta V $$

$$ Force=\,\rho AV.\,\sqrt{3}\,V=\sqrt{3}\,\rho A{{V}^{2}} $$

Hence, force on tube is $$\sqrt{3}\,\rho A{{V}^{2}}$$ 

The pressure is force exerted by a body per unit area.

$$P = \dfrac{F}{A}$$

From the above equation we can see that the force is inversely proportional to that of the area. We can see by physical example how the pressure exerted by the body depends on the area.

When we walk in high heels the area of the contact of the heels with the surface is very small and hence the pressure exerted will be very high so is the pressure exerted on the body.

Rate of infusion$$ = 30\;{\rm{mg/h}}$$

We have a bag of $$200\;{\rm{mg}}$$ in $$50\;{\rm{ml}}$$.

Rate of pump setting$$ = 30 \times \dfrac{{50}}{{200}} = 7.5\;{\rm{ml/h}}$$

The liquid is less dense s compared to the solid. They have less tensile strength in them and hence they can’t keep themselves in shape as solid do. Liquid tend to flow from higher level to lower level to the gravity.

If the liquid is stored in the cylinder it requires an external force.

Also the molecules of the liquid keep on vibrating and thus exert pressure on the wall of the container.According to the Newton’s law every action has equal and opposite reaction.

The pressure difference is given as,

$$\Delta P = \frac{1}{2}\rho \left( {{V_2}^2 - {V_1}^2} \right)$$

$$\Delta P = \frac{1}{2} \times 0.29\left( {{{\left( {90} \right)}^2} - {{\left( {72} \right)}^2}} \right)$$

$$\Delta P = 422.82\;{\rm{N/}}{{\rm{m}}^2}$$

The lift on the wing is given as,

$$F = \Delta P \times A$$

$$F = 422.82 \times 3.2$$

$$F = 1353.024\;{\rm{N}}$$

Given that,

velocity of flow of water  $$v{\rm{ }} = {\rm{ }}10.0{\rm{ }}m/s$$

$${V_h} = \dfrac{{{v^2}}}{{2g}}$$

Here, g = acceleration due to gravity = $$9.8m/{s^2}$$

Put the value into the formula

$${V_h} = \dfrac{{{{(10m/s)}^2}}}{{2 \times 9.8m/s^2}}$$

$${V_h} = 5.10m$$

Hence,  Velocity head of water $${V_h} = 5.10m$$