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

Its SI unit is kg $$m-3$$. 

                      = $${ 10 }^{ -2 }\times \dfrac { 6 }{ 6 } $$

                      = $$-{ 10 }^{ -2 }\dfrac { poise }{ sec. } $$

$$P_0+\rho_{oil}g(15)+\cfrac{1}{2}\rho_w V^2=P_0+\rho_w g(10)+\rho_{oil}g(5)\\ \therefore V=\sqrt{40}=6.3m/s$$

$$(b)$$ Let h be the light when flow through orifice stops.

$$\therefore P_0+\rho_{oil}g(5)+\rho_w gh=P_0+\rho_{oil}g(15)\\ \Rightarrow h=\cfrac{\rho_{oil}}{\rho_w}10=8m\\(c) P_0+\rho_{oil}g(5)+\rho_w gh=P_0+\rho_{oil}g(15)+\cfrac{1}{2}\rho_w V^2\\ \Rightarrow V=\sqrt{20(h-8)}\\ \Rightarrow -\cfrac{dh}{dt}=\sqrt{20(h-8)}$$

$$\Rightarrow-\int _{ 10 }^{ 8 }{ \cfrac { dh }{ \sqrt { 20\left( h-8 \right)  }  }  } =\int _{ 0 }^{ t }{ dt } $$

$$v=\dfrac{2r^2}{9\eta}(P-\sigma)$$

now mass = volume $$\times$$ density

for a spherical body $$v=\dfrac{4}{3}\pi r^3$$, So,

$$m=\dfrac{4}{3}\pi P r^3$$ 

$$=r^2=\left(\dfrac{3m}{4\pi p}\right)^{2/3}$$.....2

Hence ratio form $$1$$ and $$2$$

$$\dfrac{V_1}{V_2}=\left(\dfrac{r_1}{r_2}\right)=\left(\dfrac{m_1}{m_2}\right)^{2/3}$$

Using     $$A_1v_1 = A_2v_2$$

Or     $$8\times 1.5 = 4\times v_2$$

$$\implies \ v_2 = 3 \ m/s$$

Terminal velocity is calculated by

$$V = \dfrac{2gr^2(\rho_b-\rho_l)}{9\eta}$$

where  $$\rho_b$$ and $$\rho_l$$ is the density of body and liquid respectively

           $$\eta$$ is the coefficient of viscosity of liquid.

           $$r$$ is the radius of the body

$$d=2$$mm

$$=2\times 10^{-3}$$m

$$r=10^{-3}$$m

$$\eta =8.3$$ poise$$=8.3\times \dfrac{1}{10}$$ deca poise $$=0.83$$

$$g=9.8$$

$$s=10^3\times 8$$kg

$$6g=1.3\times 10^3$$ $$kg/m^3$$

$$\theta_t=\dfrac{2r^2(s-6g)g}{9\eta}$$

$$=\dfrac{2\times 10^{-3}\times 10^{-3}\times 9.8\times 8\times 10^3-1.3\times 10^3}{9\times 0.83}$$

$$=\dfrac{19.6\times 10^{-6}\times 10^3\times 6.7\times 10^2}{242}=0.17\times 10^{-1}$$

$$\theta_t=1.7\times 10^{-2}$$ m/s.

A mercury barometer works by stabilizing the level of mercury in the sealed glass tube with the atmospheric pressure on the open mercury chamber. The top of the barometer is filled with nothing (vacuum) because the mercury must have the space to expand and contract without any other force affecting it; a gas or liquid in the top would have its own pressure against the mercury and give inaccurate readings. Since the level of mercury in the barometer is dependent on the pressure of the actual atmosphere, there must be no other factor that inhibits the ability of the mercury to rise. High atmospheric pressure pushes the mercury up; low atmospheric pressure allows the mercury to drop. The barometer markings are also adjusted for temperature, as this affects the mercury's density and also its level in the barometer.

Greater the atmospheric pressure, greater is the vertical height of the mercury column. 

The space above mercury is a vacuum.    

E. Torricelli was the first to explain this phenomenon; hence this empty space is called 'Torricellian vacuum'. 

The velocity of flow of water $$v = \sqrt{2gD}$$

After coming out of orifice, time taken (time of flight) to reach the ground.

$$T = \sqrt{\dfrac{2 \left(H - D \right)} {g}}$$

$$\therefore$$ Range of water

$$x = velocity \times time \space  of \space light$$

$$= \sqrt{2gD} \times \sqrt{\dfrac{2 \left(H - D \right)} {g}}$$

or $$x = 2 \sqrt{D \left(H - D \right)}$$

When a body falls in a viscous medium, it carries with its layers of fluid which are in body's contact where as the layers of fluid in contact with the stationary surface remain almost at rest. The layers of fluid destroys the relative motion and motion of the body is thus opposed. The viscous drag increases with velocity of the body till viscous drag and upthrust of the body are together equal to the weight of the body which acts downwards.

When there is no net force, the body moves with the uniform velocity. This velocity is called terminal velocity.

Stoke showed that the retarding force F due to viscous drag for a spherical body of radius r that moves with a velocity v in a fluid, with coefficient of viscosity $$\eta $$, is given as :

$$F = 6 \pi \eta r v$$

This expression is known as Stoke's Law.

Derivation with help dimensions :

$$F \space  \alpha \space  v$$, velocity

$$F \space  \alpha \space  r$$, radius of the body;

$$F \space  \alpha \space \eta$$, coefficient of viscosity of the fluid.

$$\Rightarrow F = A \eta^{a} r^{b} v^{c}$$

where A is a constant with no dimensions. Putting the dimensions, 

$$\left[MLT^{-2} \right] = \left[ML^{-1}T^{-2} \right]^{a} \left[L \right]^{b} \left[LT^{-1} \right]^{c}$$

or $$\left[MLT^{-2} \right] = \left[M^{a} L^{-a + b + c} T^{-a -c} \right]$$

Comparing the two sides, we get

a = 1, b = 1 and c = 1

$$\therefore F = A \eta r v$$

A was determined to be $$6\pi$$.

$$\therefore F = 6 \pi \eta r v$$

Derivation for terminal velocity :

Let $$p$$ be the density of the body and $$\rho$$ be the density of the medium. Then,

Weight of body = $$\frac{4} {3} \pi r^{3} p g$$ ........(11.4)

Upthrust of body = $$\frac{4} {3} \pi r^{3} \sigma g$$ ........(11.5)

Resultant force (downwards) = $$\frac{4} {3} \pi r^{3} \left( p - \sigma\right)  g$$

In equilibrium,

$$6\pi \eta r v = \frac{4} {3} \pi r^{3} \left( p - \sigma \right)  g$$ ......(11.6)

$$\Rightarrow v = \frac{2} {9} r^{2} \frac{\left(p - \sigma \right)} {\eta} g$$  .....(11.7)

This is the formula for terminal velocity.

It is apparent from the above expression that the terminal velocity is (a) directly proportional to the square of the radius of body, (b) directly proportional to the densities of the body and the medium, (conversely proportional to the coefficient of viscosity of the medium.

Variation of viscosity with temperature and pressure effect of temperature on viscosity :

(i) When a liquid is heated then the kinetic energy of it's molecules increases and the intermolecular force between them is decreases. Hence the viscosity of a liquid decreases with the increase in it's temperature.

(ii) Viscosity of gases is due to the diffusion of molecules from one moving layer to another. But the rate of diffusion of a gas is directly proportional to the square root of the temperature. So the viscosity of a gas increase with it's temperature.

Effect of pressure :

(i) Except the water of the viscosity of liquid increases with the increase in pressure. In case of water, the viscosity decrease with increase in pressure.

(ii) The viscosity of gas is independent of pressure.

Given, $$r_1 = 2 \space cm; v_1 = 20 \space cm/s; r_2 = 6 \space cm; v_2 = ?$$

From the equation of continuity

$$A_2v_2 = A_1v_1$$

or $$\pi r^{2}_2 v_2 = \pi r^{2}_1 v_1 $$

$$r^{2}_2 v_2 = r^{2}_1 v_1 $$

$$v_2 = \left ( \dfrac{r_1} {r_2} \right)^{2} v_1$$

$$\therefore v_2 = \left ( \dfrac{2} {6} \right)^{2} \times 20$$

$$= \dfrac{1} {9} \times 20$$

or $$v_2 = 2.22 \space cms^{-1}$$

Consider a rectangular frame (see fig. 11.20), having a sliding wire on one of its arms. Dip the frame in a soap solution and take it out. A soap film is formed on the frame and have two surfaces. Both the surfaces are in contact with the sliding wire. So, surface tension acts on the wire due to both the surface.

Let T be the surface tension of the soap solution and L be the length of the wire.

The force exerted by each surface on the wire = $$T \times L$$

So, total force on the wire = $$2TL$$  ......(11.23)

Let the surfaces contract by $$\Delta x$$.

Work done on the film, $$W = F \times \Delta x$$

$$= 2T L \Delta x$$ ......(11.24)

Here $$2L \times \Delta x \rightarrow$$ total increase in the area of both of the surfaces of film.

Let $$2L \times \Delta x = \Delta A$$

$$\therefore   W = T \Delta A$$

or  $$T = \frac{W} {\Delta A}$$  .........(11.25)

So, surface tension of a liquid is equal to the work done in increasing the surface area of its free surface by one unit. In other words, surface tension is equal to surface energy per unit area.

$$E_s = T \Delta A$$

The unit of T is $$Jm^{-2}$$ and its dimensional formula is $$\left[M^{1} L^{0} T^{-2} \right]$$

Consider the surface of a liquid in a wide container at a height h above the orifice. The velocity of the liquid at the surface is taken zero and the atmospheric pressure at P and the exist is the same.

Apply Bernoulli's theorem for the unit mass of the fluid.

$$0 + gh + 0 = 0 + 0 + \frac{1} {2} v^{2}$$

At P, only potential energy is there; and , at exit, only kinetic energy is there.

So, $$U = \sqrt {2gh}$$ ..... (11.20)

According to Bernoulli's principle, the sum of total energy and pressure of unit volume of liquid at its free surface and at hole must remain constant.

So $$P + 0 + pgh = P + \frac{1} {2} pU^{2} + pg\left(H - h \right)$$

$$U = \sqrt {2gh}$$ ..... (11.21)

The above expression represents the speed of a freely falling body and is known as Toricelli's law.

The horizontal range of fluid :

The height (sy) = $$H - h$$

The horizontal velocity $$U_x = \sqrt{2gh}$$

The vertical velocity = $$Uy = 0 $$

The horizontal acceleration = $$(aX) = 0$$

The vertical acceleration $$ay = +g$$

from the vertical motion :

$$sy = uyt + \frac{1} {2} ayf^{2}$$

$$H - h = 0 + \frac{1} {2} gt^{2}$$

$$ t = \sqrt{\frac{2 \left(H - h \right)} {g}}$$

The range (R) = Horizontal distance

$$sy = uyt + \frac{1} {2} axtf^{2}$$

$$R = \sqrt{2gh} \times \sqrt{\frac{2 \left(H - h \right)} {g}} + 0$$

$$R = 2 \sqrt{h\left(H - h \right)}$$

Given, $$ r = 1 \space mm = 1 \times 10^{-3} \space  m$$;

$$\eta = 1.8 \times 10^{-2} \space Ns/m^{-2}$$

$$p = 10^{3} \space k \space gm^{-3}; \sigma = 1.2 \space k \space gm^{-3}; g = 10 \space ms^{-2}$$

$$v_t = ?$$

$$\therefore v_t = \dfrac{2r^{2} \left(p - \sigma \right) g} {9 \eta} = \dfrac{2 \times 10^{-6} \left(1000 - 1.2 \right) \times 10 } {9 \times 1.8 \times 10^{-2}}$$

$$= \dfrac{2 \times 998.8} {9 \times 1.8} \times 10^{-3}$$

$$= 123.3 \times 10^{-3} = 0.123 \space m/s = 12.3 \space cm/s$$

Terminal velocity $$v_t = \dfrac{2r^{2} \left(P - \sigma \right) g} {9 \eta}$$

$$v_t \propto r^{2}$$

$$\therefore \dfrac{v_2} {v_1} =  \dfrac{r^{2}_2} {r^{2}_1} = \left(\dfrac{r_2} {r_1}\right)^{2}$$

or $$\therefore \dfrac{v_2} {v_1}  = \left(\dfrac{r_2} {r_1}\right)^{2}$$

$$\therefore M = \dfrac{4} {3} \pi r^{3}_1 p$$

$$\therefore 8M = \dfrac{4} {3} \pi r^{3}_2 p$$

$$\therefore \dfrac{1} {8} = \left(\dfrac{r_1} {r_2}\right)^{3} \Rightarrow \frac{1} {2} =\dfrac{r_1} {r_2}$$

$$\therefore \dfrac{r_2} {r_1} = 2$$

From equation (1)

$$\dfrac{v_2} {v_1} = (2)^{2} = 4 \Rightarrow v_2 = 4v_1$$

$$ \therefore v_2 = 4 V$$

dividing by $$\cos { \alpha  } $$ to both numerator and denominator we will get,

     $$\cfrac { { d }_{ 1 } }{ { d }_{ 2 } } =\cfrac { 1+\tan { \alpha  }  }{ 1-\tan { \alpha  }  } $$

$$T\quad is\quad surface\quad tensionT=0.465N/m$$

$$\theta \quad contact\quad angle\quad 135{  }^{ 0 }$$

$$\rho \quad density\quad 13600\quad kg/{ m }^{ 3 }$$

$$g\quad gravitational\quad 9.8m/{ s }^{ 2 }$$

$$r\quad radius\quad 0.001m$$

$$h=\left( 2\times 0.465\times cos135 \right) /\left( 13600\times 9.8\times 0.001 \right) $$

$$h\leadsto 0.005m=75.5cm$$

Let the radius of the smaller ball $$= r$$

the radius of the bigger ball$$ = 2r$$

Density of the ball's material$$ =\rho _b$$

Density of liquid$$ =\rho $$

Terminal velocity of the balls is v1 and v2 for the smaller and bigger respectively. 

Volume of the smaller ball $$V =\dfrac{ 4}{3}\pi r^3$$

The mass of the ball will be $$=\rho_bV$$

Volume of the bigger ball $$=\dfrac{ 4}{3}\pi(2r)^3=8V$$

The mass of the ball will be$$ = 8\rho_bV$$

Upthrust on the smaller ball$$ =\rho V_g$$

Viscous drag force on smaller ball $$= 6\pi \eta rv_1$$

Weight of the smaller ball $$= \rho _bV_g$$

Now, since the balls are moving with terminal velocity, the total upward force will balance its weight 

$$\rho V_g+6\pi \eta rv_1=\rho _bV_g      ----1. 

Upthrust on the bigger ball $$= 8 \rho V_g$$

Viscous drag force on bigger ball $$= 6\pi \eta 2rv_2$$

Weight of the bigger ball $$= 8\rho _bV_g$$

Similarly,

$$8\rho V_g+6\pi\eta 2rv_2=8\rho _bV_g$$----2.

 Dividing eqn. 1 by eqn. 2

$$\dfrac{ρVg+6πηrv1}{8ρVg+12πηrv2}=\dfrac{1}{8}$$

=>$$8\rho V_g+12\pi \eta rv_2=8\rho V_g+48\pi \eta rv_1$$

=>$$12\pi \eta rv_2=48\pi \eta rv_1$$

=>$$v_2=4v_1$$

=>$$\dfrac{v_1}{v_2}=\dfrac{1}{4}$$

Thus, the ratio of their terminal velocity $$= 1:4$$