# Fluid Mechanics

## Concept Of Physics

### H C Verma

1   The surface of water in a water tank on the top of a house is $4$ m above the tap level. Find the pressure of water at the tap when the tap is closed. Is it necessary to specify that the tap is closed ? Take g = $10$ $m/s^2$.

##### Solution :

p = h $\rho$ g $\\$ it is necessary to specify that the tab is closed.Otherwise pressure will gradually decrease, as h decrease, because, of the tap is open, the pressure at the tap is atmospheric.

2   The heights of mercury surfaces in the two arms of the manometer shown in figure (13-E1) are $2$ cm and $8$ cm. Atmospheric pressure = $1.01$ x $10$ S $N/m^{2}.$ Find (a) the pressure of the gas in the cylinder and (b) the pressure of mercury at the bottom of the U tube

Pressure at the bottom of the tab should be same when considered for both limbs.$\\$ From the figure are shown, $\\$ $p_g$ + $\rho_{Hg}$ $\times$ $h_2$ $\times$ g = $p_a$ + $\rho_{Hg}$ $\times$ $\times$ $h_1$ $\times$ g $\\$ $p_g$ = $p_a$ + $\rho_{Hg}$ $\times$ $g(h_1 - h_2)$ $\\$ b) Pressure of mercury at the bottom of u tube $\\$ $p$ = $p_a$ +$\rho_{Hg}h_1$ $\times$$g 3 A glass full of water has a bottom of area 20 cm^2, top of area 20 cm^2, height 20 cm and volume half a litre. (a) Find the force exerted by the water on the bottom. (b) Considering the equilibrium of the water, find the resultant force exerted by the sides of the glass on the water. Atmospheric pressure = 1.0 \times 10 N/m^2. Density of water = 1000 kg/m^3 and g = 10 m/s^2. Take all numbers to be exact. ##### Solution : a)Force exerted at the bottom.\\ =Force due to the cylindrical water colum + atm.Force \\ = A \times h \times \rho_w \timesg + p_a \timesA \\ = A(h\rho_wg + p_a)\\ b)To find out the resultant force exerted by the sides of the glass,from the freebody, diagram of water inside the glass \\ p_a \timesA + mg = A \times h \times$$\rho_w$$\timesg+F_s+p_a$$\times$A $\\$ $\Rightarrow$ mg = A $\times$ h $\times$$\rho_w$$\times$g+$F_s$ $\\$ This force is provided by the sides of the glass

4   The area of cross-section of the wider tube shown in figure (13-E2) is $900$ $cm^{2}$. If the boy standing on the piston weighs $45$ kg, find the difference in the levels of water in the two tubes.

##### Solution :

From the figure shown $\\$ $p_a$ +h$\rho$g = $p_a$ +mg\A $\\$ $\Rightarrow$ h$\rho$g = mg\A $\\$ $\Rightarrow$ h = $\frac{m}{A\rho}$

5   Suppose the glass of the previous problem is covered by a jar and the air inside the jar is completely pumped out. (a) What will be the answers to the problem ? (b) Show that the answers do not change if a glass of different shape is used provided the height, the bottom area and the volume are unchanged.

##### Solution :

If the glass will be covered by a jar and the air is pumped out,the atmospheric pressure has no effect.$\\$ So,$\\$ a)Force exerted on the bottom.$\\$ =(h$\rho_w$g)$\times$A $\\$ b)mg = h $\times$ $\rho_w$ $\times$g $\times$ + $F_s$ $\\$ c)It glass of different shape is used provided the volume,height and area remain same,no change in answer wil occur

6   If water be used to construct a barometer, what would be the height of water column at standard atmospheric pressure ($76$ cm of mercury) ?

##### Solution :

Standard atmospheric pressure is always pressure exerted by $76$cm Hg column $\\$ = ($76$ $\times$ $13.6$ $\times$ g)Dyne/$cm^2$.$\\$ If water is used in the barometer.$\\$ Let h $\longrightarrow$height of water column. $\\$ $\therefore$ h $\times$ $\rho_w$ $\times$g

7   Find the force exerted by the water on a $2$$m^2 plane surface of a large stone placed at the bottom of a sea 500 m deep. Does the force depend on the orientation of the surface ? ##### Solution : a) F = P \times A = (h\rho_w \times g)A \\ b)The force does not depend on the orientation of the rock as long as the surface area remains same 8 Water is filled in a rectangular tank of size 3m x 2m x 1 m. (a) Find the total force exerted by the water on the bottom surface of the tank. (b) Consider a vertical side of area 2 m x 1 m. Take a horizontal strip of width ox metre in this side, situated at a depth of x metre from the surface of water. Find the force by the water on this strip. (c) Find the torque of the force calculated in part (b) about the bottom edge of this side. (d) Find the total force by the water on this side. (e) Find the total torque by the water on the side about the bottom edge. Neglect the atmospheric pressure and take g = 10 m/s^2. ##### Solution : a)F = A h \rho g \\ b)The force exerted by water on the strip of width \deltaX as shown,\\ dF = p \times A \\ = (x\rhog) \times A \\ c) Inside the liquid force act in every direction due to adhesion. \\ di = F \times r \\ d)The total force by the water on that side is given by F = \int_{0}^{1} 20000 X\deltax \Rightarrow F= 20,000[\frac{x^2}{2}]^{1}_{0} \\ e)The torque by the water on that side is given by \\ i= \int_{0}^{1} 20000 X\deltax (1-x)\Rightarrow 20,000[\frac{x^2}{2} - \frac{x^3}{3}]^{1}_{0} 9 An ornament weighing 36 g in air, weighs only 34 g in water. Assuming that some copper is mixed with gold to prepare the ornament, find the amount of copper in it. Specific gravity of gold is 19.3 and that of copper is 8.9. ##### Solution : Here, m_0 = m_{au} + m_{cu} = 36 g \\ Let V be the volume of the ornament in cm^3 \\ So, V \times \rho_w x g = 2 x g \\ \Rightarrow (V_{au} + V_{cu}) \times \rho_w x g = 2 x g \\ \Rightarrow \big(\frac{m}{\rho_{au}} + \frac{m}{p_{au}}\big) \rho_w x g = 2 x g \\ \Rightarrow \big(\frac{m_{Au}}{19.3} + \frac{m_{Au}}{8.9}\big) \times 1 = 2 \\ 8.9 m_{Au} + 19.3 m_{cu} = 2 \times 19.3 \times 8.9 = 343.54 .....(2) \\ From equation (1) and (2), 8.9 m_{Au} + 19.3 m_{cu} =343.54 \\ \Rightarrow \frac{8.9(m_{Au} + m_{cu}) = 8.9 \times 36}{m_{cu} = 2.225g} \\ So, the amount of copper in the ornament is 2.2 g. 10 Refer to the previous problem. Suppose, the goldsmith argues that he has not mixed copper or any other material with gold, rather some cavities might have been left inside the ornament. Calculate the volume of the cavities left that will allow the weights given in that problem. ##### Solution : \big(\frac{m_{Au}}{p_{Au}} + V_c \big) \rho_w \times g = 2 \times g (where V_c = Volume Of Cavity) 11 A metal piece of mass 160 g lies in equilibrium inside a glass of water (figure 13•E4). The piece touches the bottom of the glass at a small number of points. If the density of the metal is 8000 kg/m^3, find the normal force exerted by the bottom of the glass on the metal piece. ##### Solution : mg = U + R (Where U = Upward thrust) \\ \Rightarrow mg -U = R \\ \Rightarrow R = mg - v \rho_w g (because, U=v\rho_wg) \\ = mg - \frac{m}{p} \times \rho_w \timesg 12 A ferry boat has internal volume 1 m^3 and weight 50 kg. (a) Neglecting the thickness of the wood, find the fraction of the volume of the boat immersed in water. (b) If a leak develops in the bottom and water starts coming in, what fraction of the boat's volume will be filled with water before water starts coming in from the sides ? ##### Solution : a)Let V_i \rightarrow volume of boat inside water = volume of water displace in m^3. \\ Since, Weight of the boat is balanced by the buoyant force.\\ \Rightarrow mg = V_i \times \rho_w \times g \\ b)Let, v^1 \rightarrow volume of boat filled with water before water starts coming in from the sides. \\ mg + v^1 \rho_w \times g = V \times \rho_w \times g 13 A cubical block of ice floating in water has to support a metal piece weighing 0.5 kg. What can be the minimum edge of the block so that it does not sink in water ? Specific gravity of ice = 0.9. ##### Solution : Let X \rightarrow minimum edge of the ice block in cm. \\ So, mg +W_{ice} = U.(Where U = Upward thrust) \\ \Rightarrow 0.5 \times g + X^3 \times \rho_{ice} \times g = X^3 \times \rho_w \times g 14 A cube of ice floats partly in water and partly in K.oil (figure 13-E5). Find the ratio of the volume of ice immersed in water to that in K.oil. Specific gravity of K.oil is 0.8 and that of ice is 0.9. ##### Solution : V_ice = V_k + V_w \\ V_ice \times \rho_ice \times g = V_k \times \rho_k \times g + V_w \times \rho_w \times g \\ \Rightarrow(V_k + V_w) \times \rho_{ice} = V_k \times \rho_k + V_w \times \rho_w \\ \Rightarrow \frac{V_w}{V_k}=1. 15 A cubical box is to be constructed with iron sheets 1 mm in thickness. What can be the minimum value of the external edge so that the cube does not sink in water ? Density of iron = 8000 kg/m^3 and density of water = 1000 kg/m^3. ##### Solution : V_ig =V \rho_w g 16 A cubical block of wood weighing 200 g has a lead piece fastened underneath. Find the mass of the lead piece which will just allow the block to float in water. Specific gravity of wood is 0.8 and that of lead is 11.3. ##### Solution : (m_w + m_{pb})g = (V_w + V_{pb})\rho \times g \\ \Rightarrow (m_w + m_{pb}) = \big(\frac{m_{w}}{\rho_{w}} + \frac{m_{pb}}{\rho_{pb}}\big) \rho 17 A cubical metal block of edge 12 cm floats in mercury with one fifth of the height inside the mercury. Water is poured till the surface of the block is just immersed in it. Find the height of the water column to be poured. Specific gravity of mercury = 13.6. ##### Solution : Given, x = 12 cm \\ Length of the edge of the block \rho_{Hg} = 13.6 gm/cc \\ Given that , initially \frac{1}{5} of block is inside mercury \\ Let \rho_b \rightarrow density of block in gm/cc.\\ \therefore (X)^3 \times \rho_b \times g = (X)^2 \times (\frac{X}{5}) \times \rho_{Hg} \times g \\ \Rightarrow 12^3 \times \rho_b = 12^2 \times \frac{12}{5} \times 13.6 \\ \rho = \frac{13.6}{5} gm/cc \\ After water poured , let x = hight of water column.\\ V_b= V_{Hg}+V_w = 12^3 \\ Where V_{Hg} and V_w are volume of block inside mercury and water respectively \\ \therefore (V_b \times \rho_b \times g ) = (V_{Hg} \times \rho_{Hg} \times g ) + (V_w \times \rho_w \times g ) \\ \Rightarrow ( V_{Hg} + V_w) \rho_b = V_{Hg} \times \rho_{Hg} + V_w \times \rho_w \\ \Rightarrow ( V_{Hg} + V_w) \times \frac{13.6}{5} = V_{Hg} \times 13.6 + V_w \times 1 \\ \Rightarrow (12)^3 \times \frac{13.6}{5} = (12-x) \times (12)^2 \times 13.6 +(x) \times (12)^2 \times 1 \\ \Rightarrow x = 10.4 cm 18 Solve the previous problem if the lead piece is fastened on the top surface of the block and the block is to float with its upper surface just dipping into water. ##### Solution : Mg = w \Rightarrow (m_{w} + m_{pb})g = V_w \times \rho \times g 19 A hollow spherical body of inner and outer radii 6 cm and 8 cm respectively floats half submerged in water. Find the density of the material of the sphere. ##### Solution : Here, Mg = Upward thrust \\ \Rightarrow V_{pg} = \frac{V}{2}(\rho_{w}) \times g (where \rho_{w} = density of water) \\ \Rightarrow \big(\frac{4}{3}\pi r^{3}_{2} - \frac{4}{3}\pi r^{3}_{1} \big) \rho \\ \Rightarrow (r^{3}_{2} - r^{3}_{1}) \times \rho = \frac{1}{2}r^{3}_{2} \times 1 = 865 kg/ m^3 20 A solid sphere of radius 5 cm floats in water. If a maximum load of 0.1 kg can be put on it without wetting the load, find the specific gravity of the material of the sphere. ##### Solution : W1 +W2 = U \\ \Rightarrow mg+ V \times \rho_s \times g = V \times \rho_w \times g (where \rho_s = density of sphere in gm/cc) \\ \Rightarrow 1 - \rho_s = 0.19 \\ \Rightarrow \rho_s = 1-(0.19) = 0.8gm/cc \\ So,specific gravity of the material is 0.8 21 Find the ratio of the weights, as measured by a spring balance, of a 1 kg block of iron and a 1 kg block of wood. Density of iron = 7800 kg/m^3, density of wood = 800 kg/m^3 and density of air = 1.293 kg/m^3. ##### Solution : W_i = mg - V_i \rho_{air} \times g = \big(m - \frac{m}{p_i}\rho_{air}\big)g \\ W_w = mg - V_w \rho_{air} \times g = \big(m - \frac{m}{p_w}\rho_{air}\big)g 22 A cylindrical object of outer diameter 20 cm and mass 2 kg floats in water with its axis vertical. If it is slightly depressed and then released, find the time period of the resulting simple harmonic motion of the object. ##### Solution : Driving force U = V\rho_{w}g \\ \Rightarrow a = \pi r^2 (X) \times \rho_{w}g \Rightarrow T = 2 \pi \sqrt \frac{displacement}{Acceleration} 23 A cylindrical object of outer diameter 10 cm, height 20 cm and density 8000 kg/m^3 is supported by a vertical spring and is half dipped in water as shown in figure(13-E6). (a) Find the elongation of the spring in equilibrium condition. (b) If the object is slightly depressed and released, find the time period of resulting oscillations of the object. The spring constant = 500 N/m. ##### Solution : a) F + U = mg(Where F= kx) \\ \Rightarrow =kx + V\rho_{w}g = mg \\ b) F= kx + V\rho_{w} \times g \\ \Rightarrow ma = KX + \pi r^2 \times (X) \times \rho_{w} \times g = (K + \pi r^2 \times \rho_{w} \times g)X \\ \Rightarrow w^2 \times (X) = \frac{K + \pi r^2 \times \rho_w \times g}{m} \times (X) \\ \Rightarrow T =2 \pi$$\sqrt \frac{m}{K + \pi r^2 \times \rho_{w} \times g}$

24   A wooden block of mass $0.5$ kg and density $800$ $kg/m^3$ is fastened to the free end of a vertical spring of spring constant $50$ N/m fixed at the bottom. If the entire system is completely immersed in water, find (a) the elongation (or compression) of the spring in equilibrium and (b) the time-period of vertical oscillations of the block when it is slightly depressed and released.

##### Solution :

A) $mg =kX + V\rho_{w}g$ $\\$ b) $a = kX/m$ $\\$ $w^2$X = $kX/m$ $\\$ $T$ = 2 $\pi$ $\sqrt \frac{m}{k}$

25   A cube of ice of edge $4$ cm is placed in an empty cylindrical glass of inner diameter $6$ cm. Assume that the ice melts uniformly from each side so that it always retains its cubical shape. Remembering that ice is lighter than water, find the length of the edge of the ice cube at the instant it just leaves contact with the bottom of the glass.

##### Solution :

Let x $\rightarrow$ edge of ice block $\\$ When it just leaves contact with bottom of the glass .$\\$ h $\rightarrow$ height of water melted from ice $\\$ W = U $\\$ $\Rightarrow x^3 \times \rho_{ice} \times g = x^2 \times h \times \rho_{w} \times g$ $\\$ Again, volume of water formed, from melting of ice is given by, $\\$ $4^3 - x^3 = \pi \times r^2 \times h -x^2h$ $\\$ $\Rightarrow 4^3 - x^3 = \pi \times 3^2 \times h -x^2h$ $\\$ Putting h =$0.9$ x $\Rightarrow$ x = $2.26$cm.

26   A U-tube containing a liquid is accelerated horizontally with a constant acceleration $a_o$. If the separation between the vertical limbs is $l$, find the difference in the heights of the liquid in the two arms.

##### Solution :

If $p_a \rightarrow$ atm.Pressure $\\$ A $\rightarrow$ area of cross secton $\\$ h $\rightarrow$ increase in hright $\\$ $p_aA + A \times L \times \rho \times a_0 = p a^A + h \rho g \times A$ $\\$ $\Rightarrow hg =a_0L \Rightarrow a_0 \frac{L}{g}$

27   At Deoprayag (Garhwal, UP) river Alaknanda mixes with the river Bhagirathi and becomes river Ganga. Suppose Alaknanda has a width of $12$ m, Bhagirathi has a width of $8$ m and Ganga has a width of $16$ m. Assume that the depth of water is same in the three rivers. Let the average speed of water in Alaknanda be $20$ km/h and in Bhagirathi be $16$ km/h. Find the average speed of water in the river Ganga.

##### Solution :

Volume of water, discharged from Alkananda + vol are of water discharged from bhagirathi = Volume of water flow in Ganga

28   Water flows through a horizontal tube of variable cross-section (figure 13-E7). The area of cross-section at A and B are 4 mm 2 and 2 mm 2 respectively. If 1 cc of water enters per second through A, find (a) the speed of water at A, (b) the speed of water at B and (c) the pressure difference PA - Pi.

##### Solution :

a) $a_A \times V_A = Q_A$ $\\$ b) $a_A \times V_A = a_B \times V_b$ $\\$ c) $\frac{1}{2} \rho V_A^{2} + P_A =\frac{1}{2} \rho V_B^{2} + P_B$ $\\$ $\Rightarrow (P_A -P_B) = \frac{1}{2} \rho (V_B^{2} - V_A^{2})$

29   Suppose the tube in the previous problem is kept vertical with A upward but the other conditions remain the same. The separation between the cross-sections at A and B is $15/16$ cm. Repeat parts (a), (b) and (c) of the previous problem. Take g = $10$ $m/s^2$.

##### Solution :

From Bernouli's equation, $\frac{1}{2} \rho V_A^{2} + \rho g h_A + p_A = \frac{1}{2} \rho V_B^{2} + \rho g h_B + p_B.$ $\\$ $\Rightarrow (P_A - P_B) = (\frac{1}{2}) \rho (V_B^{2} -V_A^{2})$

30   Suppose the tube in the previous problem is kept vertical with B upward. Water enters through B at the rate of $1$ $cm^3/s$ . Repeat parts (a), (b) and (c). Note that the speed decreases as the water falls down.

##### Solution :

$\frac{1}{2} \rho V_B^{2} +\rho g h_B + p_{B} = \frac{1}{2} \rho V_A^{2} + \rho g h_A +p_{A}$

31   Water flows through a tube shown in figure $(13-E8)$. The areas of cross-section at A and B are $1$ cm $2$ and $0.5$ $cm^2$ respectively. The height difference between A and B is $5cm$. If the speed of water at A is $10 cm/s$ find (a) the speed at B and (b) the difference in pressures at A and B.

##### Solution :

$\frac{1}{2} \rho V_A^{2} + \rho g h_A + P_A = \frac{1}{2} \rho V_B^{2} + \rho g h_B + P_B$ $\\$ $\Rightarrow P_B - P_A = \frac{1}{2} \rho(V_A^{2} - V_B^{2}) + \rho g (h_A - h_B)$

32   Water flows through a horizontal tube as shown in figure $(13-E9$). If the difference of heights of water column in the vertical tubes is $2 cm$, and the areas of cross-section at A and B are $4 cm^2$ and $2 cm^2$ respectively, find the rate of flow of water across any section.

##### Solution :

$\overline{V_A}a_A = \overline{V_B} \times a_B$ $\\$ $\Rightarrow \frac{1}{2} \rho V_{A}^2 + \rho g h_A +p_A =\frac{1}{2} \rho V_{B}^2 + \rho g h_B +p_B$ $\\$ $\Rightarrow \frac{1}{2} \rho V_{A}^2 +p_A =\frac{1}{2} \rho V_{B}^2 +p_B$ $\\$ $\Rightarrow p_A -p_B = \frac{1}{2} \times \rho(V_{B}^2 - V_{A}^2)$ $\\$ Rate of flow = $V_a \times a_A$

33   Water flows through the tube shown in figure $(13-E10)$. The areas of cross-section of the wide and the narrow portions of the tube are $5 cm^2$ and $2 cm^2$ respectively. The rate of flow of water through the tube is $500 cm^3/s$. Find the difference of mercury levels in the U-tube.

##### Solution :

$V_Aa_A = V_B a_B \times \frac{V_A}{B} = \frac{a_B}{a_A}$ $\\$ $5V_A =2V_B \Rightarrow V_B = \frac{5}{2}V_A$ $\\$ $\frac{1}{2} \rho V_{A}^2 + \rho g h_A +p_A =\frac{1}{2} \rho V_{B}^2 + \rho g h_B +p_B$ $\\$ $\Rightarrow p_A -P_B =\frac{1}{2} \rho (V_{B}^2 -V_{B}^2)$ (because $p_A - p_B = h \rho_m g$

34   Water leaks out from an open tank through a hole of area $2 mm^2$ in the bottom. Suppose water is filled up to a height of $80 cm$ and the area of cross-section of _the tank is $0.4 m^2$. The pressure at the open surface and at the hole are equal to the atmospheric pressure. Neglect the small velocity of the water near the open surface in the tank. (a) Find the initial speed of water coming out of the hole. (b) Find the speed of water coming out when half of water has leaked out. (c) Find the volume of water leaked out during a time interval dt after the height remained is h. Thus find the decrease in height dh in terms of h and dt. (d) From the result of part (c) find the time required for half of the water to leak out.

##### Solution :

$P_A + \frac{1}{2} \rho V_{A}^2 = P_B +\frac{1}{2} \rho V_{B}^2 \Rightarrow P_A -P_B =\frac{1}{2} \rho V_{B}^2$ {$V_A$ = 0} $\\$ $\Rightarrow \rho gh =\frac{1}{2} \rho V_{B}^2$ {$P_A = P_{atm} + \rho gh$} $\\$ $\Rightarrow V_B = \sqrt {2gh}$ $\\$ a) V = $\sqrt {(2gh)}$ $\\$ b) V = $\sqrt {2g \frac{(h}{2}} = \sqrt gh$ $\\$ c) V = $\sqrt {2gh}$ $\\$ V = aV $\times$ dt $\\$ AV = aV $\\$ $\Rightarrow A \times \frac{dh}{dt} = a \times \sqrt {2gh} \Rightarrow dh = \frac{a \times \sqrt {2gh} \times dt}{A}$ $\\$ d) dh = $\frac{a \times \sqrt 2gh \times dt}{A} \Rightarrow T = \frac{A}{a} \sqrt {\frac{2}{g}}[\sqrt H_1 - \sqrt H_2]$

35   Water level is maintained in a cylindrical vessel upto a fixed height H. The vessel is kept on a horizontal plane. At what height above the bottom should a hole be made in the vessel so that the water stream coming out of the hole strikes the horizontal plane at the greatest distance from the vessel (figure $13-E11$).

##### Solution :

V = $\sqrt {2g(H-h)}$ $\\$ t =$\sqrt {(\frac{2h}{g})}$ $\\$ x = v $\times$ t = $\sqrt{(2g(H-h) \times \frac{2h}{g}} = 4\sqrt {(Hh - h^2)}$ $\\$ So, $\Rightarrow \frac{d}{dh}(Hh -h^2) = 0 \Rightarrow 0 = H- 2h \Rightarrow h = \frac{H}{2}$.