Q: A gardener uses a water pipe of $2.5\text{ cm}$ diameter. It takes $1\text{ min}$ to fill $30\text{ litre}$ bucket. What is the initial speed of water coming out of the hose? Then, nozzle with an opening $0.5\text{ cm}^2$ is attached to the hose. Find the water speed coming out from the nozzle.
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
Ans: For pipe (hose): $d_1 = 2.5\text{ cm}$, $r_1 = \frac{2.5}{2} = 1.25\text{ cm}$
$A_1 = \pi r_1^2 = 3.142 \times (1.25)^2 = 4.91\text{ cm}^2$, $v_1 = ?$
For nozzle: $A_2 = 0.5\text{ cm}^2$, $v_2 = ?$
$V = 30\text{ litre} = 30 \times 1000\text{ cm}^3$, $t = 1\text{ min} = 60\text{ s}$
The volume rate $\frac{V}{t} = A_1 v_1$
$\frac{30 \times 1000}{60} = 4.91 v_1$
$v_1 = \frac{30 \times 1000}{60 \times 4.91} = \mathbf{101.83\text{ cm s}^{-1}}$
Using continuity equation,
$A_2 v_2 = A_1 v_1 \implies v_2 = \frac{A_1 v_1}{A_2}$
$= \frac{4.91 \times 101.83}{0.5} = \mathbf{999.97\text{ cm s}^{-1}}$
[OR]
The volume rate $\frac{V}{t} = A_2 v_2$
$\frac{30 \times 1000}{60} = 0.5 v_2$
$v_2 = \frac{30 \times 1000}{60 \times 0.5} = \mathbf{1000\text{ cm s}^{-1}}$
Q: What area must a heating duct have, if air moving $3\text{ m s}^{-1}$ along it can replenish the air every $15\text{ min}$ in a room of volume $300\text{ m}^3$? Assume the density of air remains constant.
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
Ans: $v = 3\text{ m s}^{-1}$, $V = 300\text{ m}^3$, $t = 15\text{ min} = 15 \times 60 = 900\text{ s}$
Volume flow rate $\frac{V}{t} = A v$
$\frac{300}{900} = A \times 3$
$A = \frac{300}{900 \times 3} = \mathbf{0.11\text{ m}^2}$
Q: Water runs into a fountain, filling all the pipes, at a steady rate of $0.75\text{ m}^3\text{ s}^{-1}$. (i) How fast will it shoot out of a hole $4.5\text{ cm}$ in diameter? (ii) At what speed will it shoot out if the diameter of the hole is three times as large?
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
Ans: $\frac{V}{t} = 0.75\text{ m}^3\text{ s}^{-1}$, $d = 4.5\text{ cm}$
$r = \frac{4.5\text{ cm}}{2} = 2.25\text{ cm} = 2.25 \times 10^{-2}\text{ m}$
$A = \pi r^2 = 3.142 \times (2.25 \times 10^{-2})^2\text{ m}^2$, $v = ?$
(i) Volume flow rate $\frac{V}{t} = A v$
$0.75 = 3.142 \times (2.25 \times 10^{-2})^2 v$
$v = \frac{0.75}{3.142 \times (2.25 \times 10^{-2})^2} = \mathbf{471.51\text{ m s}^{-1}}$
(ii) If the diameter of the hole is three times larger, $d_2 = 3 d_1$, $r_2 = 3 r_1$, $A_2 = 9 A_1$
Using continuity equation, $A_2 v_2 = A_1 v_1 \implies 9 A_1 v_2 = A_1 v_1$
$v_2 = \frac{v_1}{9} = \frac{471.51}{9} = \mathbf{52.39\text{ m s}^{-1}}$
[OR]
$d_2 = 3 \times 4.5 = 13.5\text{ cm} \implies r_2 = 6.75\text{ cm} = 6.75 \times 10^{-2}\text{ m}$
$A_2 = \pi r_2^2$
Using continuity equation, $A_2 v_2 = A_1 v_1 \implies v_2 = \frac{A_1 v_1}{A_2} = \frac{\pi r_1^2 v_1}{\pi r_2^2}$
$= \frac{(2.25 \times 10^{-2})^2 \times 471.51}{(6.75 \times 10^{-2})^2} = \mathbf{52.39\text{ m s}^{-1}}$
Q: Water tank of dimension $3\text{ m} \times 3\text{ m} \times 3\text{ m}$ and the base of the tank is $10\text{ m}$ above the ground. It takes $1\text{ h}$ to fill. (i) Find the power output of the pump to fill the tank. (ii) Find the flow rate through the pipe and the speed of the water flow in the pipe which has $5\text{ cm}$ radius.
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
Ans: $V = 3 \times 3 \times 3 = 27\text{ m}^3$, $h = 10 + 3 = 13\text{ m}$, $t = 1\text{ h} = 3600\text{ s}$
(i) $P = \frac{W}{t} = \frac{mgh}{t} = \frac{\rho V g h}{t}$
$P = \frac{1000 \times 27 \times 9.8 \times 13}{3600} = \mathbf{955.5\text{ W}}$
(ii) $r = 5\text{ cm} = 5 \times 10^{-2}\text{ m}$
Flow rate $\frac{V}{t} = \frac{27}{3600} = 7.5 \times 10^{-3}\text{ m}^3\text{ s}^{-1}$
$\frac{V}{t} = A v \implies 7.5 \times 10^{-3} = 3.142 \times (5 \times 10^{-2})^2 v$
$v = \frac{7.5 \times 10^{-3}}{3.142 \times (5 \times 10^{-2})^2} = \mathbf{0.96\text{ m s}^{-1}}$
Q: A shower head has $20$ circular openings, each with radius $1\text{ mm}$. The shower head is connected to a pipe with radius $0.8\text{ cm}$. If the speed of water in the pipe is $3\text{ m s}^{-1}$, what is its speed as it exits the shower-head openings?
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
Ans: For pipe: $r_1 = 0.8\text{ cm} = 0.8 \times 10^{-2}\text{ m}$, $v_1 = 3\text{ m s}^{-1}$, $n_1 = 1$
For shower head: $r_2 = 1\text{ mm} = 1 \times 10^{-3}\text{ m}$, $v_2 = ?$, $n_2 = 20$
Using continuity equation, $n_2 A_2 v_2 = n_1 A_1 v_1 \implies n_2 (\pi r_2^2) v_2 = n_1 (\pi r_1^2) v_1$
$v_2 = \frac{n_1 r_1^2 v_1}{n_2 r_2^2} = \frac{1 \times (0.8 \times 10^{-2})^2 \times 3}{20 \times (1 \times 10^{-3})^2} = \mathbf{9.6\text{ m s}^{-1}}$
Q: In an adult, the radius of aorta is normally $\sim 1.5\text{ cm}$ and blood moves through it at an average speed of $30\text{ cm s}^{-1}$. If typical capillary has a radius of $5 \times 10^{-6}\text{ m}$, and blood passes through them with a velocity of $0.1\text{ cm s}^{-1}$ approximately how many capillaries are in the body?
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
Ans: For aorta: $r_1 = 1.5\text{ cm}$, $v_1 = 30\text{ cm s}^{-1}$, $n_1 = 1$
For capillary: $r_2 = 5 \times 10^{-6}\text{ m} = 5 \times 10^{-4}\text{ cm}$, $v_2 = 0.1\text{ cm s}^{-1}$, $n_2 = ?$
Using continuity equation, $n_2 A_2 v_2 = n_1 A_1 v_1 \implies n_2 (\pi r_2^2) v_2 = n_1 (\pi r_1^2) v_1$
$n_2 = \frac{n_1 r_1^2 v_1}{r_2^2 v_2} = \frac{1 \times (1.5)^2 \times 30}{(5 \times 10^{-4})^2 \times 0.1} = \mathbf{2.7 \times 10^9\text{ capillaries}}$
Q: Water flowing through a restriction in a horizontal pipe is shown in the given figure. The radius of the pipe at the left end and right end are $6\text{ cm}$ and $2\text{ cm}$ respectively. If the velocity and pressure of water at the left end of pipe are $1\text{ m s}^{-1}$ and $200\text{ kPa}$, find the velocity and pressure of water at the right end of the pipe.
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
Ans: At left end: $r_1 = 6\text{ cm} = 6 \times 10^{-2}\text{ m}$, $p_1 = 200\text{ kPa} = 200 \times 10^3\text{ Pa}$, $v_1 = 1\text{ m s}^{-1}$
At right end: $r_2 = 2\text{ cm} = 2 \times 10^{-2}\text{ m}$, $p_2 = ?$, $v_2 = ?$
$\rho = 1000\text{ kg m}^{-3}$, $h_1 = h_2$ (horizontal pipe)
Using continuity equation, $v_2 = \frac{r_1^2 v_1}{r_2^2} = \frac{(6 \times 10^{-2})^2 \times 1}{(2 \times 10^{-2})^2} = \mathbf{9\text{ m s}^{-1}}$
Using Bernoulli's theorem, $p_2 = p_1 + \frac{1}{2}\rho (v_1^2 - v_2^2)$
$= [200 \times 10^3] + \frac{1}{2} \times 1000 \times (1^2 - 9^2) = \mathbf{160\,000\text{ Pa (or) } 160\text{ kPa}}$
Q: Calculate the pressure and speed of water at points B and C shown below. (density of water is $1000\text{ kg m}^{-3}$)
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
Ans: At A: $A_1 = 3\text{ m}^2$, $v_1 = 10\text{ m s}^{-1}$, $p_1 = 300\text{ kPa} = 300 \times 10^3\text{ Pa}$, $h_1 = 20\text{ m}$
At B: $A_2 = 3\text{ m}^2$, $v_2 = ?$, $p_2 = ?$, $h_2 = 0\text{ m}$
At C: $A_3 = 6\text{ m}^2$, $v_3 = ?$, $p_3 = ?$, $h_3 = 0\text{ m}$
$v_2 = v_1 = \mathbf{10\text{ m s}^{-1}}$ (Since $A_1 = A_2$)
Using Bernoulli's theorem for A and B:
$p_2 = p_1 + \rho g (h_1 - h_2) + \frac{1}{2}\rho(v_1^2 - v_2^2) = 300 \times 10^3 + 1000 \times 9.8 \times 20 = \mathbf{496\text{ kPa}}$
Using continuity equation for A and C: $A_3 v_3 = A_1 v_1 \implies v_3 = \frac{3 \times 10}{6} = \mathbf{5\text{ m s}^{-1}}$
Using Bernoulli's theorem for A and C, $p_3 = \mathbf{533.5\text{ kPa}}$
Q: A sealed tank containing seawater to a height of $11\text{ m}$ also contains air above the water at a pressure of $4\text{ atm}$. Water flows out from the bottom through a small hole. How fast is this water moving?
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
Ans: Top: $h_1 = 11\text{ m}$, $p_1 = 4\text{ atm} = 4 \times 1.01 \times 10^5\text{ Pa}$, $v_1 \approx 0$
Bottom: $h_2 = 0\text{ m}$, $p_2 = 1\text{ atm} = 1.01 \times 10^5\text{ Pa}$, $v_2 = ?$
$\rho = 1025\text{ kg m}^{-3}$ (sea water)
Using Bernoulli's theorem, $v_2 = \sqrt{\frac{2(p_1 - p_2)}{\rho} + 2g h_1} = \mathbf{28.4\text{ m s}^{-1}}$
Q: A small circular hole $6.00\text{ mm}$ in diameter is cut in the side of a large water tank, $14.0\text{ m}$ below the water level in the tank. The top of the tank is open to the air. Find (i) the speed of efflux of the water, and (ii) the volume discharged per second.
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
Ans: $h = 14\text{ m}$, $d = 6\text{ mm} \implies r = 3 \times 10^{-3}\text{ m}$, $\rho = 1000\text{ kg m}^{-3}$
(i) Using Torricelli's law (derived from Bernoulli's theorem):
$v_2 = \sqrt{2 g h} = \sqrt{2 \times 9.8 \times 14} = \mathbf{16.57\text{ m s}^{-1}}$
(ii) Volume discharged per second ($\frac{V}{t} = A_2 v_2$):
$\frac{V}{t} = \pi r^2 v_2 = 3.142 \times (3 \times 10^{-3})^2 \times 16.57 = \mathbf{4.68 \times 10^{-4}\text{ m}^3\text{ s}^{-1}}$
Q: One hypodermic syringe contains medicine with density $1010\text{ kg m}^{-3}$. The barrel of the syringe has a cross-sectional area $2.5 \times 10^{-5}\text{ m}^2$, and the needle has a cross-sectional area $1 \times 10^{-8}\text{ m}^2$. The syringe is in horizontal position and injection is forced by $2\text{ N}$ acting on the plunger. Find the speed of the injection that enters into the patient. (The volume of syringe is $3\text{ cc}$ and injection time is $2\text{ s}$)
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
Ans: Barrel: $A_1 = 2.5 \times 10^{-5}\text{ m}^2$, $F = 2\text{ N}$, $p_1 = 1\text{ atm} + \frac{F}{A_1}$
Needle: $A_2 = 1 \times 10^{-8}\text{ m}^2$, $p_2 \approx 1\text{ atm} = 1.01 \times 10^5\text{ Pa}$, $v_2 = ?$
$\rho = 1010\text{ kg m}^{-3}$, $h_1 = h_2$
Using continuity equation, $A_1 v_1 = A_2 v_2 \implies v_1 = \frac{A_2 v_2}{A_1} = \frac{v_2}{2500}$ ($v_1 \ll v_2$, so $v_1$ can be neglected)
Using Bernoulli's theorem:
$p_2 + \frac{1}{2}\rho v_2^2 = p_1 \implies \frac{1}{2}\rho v_2^2 = p_1 - p_2 = 1\text{ atm} + \frac{F}{A_1} - 1\text{ atm} = \frac{F}{A_1}$
$\frac{1}{2} \times 1010 \times v_2^2 = \frac{2}{2.5 \times 10^{-5}} = 80\,000$
$v_2 = \sqrt{\frac{80\,000}{505}} = \mathbf{12.59\text{ m s}^{-1}}$
Q: Air is streaming past a horizontal aeroplane's wings such that its speed is $120\text{ m s}^{-1}$ at the upper surface and $90\text{ m s}^{-1}$ at the lower surface. If the wing is $10\text{ m}$ long and $2\text{ m}$ wide and density of air is $1.3\text{ kg m}^{-3}$, find the net lift on it.
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
Ans: $v_1 = 120\text{ m s}^{-1}$, $v_2 = 90\text{ m s}^{-1}$, $\rho = 1.3\text{ kg m}^{-3}$
Area $A = 10 \times 2 = 20\text{ m}^2$, $h_1 = h_2$
$\Delta p = \frac{1}{2}\rho (v_1^2 - v_2^2) = \frac{1}{2} \times 1.3 \times [120^2 - 90^2] = 4095\text{ Pa}$
$F_{\text{net}} = \Delta p \times A = 4095 \times 20 = \mathbf{81\,900\text{ N}}$
Q: An aeroplane's wings have a total surface area of $480\text{ m}^2$. The pressure difference upper and lower surfaces of each wing is $6500\text{ Pa}$. Calculate the lift created.
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
Ans: $A = 480\text{ m}^2$, $\Delta p = 6500\text{ Pa}$, lift $F = ?$
$F = \Delta p \times A = 6500 \times 480 = \mathbf{3.12 \times 10^6\text{ N}}$
Q: Air streams horizontally past a small aeroplane's wings such that the speed is $70\text{ m s}^{-1}$ over and $60\text{ m s}^{-1}$ past the bottom surface. If the plane has a wing area of $16.2\text{ m}^2$ on the top and on the bottom, what is the net vertical force that the air exerts on the airplane? (Density of air = $1.2\text{ kg m}^{-3}$)
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
Ans: $v_1 = 70\text{ m s}^{-1}$, $v_2 = 60\text{ m s}^{-1}$, $A = 16.2\text{ m}^2$, $\rho = 1.2\text{ kg m}^{-3}$
$\Delta p = \frac{1}{2}\rho (v_1^2 - v_2^2) = \frac{1}{2} \times 1.2 \times [70^2 - 60^2] = 780\text{ Pa}$
$F_{\text{net}} = \Delta p \times A = 780 \times 16.2 = \mathbf{12\,636\text{ N}}$
Q: Derive expression for terminal velocity when a ball of radius $r$ is dropped through a liquid of viscosity $\eta$ and density $\rho$.
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
Ans: The forces acting on the object are:
- Weight of the sphere ball, $w = mg = \rho_1 V g = \frac{4}{3}\pi r^3 \rho_1 g$
- Upward thrust, $F_1 = mg = \rho_2 V g = \frac{4}{3}\pi r^3 \rho_2 g$
- Drag force, $F = 6\pi \eta r v$
At a constant terminal velocity, upward and downward forces are in balance:
$F_1 + F = w$
$\frac{4}{3}\pi r^3 \rho_2 g + 6\pi \eta r v = \frac{4}{3}\pi r^3 \rho_1 g$
$6\pi \eta r v = \frac{4}{3}\pi r^3 (\rho_1 - \rho_2) g$
$v = \frac{2 g r^2 (\rho_1 - \rho_2)}{9\eta}$
Q: The shear stress at a point in a liquid is found to be $0.03\text{ N m}^{-2}$. The velocity gradient at the point is $0.15\text{ s}^{-1}$. What will be its viscosity (in poise)? ($1\text{ Pa s} = 10\text{ poise}$)
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
Ans: Shear stress $\tau = \frac{F}{A} = -0.03\text{ N m}^{-2}$
Velocity gradient $\frac{dv}{dy} = \frac{\Delta v}{\Delta y} = 0.15\text{ s}^{-1}$
$\eta = \frac{-\tau}{dv/dy} = \frac{-(-0.03)}{0.15} = 0.2\text{ Pa s}$
$= 0.2\text{ Pa s} \times \frac{10\text{ poise}}{1\text{ Pa s}} = \mathbf{2\text{ poise}}$
Q: A square plate $0.1\text{ m}$ side moves parallel to second plate with a velocity of $0.1\text{ m s}^{-1}$, both plates being immersed in water. If the viscous force is $0.002\text{ N}$ and the coefficient of viscosity is $0.01\text{ poise}$, what is the distance between the plates? ($1\text{ Pa s} = 10\text{ poise}$)
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
Ans: $A = 0.1 \times 0.1 = 0.01\text{ m}^2$, $\Delta v = 0.1\text{ m s}^{-1}$, $F = -0.002\text{ N}$
$\eta = 0.01\text{ poise} = 0.01 \times \frac{1\text{ Pa s}}{10\text{ poise}} = 0.001\text{ Pa s}$
$F = -\eta A \frac{\Delta v}{\Delta y} \implies \Delta y = \frac{\eta A \Delta v}{-F} = \frac{0.001 \times 0.01 \times 0.1}{0.002} = \mathbf{5 \times 10^{-4}\text{ m (or) } 0.5\text{ mm}}$
Q: A metal plate of area $2.5 \times 10^{-4}\text{ m}^2$ is placed on a $0.25 \times 10^{-3}\text{ m}$ thick layer of castor oil. If a force of $50\text{ N}$ is needed to move the plate with a velocity $3 \times 10^{-2}\text{ m s}^{-1}$, calculate the coefficient of viscosity of castor oil.
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
Ans: $A = 2.5 \times 10^{-4}\text{ m}^2$, $\Delta y = 0.25 \times 10^{-3}\text{ m}$, $F = 50\text{ N}$
$\Delta v = 3 \times 10^{-2}\text{ m s}^{-1}$, $\eta = ?$
Velocity gradient $\frac{dv}{dy} = \frac{3 \times 10^{-2}}{0.25 \times 10^{-3}} = 120\text{ s}^{-1}$
Shear stress $\tau = \frac{F}{A} = \frac{50}{2.5 \times 10^{-4}} = 20 \times 10^4\text{ N m}^{-2}$
$\eta = \frac{\tau}{dv/dy} = \frac{20 \times 10^4}{120} = \mathbf{1.67 \times 10^3\text{ Pa s}}$
Q: Determine the radius of a rain drop falling through air with terminal velocity $9\text{ m s}^{-1}$. Viscosity of air is $1.8 \times 10^{-5}\text{ Pa s}$, density of water is $1000\text{ kg m}^{-3}$, density of air is $1.21\text{ kg m}^{-3}$.
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
Ans: $v = 9\text{ m s}^{-1}$, $\eta = 1.8 \times 10^{-5}\text{ Pa s}$, $\rho_1 = 1000\text{ kg m}^{-3}$, $\rho_2 = 1.21\text{ kg m}^{-3}$
$r = \sqrt{\frac{9 \eta v}{2 g (\rho_1 - \rho_2)}} = \sqrt{\frac{9 \times 9 \times 1.8 \times 10^{-5}}{2 \times 9.8 \times (1000 - 1.21)}} = \mathbf{2.729 \times 10^{-4}\text{ m}}$
Q: Assume that a spherical object is flowing through water. Viscosity of water is $0.001\text{ Pa s}$, radius of spherical object is $2\text{ mm}$ and the velocity of the object at particular instant is $2\text{ m s}^{-1}$. Find the drag force on the object due to the fluid. Assume that Stokes' law is valid.
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
Ans: $\eta = 0.001\text{ Pa s}$, $r = 2\text{ mm} = 2 \times 10^{-3}\text{ m}$, $v = 2\text{ m s}^{-1}$
$F = 6\pi \eta r v = 6 \times 3.142 \times 0.001 \times 2 \times 10^{-3} \times 2 = \mathbf{7.54 \times 10^{-5}\text{ N}}$
Q: A horizontal circular loop of wire has a diameter of $5\text{ cm}$ and is lowered in a sample of crude oil. The additional force required to pull the loop out of the oil is $0.04\text{ N}$; calculate the surface tension of the crude oil.
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
Ans: $d = 5\text{ cm}$, $r = 2.5\text{ cm} = 2.5 \times 10^{-2}\text{ m}$, $F = 0.04\text{ N}$
Circumference $l = 2\pi r = 2 \times 3.142 \times 2.5 \times 10^{-2} = 0.1571\text{ m}$
$\gamma = \frac{F}{2l} = \frac{0.04}{2 \times 0.1571} = \mathbf{0.127\text{ N m}^{-1}}$
Q: Calculate the force required to pull a flat circular plate of radius $15\text{ cm}$ from the surface of a liquid which has a surface tension of $0.053\text{ N m}^{-1}$.
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
Ans: $r = 15\text{ cm} = 15 \times 10^{-2}\text{ m}$, $\gamma = 0.053\text{ N m}^{-1}$, $F = ?$
$l = 2\pi r = 2 \times 3.142 \times 15 \times 10^{-2} = 94.26 \times 10^{-2}\text{ m}$
$\gamma = \frac{F}{l} \implies F = \gamma l = 0.053 \times 94.26 \times 10^{-2} = \mathbf{0.0499\text{ N}}$
Q: A needle has a length of $3.2\text{ cm}$. When placed gently on the surface of the water in a glass, this needle will float if it is not too heavy. What is the weight of the heaviest needle? Assume that the surface tension of water is $0.073\text{ N m}^{-1}$.
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
Ans: $l = 3.2\text{ cm} = 3.2 \times 10^{-2}\text{ m}$, $\gamma = 0.073\text{ N m}^{-1}$, $w = ?$
$\gamma = \frac{F}{2l} \implies w = F = \gamma (2l) = 0.073 \times 2 \times 3.2 \times 10^{-2} = \mathbf{4.672 \times 10^{-3}\text{ N}}$
Q: A soap film is formed on a rectangular frame of length $7\text{ cm}$ side dipping into soap solution. The frame hangs from the arm of a balance. An extra weight of $0.4\text{ g}$ is to be placed in the opposite pan to balance the pull on the frame. Calculate the surface tension of soap solution.
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
Ans: $l = 7\text{ cm} = 7 \times 10^{-2}\text{ m}$, $m = 0.4\text{ g} = 0.4 \times 10^{-3}\text{ kg}$, $g = 9.8\text{ m s}^{-2}$
$F = w = mg = 0.4 \times 10^{-3} \times 9.8 = 0.00392\text{ N}$
Since the rectangular frame has two edges in contact with the soap solution:
$\gamma = \frac{F}{2l} = \frac{0.00392}{2 \times 7 \times 10^{-2}} = \mathbf{0.028\text{ N m}^{-1}}$
Q: When a capillary tube stands upright in a beaker of water, the water rises $h$ in the tube. If the radius of the tube is $2\text{ mm}$, contact angle is $30^\circ$, density of water is $1000\text{ kg m}^{-3}$ and surface tension is $0.072\text{ N m}^{-1}$, calculate $h$.
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
Ans: $r = 2\text{ mm} = 2 \times 10^{-3}\text{ m}$, $\gamma = 0.072\text{ N m}^{-1}$, $\theta = 30^\circ$, $\rho = 1000\text{ kg m}^{-3}$
$h = \frac{2\gamma \cos\theta}{\rho g r} = \frac{2 \times 0.072 \times \cos 30^\circ}{1000 \times 9.8 \times 2 \times 10^{-3}} = \mathbf{0.0064\text{ m (or) } 0.64\text{ cm}}$
Q: At a certain temperature, water has a surface tension of $0.072\text{ N m}^{-1}$. In a $3\text{ mm}$ diameter vertical tube if the liquid rises $6\text{ mm}$ above the liquid outside the tube, calculate the contact angle. Assume density of water $\rho = 1000\text{ kg m}^{-3}$.
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
Ans: $d = 3\text{ mm} \implies r = 1.5 \times 10^{-3}\text{ m}$, $\gamma = 0.072\text{ N m}^{-1}$, $h = 6\text{ mm}$
$\cos\theta = \frac{\rho g r h}{2\gamma} = \frac{1000 \times 9.8 \times 1.5 \times 10^{-3} \times 6 \times 10^{-3}}{2 \times 0.072}$
$\theta = \mathbf{52.23^\circ \text{ (or) } 52^\circ 13'}$
Q: Calculate the height to which water will rise in a capillary tube of diameter $1\text{ mm}$. The surface tension of water is $0.072\text{ N m}^{-1}$, density of water is $1000\text{ kg m}^{-3}$. (Assume angle of contact is $0^\circ$.)
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
Ans: $d = 1\text{ mm} \implies r = 0.5 \times 10^{-3}\text{ m}$, $\gamma = 0.072\text{ N m}^{-1}$, $\theta = 0^\circ$
$h = \frac{2\gamma \cos\theta}{\rho g r} = \frac{2 \times 0.072 \times \cos 0^\circ}{1000 \times 9.8 \times 0.5 \times 10^{-3}} = \mathbf{0.0294\text{ m (or) } 2.94\text{ cm}}$