1. [2074 ‘B’] An iceberg having a volume of 2060 cc floats in sea water of density 1030 kgm^-3 with a portion of 224 cc above the surface. Calculate the density of ice.
   Solution:
   Volume of iceberg [latex](V_i)[/latex] = 2060 cc
   Density of water [latex](\rho_w) = 1030 kgm^{-3}[/latex]
   Portion of ice above the surface = 224 cc
   [latex]\therefore \text{Portion of ice below the surface} = \text{Volume of water displaced}(V_w)[/latex]
   = 2060 – 224 = 1836 cc
   Density of ice [latex](\rho_i)[/latex] =?
   We know that, from principle of floatation
   Weight of ice = Weight of water displaced
   Or, [latex]m_ig = m_wg[/latex]
   Or, [latex]\rho_i V_i = \rho_wV_w[/latex]
   Or, [latex]\rho_i \times 2060 = 1030 \times 1836[/latex]
   Or, [latex]\rho_i = \frac{1891080}{2060} = 918 kgm^{-3}[/latex]
2. [2073 ‘S’] A boy can lift a maximum load of 150 N of water. How many litres of mercury of density 13600 kgm^-3 he can lift in an identical vessel?
   Solution:
   Weight of water a boy can lift = 150 N
   [latex]\therefore \text{Weight of mercury he can lift in identical vessel}[/latex] = 150 N
   Volume of mercury (V) =?
   Density of mercury [latex](\rho) = 13600 kgm^{-3}[/latex]
   We know that,
   Weight of mercury = mg
   Or, 150 = [latex]\rho \times V \times g[/latex]
   Or, 150 = 13600 [latex]\times V \times g[/latex]
   Or, V = [latex]\frac{150}{13600\times 10} = 1.103\times 10^{-3}\ m^3[/latex] = 1.103 liter

1. [2082] Water flows steadily through a horizontal pipe of non-uniform cross section. If the pressure of water is [latex]4\times 10^4 Nm^{-2}[/latex] at a point where the velocity of flow is 2 m/s and cross section is 200 cm². Calculate the pressure at a point where cross section reduces to 50 cm². (density of water = 1000 kg/m³). [3]
   Solution:
   Water pressure at first position ([latex]P_1[/latex]) = [latex]4\times 10^4 Nm^{-2}[/latex]
   Speed at 1^st position ([latex]v_1[/latex]) = 2 m/s
   Area at 1^st position ([latex]A_1[/latex]) = 200 cm² = 0.02 m²
   Water pressure at 2^nd position ([latex]P_2[/latex]) =?
   Area at 2^nd position [latex](A_2)[/latex] = 50 cm² = 0.005 m²
   Density of water = 1000 kg/m³
   From equation of continuity, [latex]A_1v_1 = A_2v_2[/latex]
   Or, [latex]v_2 = \frac{A_1}{A_2}v_1 = \frac{0.02}{0.005} = 8 ms^{-1}[/latex]
   From Bernoulli’s principle,
   [latex]P_1 + \rho gh + \frac{1}{2}\rho v_1^2 = P_2 + \rho gh + \frac{1}{2}\rho v_2^2[/latex]           [For same horizontal, [latex]h_1 = h_2[/latex]]
   Or, [latex]P_2 = P_1 + \frac{1}{2}\rho (v_1^2 – v_2^2) = (4\times 10^4)+\frac{1}{2}\times 1000\times (2^2 – 8^2)[/latex]
   = [latex](4\times 10^4) – 30000 = 1\times 10^4 N/m^2[/latex]
   So, [latex]P_2 = 1\times 10^4 N/m^2[/latex]
2. [2081 GIE ‘B’] Caster oil at 20^oC has a coefficient of viscosity 2.42 [latex]NSm^{-2}[/latex] and density 940 [latex]kgm^{-3}[/latex]. Calculate the terminal velocity of a steel ball of radius 1 mm falling under gravity in the oil, taking the density of steel as 7800 [latex]kgm^{-3}[/latex]. (Take g = 10 [latex]ms^{-2})[/latex]. [2]
   Solution:
   Coefficient of viscosity ([latex]\eta) = 2.42\ NSm^{-2}[/latex]
   Density of liquid ([latex]\sigma) = 940\ kgm^{-3}[/latex]
   Terminal velocity (v) =?
   Radius of air bubble (r) = 1 mm = [latex]1\times 10^{-3}[/latex] m
   Density of steel [latex](\rho)[/latex] = 7800 kg/m³
   Acceleration due to gravity (g) = 10 [latex]ms^{-2}[/latex]
   We know,
   [latex]v = \frac{2r^2(\rho – \sigma)g}{9\eta}[/latex]
   = [latex]\frac{2\times (1\times 10^{-3})^2\times (7800 – 940)\times 10}{9\times 2.42} = 6.3 \times 10^{-3}\ ms^{-1}[/latex]
3. [2081 ‘B/C’] N identical spherical drops of water falling with a terminal velocity  were to coalesce to form a bigger drop and fall with a new terminal velocity  through air. (Neglect air resistance)

1. Obtain a relation between [latex]v_2[/latex] and [latex]v_1[/latex].                   [2]
2. If N = 2 and [latex]v_1[/latex] = [latex]0.4 ms^{-1}[/latex] then calculate [latex]v_2[/latex]     [1]
   Solution:
   (i) Terminal velocity of each small drop = [latex]v_1[/latex]
   We have,
   [latex]v_1 = \frac{2r^2(\rho – \sigma)g}{9\eta}[/latex] ………………….. (i) where, r = radius
   [latex]\rho[/latex] = density of water drop
   [latex]\sigma = [/latex] density of air
   [latex]\eta = [/latex] coefficient of viscosity of air
   Let, R be the radius of combined drop.
   Terminal velocity of larger drop = [latex]v_2[/latex]
   [latex]v_2 = \frac{2R^2(\rho – \sigma)g}{9\eta}[/latex] …………………………… (ii)
   Dividing equation (ii) by (i), we get,
   [latex]\frac{v_2}{v_1} = \frac{R^2}{r^2}[/latex] …………………………… (iii)
   Now, volume of large drop = N. volume of smaller drop
   Or, [latex]\frac{4}{3}\pi R^3 = N\times \frac{4}{3}\pi r^3[/latex]
   Or, [latex]R^3 = N.r^3[/latex]
   Or, [latex]R = \sqrt[3]{N}r[/latex] ……………. (iv)
   From equation (iii) & (iv),
   [latex]v_2 = v_1\times \frac{R^2}{r^2} = v_1\times \frac{r^2.N^{2/3}}{r^2}[/latex]
   Or, [latex]v_2 = v_1 \times N^{2/3}[/latex] ………………….. (v)
   (ii) Here, N = 2 and [latex]v_1 = 0.4 ms^{-1}[/latex]
   We get, [latex]v_2 = 0.4\times (2)^{2/3} = 0.635[/latex] m/s

4. [2081 ‘D’/2068] Water flows steadily through a horizontal pipe of non-uniform cross-section. If the pressure of water is  at a point where the velocity of the flow is  and cross-section is 0.02 m². What is the pressure at a point where cross-section reduced to 0.01 m²?                          [3]
   Solution:
   Water pressure at first position ([latex]P_1[/latex] = [latex]4\times 10^4 Nm^{-2})[/latex]
   Speed at 1^st position ([latex]v_1[/latex]) = 2 m/s
   Area at 1^st position ([latex]A_1[/latex]) = 0.02 [latex]m^2[/latex]
   Water pressure at 2^nd position ([latex]P_2[/latex]) =?
   Area at 2^nd position ([latex]A_2[/latex]) = 0.01 m²
   From equation of continuity, [latex]A_1v_1 = A_2v_2[/latex]
   Or, [latex]v_2 = \frac{A_1}{A_2}v_1 = \frac{0.02}{0.01}\times 2 = 4 ms^{-1}[/latex]
   From Bernoulli’s principle,
   [latex]P_1 + \rho gh + \frac{1}{2}\rho v_1^2 = P_2 + \rho gh + \frac{1}{2}\rho v_2^2[/latex]      
   [For same horizontal, [latex]h_1 = h_2[/latex]]
   Or, [latex]P_2 = P_1 + \frac{1}{2}\rho (v_1^2 – v_2^2) = 4\times 10^4 + \frac{1}{2}\times 1000 \times (2^2 – 4^2)[/latex]
   = [latex]4\times 10^4 – 6000 = 3.4\times 10^4 N/m^2[/latex]
   So, [latex]P_2 = 3.4\times 10^4 N/m^2[/latex]
5. [2080 GIE ‘A’] An air bubble of radius 1 cm is rising at a steady rate of 5 mm/s through a liquid of density 0.8 g/cm³. Calculate the coefficient of viscosity of the liquid. (Neglect the density of air)              [3]
   Solution:
   Radius of air bubble (r) = 1 cm = 0.01 m
   Terminal velocity (v) = – 5 mm/s = [latex]- 5\times 10^{-3}[/latex] m/s      [The air bubble moves upward]
   Density of liquid ([latex]\sigma)[/latex] = 0.8 g/cm³ = [latex]\frac{0.8\times 10^{-3}}{(10^{-2})^3}[/latex] = 800 kg/m³
   Density of air ([latex]\rho[/latex]) = 0
   Coefficient of viscosity ([latex]\eta)[/latex] =?
   We know,
   [latex]\eta = \frac{2r^2(\rho – \sigma)g}{9v}[/latex]
   = [latex]\frac{2\times 0.01^2\times (0-800)\times 10}{9\times (-5\times 10^{-3})} = 35.56 Nsm^{-2}[/latex]
6. [2080 ‘R’] A horizontal pipe of 25 cm² cross section carries water at a velocity of 3 m/s. The pipe feeds into a smaller pipe with a cross section of 15 cm². Determine the pressure change that occurs on going from the larger diameter pipe to the smaller pipe.                       [2]
   Solution:
   Velocity of water in larger pipe ([latex]v_1[/latex]) = 3 m/s
   Area of larger pipe ([latex]A_1[/latex]) = 25 cm² = [latex]25\times 10^{-4}[/latex] m²
   Area of smaller pipe ([latex]A_2[/latex]) = 15 cm² = [latex]15\times 10^{-4}[/latex]
   Pressure change ([latex]\Delta P[/latex]) =?
   From equation of continuity, [latex]A_1v_1 = A_2v_2[/latex]
   Or, [latex]v_2 = \frac{A_1}{A_2}v_1 = \frac{25\times 10^{-4}}{15\times 10^{-4}}\times 3 = 5 ms^{-1}[/latex]
   From Bernoulli’s principle,
   [latex]P_1 + \rho gh + \frac{1}{2}\rho v_1^2 = P_2 + \rho gh + \frac{1}{2}\rho v_2^2[/latex]       [For same horizontal, [latex]h_1 = h_2[/latex]]
   Or, [latex]P_1 – P_2 = \frac{1}{2}\rho (v_2^2 – v_1^2) = \frac{1}{2}\times 1000\times (5^2 – 3^2)[/latex]
   [latex]\Delta P = 8000 N/m^2[/latex]
7. [2079 ‘O’/2075 ‘S’/2072 ‘D’] Caster oil at 20^oC has a coefficient of viscosity 2.42 [latex]Nsm^{-2}[/latex] and density [latex]940 kgm^{-3}[/latex]. Calculate the terminal velocity of a steel ball of radius 2mm falling under the gravity in the oil, taking the density of steel as 7800 [latex]kg/m^3[/latex]. (g = 10 m/s²)   [3]
   Solution:
   Coefficient of viscosity ([latex]\eta[/latex]) = [latex]2.42 Nsm^{-2}[/latex]
   Density of liquid ([latex]\sigma[/latex]) = 940 kg/m³
   Terminal velocity (v) =?
   Radius of air bubble (r) = 2 mm = [latex]2\times 10^{-3}[/latex] m
   Density of steel ([latex]\rho[/latex]) = 7800 kg/m³
   We know,
   [latex]v = \frac{2r^2(\rho – \sigma)g}{9\eta}[/latex]
   = [latex]\frac{2\times (2\times 10^{-3})^2\times (7800-940)\times 10}{9\times 2.42}=0.025 ms^{-1}[/latex]
8. [2079 ‘V’] Two spherical rain drops of equal size are falling through air with terminal velocity 10 cm/s. If these two drops were to coalesce to form a single drop, what would be the new terminal velocity?
   Solution:
   Terminal velocity of each small drop = [latex]v_1 = 10 cm/s = 0.1 m/s[/latex]
   We have,
   [latex]v_1 = \frac{2r^2(\rho – \sigma)g}{9\eta}[/latex] ………………….. (i) where, r = radius
   [latex]\rho[/latex] = density of water drop
   [latex]\sigma = [/latex] density of air
   [latex]\eta = [/latex] coefficient of viscosity of air
   Let, R be the radius of combined drop.
   Terminal velocity of larger drop = [latex]v_2[/latex]
   [latex]v_2 = \frac{2R^2(\rho – \sigma)g}{9\eta}[/latex] …………………………… (ii)
   Dividing equation (ii) by (i), we get,
   [latex]\frac{v_2}{v_1} = \frac{R^2}{r^2}[/latex] …………………………… (iii)
   Now, volume of large drop = N. volume of smaller drop
   Or, [latex]\frac{4}{3}\pi R^3 = 2\times \frac{4}{3}\pi r^3[/latex]
   Or, [latex]R^3 = 2.r^3[/latex]
   Or, [latex]R = \sqrt[3]{2}r[/latex] ……………. (iv)
   From equation (iii) & (iv),
   [latex]v_2 = v_1\times \frac{R^2}{r^2} = v_1\times \frac{r^2.2^{2/3}}{r^2}[/latex]
   Or, [latex]v_2 = 0.1 \times 2^{2/3} = 0.1587 m/s = 15.87 cm/s[/latex]
9. [2076 ‘B’] Eight spherical rain drops of the same mass and radius are falling down with a terminal speed of 5 cm/s. If they coalesce to form one big drop, what will be its terminal speed?
   Solution:
   Terminal velocity of each small drop ([latex]v_1)[/latex] = 5 cm/s = 0.05 m/s
   We have,
   [latex]v_1 = \frac{2r^2(\rho – \sigma)g}{9\eta}[/latex] ………………….. (i) where, r = radius
   [latex]\rho[/latex] = density of water drop
   [latex]\sigma[/latex] = density of air
   [latex]\eta[/latex] = coefficient of viscosity of air
   Let, R be the radius of combined drop.
   Terminal velocity of larger drop = [latex]v_2[/latex]
   [latex]v_2 = \frac{2R^2(\rho – \sigma)g}{9\eta}[/latex] …………………………… (ii)
   Dividing equation (ii) by (i), we get,
   [latex]\frac{v_2}{v_1} = \frac{R^2}{r^2}[/latex] …………………………… (iii)
   Now, volume of large drop = N. volume of smaller drop
   Or, [latex]\frac{4}{3}\pi R^3 = 8\times \frac{4}{3}\pi r^3[/latex]
   Or, [latex]R^3 = 8.r^3[/latex]
   Or, [latex]R = \sqrt[3]{8}r[/latex] = 2r ……………. (iv)
   From equation (iii) & (iv),
   [latex]v_2 = v_1 \times \frac{R^2}{r^2} = v_1\times \frac{r^2.2^2}{r^2}[/latex]
   Or, [latex]v_2 = 0.05\times 4[/latex] = 0.2 m/s
10. [2075 ‘A’] Calculate mass of an aeroplane with the wings of area 55 m² flying horizontally. The velocity of air above the below the wings is 155 m/s and 140 m/s respectively.
    Solution:
    Area of wings ([latex]A[/latex]) = 55 m²
    Velocity of air above the wings ([latex]v_1[/latex]) = 155 m/s
    Velocity of air below the wings ([latex]v_2[/latex]) = 140 m/s
    Mass of aeroplane ([latex]m[/latex]) =?
    From Bernoulli’s principle,
    [latex]P_1 + \rho gh + \frac{1}{2}\rho v_1^2 = P_2 + \rho gh + \frac{1}{2}\rho v_2^2[/latex]
    [For same horizontal, [latex]h_1 = h_2[/latex]]
    Or, [latex]P_2 – P_1 = \frac{1}{2}\rho (v_1^2 – v_2^2) = \frac{1}{2}\times 1.293\times (155^2 – 140^2)[/latex]
    [latex]\Delta P = 2860.762 N/m^2[/latex]
    Now, [latex]\Delta P = \frac{F}{A}[/latex]
    Or, [latex]F = \Delta P.A = 2860.762\times 55 = 1.573\times 10^5 N[/latex]
    Again, F = mg
    Or, [latex]m = \frac{F}{g} = \frac{1.573\times 10^5}{10} =\ 1.573\times 10^4 \ kg[/latex] kg
11. [2073 ‘S’] Calculate the magnitude and direction of the terminal velocity of an air bubble of radius 1mm passing through an oil of viscosity 0.2  and specific gravity 0.9 if the density of air is 1.29 [latex]kgm^{-3}[/latex].
    Solution:
    Coefficient of viscosity ([latex]\eta[/latex]) = [latex]0.2 Nsm^{-2}[/latex]
    Specific gravity = 0.9
    Density of liquid ([latex]\sigma[/latex]) = 0.9 g/cm³ = [latex]\frac{0.9\times 10^{-3}}{(10^{-2})^3}[/latex] = 900kg/m³
    Terminal velocity (v) =?
    Radius of air bubble (r) = 1 mm = [latex]1\times 10^{-3}[/latex] m
    Density of steel ([latex]\rho[/latex]) = 1.29 kg/m³
    We know,
    [latex]v = \frac{2r^2(\rho – \sigma)g}{9\eta}[/latex]
    [latex]=\frac{2\times (1\times 10^{-3})^2\times (1.29 – 900)\times 10}{9\times 0.2} = – 0.0099 ms^{-1}[/latex]
    – ve sign indicates the air bubble is moving upwards.
12. [2073 ‘D’] Eight spherical raindrops of equal size are falling vertically through air with a terminal velocity of 0.15 m/s. What would be the terminal velocity, if they coalesce to form a big drop?
    Solution:
    Terminal velocity of each small drop = [latex]v_1 = 0.15 m/s[/latex]
    We have,
    [latex]v_1 = \frac{2r^2(\rho – \sigma)g}{9\eta}[/latex] ………………….. (i) where, r = radius
    [latex]\rho[/latex] = density of water drop
    [latex]\sigma = [/latex] density of air
    [latex]\eta = [/latex] coefficient of viscosity of air
    Let, R be the radius of combined drop.
    Terminal velocity of larger drop = [latex]v_2[/latex]
    [latex]v_2 = \frac{2R^2(\rho – \sigma)g}{9\eta}[/latex] …………………………… (ii)
    Dividing equation (ii) by (i), we get,
    [latex]\frac{v_2}{v_1} = \frac{R^2}{r^2}[/latex] …………………………… (iii)
    Now, volume of large drop = N. volume of smaller drop
    Or, [latex]\frac{4}{3}\pi R^3 = 8\times \frac{4}{3}\pi r^3[/latex]
    Or, [latex]R^3 = 8.r^3[/latex]
    Or, [latex]R = \sqrt[3]{8}r = 2r[/latex] ……………. (iv)
    From equation (iii) & (iv),
    [latex]v_2 = v_1\times \frac{R^2}{r^2} = v_1\times \frac{r^2.2^2}{r^2}[/latex]
    Or, [latex]v_2 = 0.15 \times 2^2 = 0.6 m/s[/latex]
13. [2069 ‘S’] Three spherical raindrops of equal size are falling vertically through air with terminal velocity of 0.2 m/s. What would be the terminal velocity if these three drops were to coalesce to form larger spherical drop?
    Solution:
    Terminal velocity of each small drop = [latex]v_1 = 0.2 m/s[/latex]
    We have,
    [latex]v_1 = \frac{2r^2(\rho – \sigma)g}{9\eta}[/latex] ………………….. (i) where, r = radius
    [latex]\rho[/latex] = density of water drop
    [latex]\sigma = [/latex] density of air
    [latex]\eta = [/latex] coefficient of viscosity of air
    Let, R be the radius of combined drop.
    Terminal velocity of larger drop = [latex]v_2[/latex]
    [latex]v_2 = \frac{2R^2(\rho – \sigma)g}{9\eta}[/latex] …………………………… (ii)
    Dividing equation (ii) by (i), we get,
    [latex]\frac{v_2}{v_1} = \frac{R^2}{r^2}[/latex] …………………………… (iii)
    Now, volume of large drop = N. volume of smaller drop
    Or, [latex]\frac{4}{3}\pi R^3 = 3\times \frac{4}{3}\pi r^3[/latex]
    Or, [latex]R^3 = 3.r^3[/latex]
    Or, [latex]R = \sqrt[3]{3}r = 3^{1/3}r[/latex] ……………. (iv)
    From equation (iii) & (iv),
    [latex]v_2 = v_1\times \frac{R^2}{r^2} = v_1\times \frac{r^2.3^{2/3}}{r^2}[/latex]
    Or, [latex]v_2 = 0.2 \times 3^{2/3} = 0.416 m/s[/latex]
14. [2067] Three spherical raindrops of equal size are falling vertically through air with a terminal velocity of 0.150 m/s. What would be the terminal velocity if these three drops were to coalesce to form a larger spherical drop?
    Solution:
    Terminal velocity of each small drop = [latex]v_1 = 0.15 m/s[/latex]
    We have,
    [latex]v_1 = \frac{2r^2(\rho – \sigma)g}{9\eta}[/latex] ………………….. (i) where, r = radius
    [latex]\rho[/latex] = density of water drop
    [latex]\sigma = [/latex] density of air
    [latex]\eta = [/latex] coefficient of viscosity of air
    Let, R be the radius of combined drop.
    Terminal velocity of larger drop = [latex]v_2[/latex]
    [latex]v_2 = \frac{2R^2(\rho – \sigma)g}{9\eta}[/latex] …………………………… (ii)
    Dividing equation (ii) by (i), we get,
    [latex]\frac{v_2}{v_1} = \frac{R^2}{r^2}[/latex] …………………………… (iii)
    Now, volume of large drop = N. volume of smaller drop
    Or, [latex]\frac{4}{3}\pi R^3 = 3\times \frac{4}{3}\pi r^3[/latex]
    Or, [latex]R^3 = 3.r^3[/latex]
    Or, [latex]R = \sqrt[3]{3}r = 3^{1/3}r[/latex] ……………. (iv)
    From equation (iii) & (iv),
    [latex]v_2 = v_1\times \frac{R^2}{r^2} = v_1\times \frac{r^2.3^{2/3}}{r^2}[/latex]
    Or, [latex]v_2 = 0.15 \times 3^{2/3} = 0.312 m/s[/latex]
15. [2065] Two drops of same liquid of same radius are falling through air with steady velocity of 2.0. If the two drops coalesce what would be the terminal velocity?
    Solution:
    Terminal velocity of each small drop = [latex]v_1 = 2.0 m/s[/latex]
    We have,
    [latex]v_1 = \frac{2r^2(\rho – \sigma)g}{9\eta}[/latex] ………………….. (i) where, r = radius
    [latex]\rho[/latex] = density of water drop
    [latex]\sigma = [/latex] density of air
    [latex]\eta = [/latex] coefficient of viscosity of air
    Let, R be the radius of combined drop.
    Terminal velocity of larger drop = [latex]v_2[/latex]
    [latex]v_2 = \frac{2R^2(\rho – \sigma)g}{9\eta}[/latex] …………………………… (ii)
    Dividing equation (ii) by (i), we get,
    [latex]\frac{v_2}{v_1} = \frac{R^2}{r^2}[/latex] …………………………… (iii)
    Now, volume of large drop = N. volume of smaller drop
    Or, [latex]\frac{4}{3}\pi R^3 = 2\times \frac{4}{3}\pi r^3[/latex]
    Or, [latex]R^3 = 2.r^3[/latex]
    Or, [latex]R = \sqrt[3]{2}r = 2^{1/3}r[/latex] ……………. (iv)
    From equation (iii) & (iv),
    [latex]v_2 = v_1\times \frac{R^2}{r^2} = v_1\times \frac{r^2.2^{2/3}}{r^2}[/latex]
    Or, [latex]v_2 = 2.0 \times 2^{2/3} = 3.17 m/s[/latex]