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]
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
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).
Solution
Water pressure at first position ([latex]P_1[/latex]) = [latex]4\times 10^4 Nm^{-2}[/latex]
Speed at 1st position ([latex]v_1[/latex]) = 2 m/s
Area at 1st position ([latex]A_1[/latex]) = 200 cm2 = 0.02 m2
Water pressure at 2nd position ([latex]P_2[/latex]) =?
Area at 2nd 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]
Caster oil at 20oC 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].
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]
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