HC Verma Concepts Of Physics Exercise Solutions Of Chapter 13 (Fluid Mechanics)

HC Verma Concepts of Physics Solutions - Part 1, Chapter 13 - Fluid Mechanics:

EXERCISE

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/s².

Sol:

Given: height, h = 4 m; g = 10 m/s²; density, ρ = 1000 kg/m³.

We know, pressure due to hydrostatic is given by

→ Pressure, p = hρg = 4 * 10 * 1000 = 40000 N/m² or Pa.

It is necessary to specify that the tap 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 * 10⁵ N/m². Find (a) the pressure of the gas in the cylinder and (b) the pressure of mercury at the bottom of the U tube.

Sol:

Given: Atmospheric pressure, p_atm = 1.01 * 10⁵ N/m²; density of mercury, ρ_Hg = 13600 kg/m³; height difference between mercury surface, Δh = (8 – 2) = 6 cm = 0.06 m.

(a) We know, pressure at the same horizontal level in a continuous fluid are same.

So, p_A = p_B

Or, p_g = ρ_Hg *g*Δh + p_atm

Or, p_g = 13600 * 10 * 0.06 + 1.01 * 10⁵

Or, p_g = (0.0816 + 1.01) * 10⁵ = 1.0916 * 10⁵ N/m².

(b) The pressure of mercury at the bottom of the U tube is

= p_atm + ρ_Hg *g*h

= 1.01 * 10⁵ + 13600 * 10 * 0.08

= (1.01 + 0.11) * 10⁵ = 1.12 * 10⁵ N/m².

3. The area of cross-section of the wider tube shown in figure (13-E2) is 900 cm². If the boy standing on the piston weighs 45 kg, find the difference in the levels of water in the two tubes.

Sol:

Given: mass of man, m = 45 kg; area of wider tube, A = 900 cm² = 0.09 m²; density of water, ρ = 1000 kg/m³.

Let the difference in the levels of water in the two tubes be Δh.

We know, pressure at the same horizontal level in a continuous fluid are same.

So, p_A = p_B

Or, ρg (Δh) = mg/A

Or, Δh = m/Aρ

Or, Δh = 45/ (0.09 * 1000) = 0.5 m = 50 cm.

So, the difference in the levels of water in the two tubes is 50 cm.

4. A glass full of water has a bottom of area 20 cm², top of area 20 cm², 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 * 10⁵ N/m². Density of water = 1000 kg/m³ and g = 10 m/s². Take all numbers to be exact.

Sol:

Given: area of top of glass, A_t = area of bottom of glass, A_b = 20 cm² = 0.002 m²; height, h = 20 cm = 0.2 m; Atmospheric pressure, p_atm = 1.0 * 10⁵ N/m²; Density of water, ρ = 1000 kg/m³; g = 10 m/s²; mass of 0.5 litre water = 0.5 * 10^-3 * 1000 = 0.5 kg.

(a) Force exerted at the bottom

= Force due to cylindrical water column + atm. Force

= A_b * h * ρ * g + p_atm * A

= A_b * (h * ρ * g + p_atm)

= 0.002 * (0.2 * 1000 * 10 + 10⁵)

= 204 N.

(b) Let the resultant force exerted by the sides of the glass be F_side.

From the free body diagram of water inside the glass

→ F_side + F_bottom – p_a * A – mg = 0

Or, F_side + 204 – 10⁵ * 0.002 – 0.5 * 10 = 0

Or, F_side = 205 – 204 = 1 N upward.

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.

Sol:

Given: area of top of glass, A_t = area of bottom of glass, A_b = 20 cm² = 0.002 m²; height, h = 20 cm = 0.2 m; Atmospheric pressure, p_atm = 1.0 * 10⁵ N/m²; Density of water, ρ = 1000 kg/m³; and g = 10 m/s².

(a) Force exerted at the bottom.

= Force due to cylindrical water column + atm. Force

= A_b * h * ρ * g

= 0.002 * 0.2 * 1000 * 10

= 4 N.

(b) Let the resultant force exerted by the sides of the glass be F_side.

From the free body diagram of water inside the glass

→ F_side + F_bottom – mg = 0

Or, F_side + 4 – 0.5 * 10 = 0

Or, F_side = 5 – 4 = 1 N upward.

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)?

Sol:

Given: Density of water, ρ_w = 1000 kg/m³; Density of mercury, ρ_Hg = 13600 kg/m³; height of mercury column, h_Hg = 76 cm.

Let the height of water column be h_w.

As the atmospheric pressure is same for both the cases.

Therefore, h_w * ρ_w * g = h_Hg * ρ_Hg * g

Or, h_w * 1000 = 76 * 13600

Or, h_w = 1033.6 cm.

7. Find the force exerted by the water on a 2 m² 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? Neglect the size of the stone in comparison to the depth of the sea.

Sol:

Given: area of the plane surface, A = 2 m²; height of water, h = 500 m; density of the water, ρ = 1000 kg/m³.

(a) The force exerted by the water, F = P * A

Or, F = (h * ρ * g) * A

Or, F = 500 * 1000 * 10 * 2

Or, F = 10⁷ N.

(b) The force does not depend on the orientation of the rock as long as the surface area remains same. [No]

8. Water is filled in a rectangular tank of size 3 m * 2 m * 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 * 1 m. Take a horizontal strip of width δx 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. (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/s².

Sol:

Given: volume of the tank, V = 3 * 2 * 1 = 6 m; density of water, ρ = 1000 kg/m³.

(a) The total force exerted by the water on the bottom surface of the tank, F = ρ * V * g = 1000 * 6 * 10 = 6000 N.

(b) The force exerted by water on the strip of width δx is

→ dF = p_{at x} * δA

Or, dF = (xρg) * 2 * δx

Or, dF = x * 1000 * 10 * 2 * δx = 20000 x δx.

(c) The torque of the force about the bottom edge is

= dF * (1 – x) = 20000 x (1 – x) δx.

(d) The total force by the water on this side (from 0 to 1) is

= ∫20000 x δx = 10000 N. 

(e) The total torque by the water on the side about the bottom edge (from 0 to 1) is

= ∫20000 x(1 – x) δx = 10000/3 N-m.

9. An ornament weighing 36 g in air, weighing 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.

Sol:

Given: mass of ornament, m₀ = 36 g (in air); mass of ornament = 34 g (in water); Specific gravity of gold, s.g_Au = 19.3; Specific gravity of copper, s.g_cu = 8.9; density of water, ρ_w = 1 g/cm³.

Here, m₀ = m_Au + m_cu = 36 -------------- (1)

Let V be the volume of the ornament in cm³.

Buoyancy force = (36 – 34) * g = 2g = Vρ_wg

Or, (V_Au + V_cu) * 1 = 2

Or, [(m_Au/ρ_Au) + (m_cu/ρ_cu)] = 2

Or, [(m_Au/19.3) + (m_cu/8.9)] = 2     [ρ = s.g in cgs unit]

Or, 8.9 m_Au + 19.3 m_cu = 2 * 19.3 * 8.9

Or, 8.9 m_Au + 19.3 m_cu = 343.54 ----------- (2)

Solving equation (1) and (2), we get m_cu = 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.

Sol:

Given: mass of ornament, m₀ = 36 g (in air); mass of ornament = 34 g (in water); Specific gravity of gold, s.g_Au = 19.3; Specific gravity of copper, s.g_cu = 8.9; density of water, ρ_w = 1 g/cm³.

Let V be the volume of the ornament in cm³.

Buoyancy force = (36 – 34) * g = 2g = Vρ_wg

Or, (V_Au + V_cavity) * 1 = 2

Or, [(m_Au/ρ_Au) + V_cavity] = 2

Or, [(36/19.3) + V_cavity] = 2     [ρ = s.g in cgs unit]

Or, V_cavity = 0.134 cm³.

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/m³, find the normal force exerted by the bottom of the glass on the metal piece.

Sol:

Given: mass of metal piece, m = 160 g = 0.16; density of the metal, ρ_m = 8000 kg/m³.

Volume of metal piece, V = m/ ρ_m = 0.16/8000 = 0.02 * 10^-3, density of water, ρ_w = 1000 kg/m³.

∴ F_Buoyancy = V ρ_w g

Or, F_Buoyancy = 0.02 * 10^-3 * 1000 * 10

Or, F_Buoyancy = 0.2 N

From the FBD of the metal piece, we have

→ R + F_Buoyancy = mg

Or, R = 0.16 * 10 – 0.2 = 1.6 – 0.2 = 1.4 N.

So, the normal force exerted by the bottom of the glass on the metal piece is 1.4 N.

12. A ferry boat has internal volume 1 m³ 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 ?

Sol:

Given: internal volume of boat, V_i = 1 m³; mass of boat, m = 50 kg.

(a) Let V_f be the volume of the boat immersed in water.

Volume of boat inside water = volume of water displace in m³.

Since, weight of the boat is balanced by the buoyant force.

So, mg = V_f ρ_w g

Or, V_f = m/ ρ_w = 50/1000 = 1/20 m³.

Therefore, the fraction of the volume of the boat immersed in water = V_f/V_i = 1/20.

(b) Let, V^’ be the volume of boat filled with water before water starts coming in from the sides.

→ Weight of water inside of the boat + weight of the boat = buoyancy force

Or, V^’ ρ_w g + mg = V_i ρ_w g

Or, V^’ * 1000 + 50 = 1 * 1000

Or, V^’ = 950/1000 = 19/20 m³.

So, fraction of the boat's volume will be filled with water before water starts coming in from the sides is = V^’/V_i = 19/20.

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.

Sol:

Given: mass of metal piece, M_m = 0.5 kg; Specific gravity of ice = 0.9; density of ice block, ρ_i = 0.9 * 1000 = 900 kg/m³.

Let the minimum edge of the ice block be L.

Volume of the ice block, V_i = L³.

→ Weight of ice block + weight of the metal = buoyancy force

Or, V_i ρ_i g + M_i g = V_i ρ_w g

Or, L³ * 900 + 0.5 = L³ * 1000

Or, L³ (1000 – 900) = 0.5

Or, L = 0.17 m = 17 cm.

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.

Sol:

Given: Specific gravity of K.oil = 0.8; density of K.oil, ρ_k = 800 kg/m³; Specific gravity of ice = 0.9; density of ice, ρ_i = 900 kg/m³.

Let we assume V_k and V_w are the volume of ice block immerse in K.oil and water respectively.

V_ice = V_k + V_w

→ V_ice * ρ_ice * g = V_k * ρ_k * g + V_w * ρ_w * g

Or, (V_k + V_w) * ρ_ice = V_k * ρ_k + V_w * ρ_w

Or, (V_k + V_w) * 900 = V_k * 800 + V_w * 1000

Or, (1 + V_w/V_k) * 9 = 8 + (10 V_w)/V_k

Or, V_w/V_k = 1.

Therefore, the ratio of the volume of ice immersed in water to that in K.oil is 1:1.

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/m³ and density of water = 1000 kg/m³.

Sol:

Given: density of iron, ρ_i = 8000 kg/m³; density of water, ρ_w = 1000 kg/m³; thickness of iron sheet = 1 mm = 0.001 m.

Let the minimum value of the external edge of the box be L.

Volume of the box, V_b = L³.

Volume of the iron sheet, V_i = 6 * L² * 0.001 = 0.006L².

Mass of iron sheet, M_i = V_i * ρ_i = 0.006L² * 8000 = 48L².

Since, weight of the iron sheet is balanced by the buoyant force.

∴ Weight of the iron sheet = (volume of water displace) ρ_w g

Or, M_i g = V_b ρ_w g

Or, 48L²g = L³ * 1000 * g

Or, L = 0.048 m = 4.8 cm.

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.

Sol:

Given: mass of wooden block, m_wb = 200 g = 0.2 kg; Specific gravity of wood = 0.8; density of wood, ρ_w = 800 kg/m³; Specific gravity of lead = 11.3; density of lead, ρ_pb = 11300 kg/m³.

Let the mass of lead be m_pb.

Since, weight of the block and lead is balanced by the buoyant force.

∴ Weight of the block and lead = (volume of water displace) ρ_water g

Or, (m_wb + m_pb) * g = (V_wb + V_pb) * ρ_water * g

Or, 0.2 + m_pb = [(m_wb/ ρ_wb) + (m_pb/ ρ_pb)] * ρ_water

Or, 0.2 + m_pb = [(0.2/ 800) + (m_pb/ 11300)] * 1000

Or, [1 – (1/11.3)] m_pb = 0.05

Or, m_pb = 0.0548 kg = 54.8 g.

17. 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.

Sol:

Given: mass of wooden block, m_wb = 200 g = 0.2 kg; Specific gravity of wood = 0.8; density of wood, ρ_w = 800 kg/m³; Specific gravity of lead = 11.3; density of lead, ρ_w = 11300 kg/m³.

Let the mass of lead be m_pb.

Since, weight of the block and lead is balanced by the buoyant force.

∴ Weight of the block and lead = (volume of water displace) ρ_water g

Or, (m_w + m_pb) * g = V_w * ρ_water * g

Or, 0.2 + m_pb = (m_w/ ρ_w) * ρ_water

Or, 0.2 + m_pb = (0.2/ 800) * 1000

Or, m_pb = 0.05 kg = 50 g.

18. 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.

Sol:

Given: edge of the block, L = 12 cm = 0.12 m; Specific gravity of mercury = 13.6; density of the mercury, ρ_Hg = 13600 kg/m³.

Let the density of metal block be ρ_mb.

Since, weight of the block is balanced by the buoyant force.

∴ Weight of the block = (volume of water displace) ρ_Hg g

Or, V_mb ρ_mb g = (1/5) * L * L² * ρ_Hg * g

Or, 0.12³ * ρ_mb = (1/5) * 0.12 * 0.12² * 13600

Or, ρ_mb = 2720 kg/m³

Let the height of the water column be h.

Now, weight of the block balanced by the buoyant force.

∴ Weight of the block = weight of displace volume of Hg and water

Or, m_mb g = V_Hg ρ_Hg g + V_w ρ_w g

Or, (V_mb * ρ_mb) = (V_Hg * ρ_Hg) + (V_w * ρ_w)

Or, (V_Hg + V_w) ρ_b = V_Hg * ρ_Hg + V_w * ρ_w

Or, 0.12³ * 2720 = (0.12 – h) * 0.12² * 13600 + h * 0.12² * 1000

Or, 0.12 * 272 = (0.12 – h) * 1360 + h * 100

Or, h = 0.104 m = 10.4 cm.

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.

Sol:

Given: inner radius of spherical body, r_i = 6 cm = 0.06 m; outer radius of spherical body, r_o = 8 cm = 0.08 m.

Let the density of the material of the sphere be ρ.

Volume of the material, V_m = (4/3) π (r_o³ – r_i³)

Or, V_m = (4/3) π (0.08³ – 0.06³) = 0.00124 m³

Volume submerged in water, V_s = ½ * volume of the sphere

Or, V_s = ½ * (4/3) * π * r_o³ = ½ * (4/3) * π * 0.08³

Or, V_s = 0.00107 m³

Since, weight of the spherical body is balanced by the buoyant force.

∴ V_m ρ_m g = V_s ρ_w g

Or, 0.001236 * ρ_m = 0.00107 * 1000

Or, ρ_m = 865 kg/m³ 

Therefore, the density of the material of the sphere is 865 kg/m³.

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.

Sol:

Given: radius of solid sphere, r = 5 cm = 0.05 m; Load, F = 0.1 kg.

Let, the specific gravity of the material of the sphere be S.g.

∴ Density of material, ρ = s.g * 1000

Volume of sphere, V = (4/3) π r³ = (4/3) π (0.05)³ = 0.00052 m³.

Since, weight of the sphere and load is balanced by the buoyant force.

∴ Vρg + Lg = V ρ_w g

Or, 0.00052 * S.g * 1000 + 0.1 = 0.00052 * 1000

Or, S.g = 0.8

Therefore, the specific gravity of the material of the sphere 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/m³, density of wood = 800 kg/m³ and density of air = 1.293 kg/m³.

Sol:

Given: mass of iron block, m_i = 1 kg; mass of wooden block, m_w = 1 kg; density of iron, ρ_i = 7800 kg/m³; density of wood, ρ_w = 800 kg/m³; density of air, ρ_a = 1.293 kg/m³.

Volume of iron block,

→ V_i = m_i/ρ_i = 1/7800 = 12.8 * 10^-5 m³.

Volume of wooden block,

→ V_w = m_w/ρ_w = 1/800 = 125 * 10^-5 m³.

Weight of the iron block,

→ W_i = m_ig – V_i ρ_a g = (m_i – V_i ρ_a) g

Weight of the wooden block,

→ W_w = m_wg – V_w ρ_a g = (m_w – V_w ρ_a) g

∴ (W_i/ W_w) = (m_i – V_i ρ_a)/ (m_w – V_w ρ_a)

Or, (W_i/ W_w) = (1 – 12.8 * 10^-5 * 1.293)/ (1 – 125 * 10^-5 * 1.293)

Or, (W_i/ W_w) = 1.0015.

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.

Sol:

Given: mass of cylinder, m = 2 kg; outer diameter of cylinder, d = 20 cm = 0.2 m; area, A = (π/4) * (0.2)² = 0.01π m².

→ Restoring force = Buoyancy force (F_B)

And, Buoyancy force = ρ_w V g

Or, Buoyancy force = 1000 * Ax * g

Equation of motion is given by

→ Ma + F_B = 0

Or, 2a + 1000 * 0.01πx * g = 0

Or, a = – 5πxg -------------------- (1)

 Comparing eqtⁿ (1) with a = – ω²x, we get

→ – ω²x = – 5πxg

Or, ω = √ (5πg) = 12.53 rad/s.

And time period, T = 2π/ω = 0.5 s.

The time period of the resulting simple harmonic motion of the object is 0.5 s.

23. A cylindrical object of outer diameter 10 cm, height 20 cm and density 8000 kg/m³ 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.

Sol:

Given: density of cylinder, ρ = 8000 kg/m³; outer diameter of cylinder, d = 10 cm = 0.1 m; height, h = 20 cm = 0.2 m; spring constant, k = 500 N/m.

Volume of cylinder, V_c = (π/4) * (d)² h = 0.0005 π m³.

Mass of cylinder, M = ρ V_c = 8000 * 0.0005 π

Or, M = 12.5 kg

(a) Let the elongation of the spring in equilibrium condition be x.

Since, weight of cylinder is balanced by spring force and buoyancy force.

∴ F_B + F_S = Mg

Or, ρ_w V_d g + kx = ρ_c V_c g

Where, V_d = displace volume

Or, 1000 * ½ * V_c * g + 500x = 8000 * V_c * g

Or, x = (7500 * 0.0005 π * 10)/500

Or, x = 0.235 m = 23.5 cm.

The elongation of the spring in equilibrium condition is 23.5 cm.

(b) → Restoring force = Buoyancy force (F_B) + spring force (F_s)

And, Buoyancy force = ρ_w V g

Or, Buoyancy force = 1000 * Ax * g

Or, Buoyancy force = 1000 * 0.0025 * 10x

Or, Buoyancy force = 25x

Equation of motion is given by

→ Ma + F_B + F_s = 0

Or, 12.5 a + 25x + 500x = 0

Or, a = – 42x -------------------- (1)

 Comparing eqtⁿ (1) with a = – ω²x, we get

→ – ω²x = – 42x

Or, ω = √ (42) = 6.5 rad/s.

And time period, T = 2π/ω = 0.95 s.

The time period of the resulting simple harmonic motion of the object is 0.95 s.

24. A wooden block of mass 0.5 kg and density 800 kg/m³ 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.

Sol:

Given: mass of wooden block, M = 0.5 kg; density of wooden block, ρ = 800 kg/m³; spring constant, k = 50 N/m.

Volume of block, V_b = M/ ρ = 0.5/800 = 0.000625 m³.

(a) Let we assume the elongation of the spring be x.

Since, the wooden block is in equilibrium.

∴ F_B = Mg + F_s

Or, V_b ρ_w g = Mg + kx

Or, 0.000625 * 1000 * 10 = 0.5 * 10 + 50k

Or, 50k = 6.25 – 5

Or, k = 0.025 m = 2.5 cm.

Therefore, the elongation of the spring is 2.5 cm.

(b) → Restoring force = spring force (F_s)

Equation of motion is given by

→ Ma + F_s = 0               

Or, 0.5 a + 50x = 0

Or, a = – 100x ------------- (1)

 Comparing eqtⁿ (1) with a = – ω²x, we get

→ – ω²x = – 100x

Or, ω = √ (100) = 10 rad/s.

And time period, T = 2π/ω = 2π/10 s = π/5 s.

The time-period of vertical oscillations of the block is π/5 s. 

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.

Sol:

Given: side of ice cube, L = 4 cm = 0.04 m; cylindrical glass diameter, d = 6 cm = 0.06 m; ρ_ice = 900 kg/m³; ρ_w = 1000 kg/m³.

Let, the length of the edge of the ice cube at the instant it just leaves contact with the bottom of the glass be x.

Height of water melted from ice be h.

→ Weight of remaining ice = buoyancy force

Or, x³ * ρ_ice * g = x² * h * ρ_w * g

Or, h = 0.9x

Again, volume of water formed, from melting of ice is given by,

4³ – x³ = π * r² * h – x²h  [because amount of water = (πr² – x²)h]

Or, 4³ – x³ = π * 3² * h – x²h

Or, 64 – x³ = 25.45x – 0.9x³        [since h = 0.9 x]

Or, x³ + 254.5x – 640 = 0

Or, x = 2.26 cm. 

26. A U-tube containing a liquid is accelerated horizontally with a constant acceleration a₀. If the separation between the vertical limbs is l, find the difference in the heights of the liquid in the two arms. 

Sol:

Given: acceleration of U-tube is a₀; separation between the vertical limbs is l.

Let angle of inclination of free the free surface be θ and increase in height be h.

We know, tan θ = a₀/g ------------ (1)

And from the figure we can write,

→ tan θ = h/l -------------------- (2)

From equation 1 and 2, we get

→ h/l = a₀/g                 

Or, h = a₀l/g.

So, the difference in the heights of the liquid in the two arms is a₀l/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.

Sol:

Given: width of Alaknanda, W_A = 12 m; width of Bhagirathi, W_B = 8 m; width of Ganga, W_G = 16 m; velocity of Alaknanda, v_A = 20 km/h; velocity of Bhagirathi, v_B = 16 km/h; depth, d_G = d_A = d_B = d.

From the equation of continuity,

→ Volume of water, discharged from (Alaknanda + Bhagirathi) = Volume of water flow in Ganga.

Or, W_A * d * v_A + W_B * d * v_B = W_G * d * v_G     

Or, W_A * v_A + W_B * v_B = W_G * v_G

Or, 12 * 20 + 8 * 16 = 16 * v_G

Or, v_G = 23 km/h.                            

Therefore, the average speed of water in the river Ganga is 23 km/h.

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 mm² and 2 mm² 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 P_A – P_B.

Sol:

Given: area of cross-section at A, A_A = 4 mm², area of cross-section at B, A_B = 2 mm², volume flow rate, V = 1 cm³/s.

From the continuity,

→ A_Av_A = A_Bv_B = rate of flow of water (V)

(a) A_Av_A = rate of flow of water (V)

Or, 4 * 10^-6 * v_A = 1 * 10^-6             

Or, v_A = 0.25 m/s = 25 cm/s.

(b) A_Bv_B = rate of flow of water (V)

Or, 2 * 10^-6 * v_B = 1 * 10^-6

Or, v_B = 0.5 m/s = 50 cm/s.

(c) Appling Bernoulli’s equation between point A and B, we get

→ P_A + ½ ρ_w v_A = P_B + ½ ρ_w v_B

Or, P_A – P_B = ½ ρ_w (v_B)² – ½ ρ_w (v_A)²

Or, P_A – P_B = ½ * 1000 * (0.5² – 0.25²)

Or, P_A – P_B = 94 N/m².

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 m²/s

Sol:

Given: area of cross-section at A, A_A = 4 mm²; area of cross-section at B, A_B = 2 mm²; volume flow rate, V = 1 cm³/s; h_A – h_B = 15/16 cm = 0.0094 m.

Now, the tube in the previous problem is kept vertical with A upward.

From the continuity,

→ A_Av_A = A_Bv_B = rate of flow of water (V)

(a) A_Av_A = rate of flow of water (V)

Or, 4 * 10^-6 * v_A = 1 * 10^-6

Or, v_A = 0.25 m/s = 25 cm/s.

(b) A_Bv_B = rate of flow of water (V)

Or, 2 * 10^-6 * v_B = 1 * 10^-6

Or, v_B = 0.5 m/s = 50 cm/s.

(c) Appling Bernoulli’s equation between point A and B, we get

→ P_A + ½ ρ_w v_A + h_A ρ_w g= P_B + ½ ρ_w v_B + h_B ρ_w g

Or, P_A – P_B = ½ ρ_w [(v_B)² – (v_A)²] + (h_B – h_A) ρ_w g

Or, P_A – P_B = ½ * 1000 * (0.50² – 0.25²) – 0.0094 * 1000 * 10

Or, P_A – P_B = 94 – 94 = 0 (zero).

30. Suppose the tube in the previous problem is kept vertical with B upward. Water enters through B at the rate of 1 cm³/s. Repeat parts (a), (b) and (c). Note that the speed decreases as the water falls down.

Sol:

Given: area of cross-section at A, A_A = 4 mm²; area of cross-section at B, A_B = 2 mm²; volume flow rate, V = 1 cm³/s; h_A – h_B = 15/16 cm = 0.0094 m.

Now, the tube in the previous problem is kept vertical with B upward.

From the continuity,

→ A_Av_A = A_Bv_B = rate of flow of water (V)

(a) A_Av_A = rate of flow of water (V)

Or, 4 * 10^-6 * v_A = 1 * 10^-6

Or, v_A = 0.25 m/s = 25 cm/s.

(b) A_Bv_B = rate of flow of water (V)

Or, 2 * 10^-6 * v_B = 1 * 10^-6

Or, v_B = 0.5 m/s = 50 cm/s.

(c) Appling Bernoulli’s equation between point A and B, we get

→ P_A + ½ ρ_w v_A + h_A ρ_w g= P_B + ½ ρ_w v_B + h_B ρ_w g

Or, P_A – P_B = ½ ρ_w [(v_B)² – (v_A)²] + (h_B – h_A) ρ_w g

Or, P_A – P_B = ½ * 1000 * (0.50² – 0.25²) + 0.0094 * 1000 * 10

Or, P_A – P_B = 94 + 94 = 188 N/m².

31. Water flows through a tube shown in figure (13-E8). The areas of cross-section at A and B are 1 cm² and 0.5 cm² respectively. The height difference between A and B is 5 cm. 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.

Sol:

Given: area of cross-section at A, A_A = 1 cm²; area of cross-section at B, A_B = 0.5 cm²; v_A = 10 cm/s = 0.1 m/s; The height difference between A and B, (h_A – h_B) = 5 cm = 0.05 m.

(a) From the continuity,

→ A_Av_A = A_Bv_B

Or, 1 * 0.1 = 0.5 * v_B

Or, v_B = 0.2 m/s = 20 cm/s

So, the speed at B is 20 cm/s.

(b) Appling Bernoulli’s equation between point A and B, we get

→ P_A + ½ ρ_w v_A + h_A ρ_w g= P_B + ½ ρ_w v_B + h_B ρ_w g

Or, P_A – P_B = ½ ρ_w [(v_B)² – (v_A)²] + (h_B – h_A) ρ_w g

Or, P_A – P_B = ½ * 1000 * (0.2² – 0.1²) – 0.05 * 1000 * 10

Or, P_A – P_B = 15 – 500 = – 485 N/m² 

So, the difference in pressures at A and B is 485 N/m².

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 cm² and 2 cm² respectively, find the rate of flow of water across any section.

Sol:

Given: area of section A, A_A = 4 cm² = 4 * 10^-4 m; area of section B, A_B = 2 cm² = 2 * 10^-4; the difference of heights of water column, Δh = 2 cm = 0.02 m.

From the continuity,

→ A_Av_A = A_Bv_B = rate of flow of water (V)

→ v_A = V/A_A and v_B = V/A_B

We know, P_A – P_B = Δh ρ_w g

Or, P_A – P_B = 0.02 * 1000 * 9.8

Or, P_A – P_B = 196 Pa

Appling Bernoulli’s equation between point A and B, we get

→ P_A + ½ ρ_w v_A = P_B + ½ ρ_w v_B

Or, P_A – P_B = ½ ρ_w (v_B)² – ½ ρ_w (v_A)²

Or, P_A – P_B = ½ * 1000 * [(V/A_B)² – (V/A_A)²]

Or, 196 = 500 * V² * [(10⁴/2)² – (10⁴/4)²]

Or, V = 0.0001446 m³/s = 144.6 cm³/s.

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 cm² and 2 cm² respectively. The rate of flow of water through the tube is 500 cm³/s. Find the difference of mercury levels in the U-tube.

Sol:

Given: volume flow rate, V = 500 cm³/s; area of section 1, A₁ = 5 cm^2; area of section 2, A₂ = 2 cm².

From the continuity,

→ A₁v₁ = A₂v₂ = 500 cm³/s

→ v₁ = 100 cm/s = 1 m/s and v₂ = 250 cm/s = 2.5 m/s

Let the difference of mercury levels in the U-tube be Δh.

Appling Bernoulli’s equation between point 1 and 2, we get

→ P₁ + ½ ρ_w v₁ = P₂ + ½ ρ_w v₂

Or, P₁ – P₂ = ½ ρ_w (v₂)² – ½ ρ_w (v₁)²

Or, P₁ – P₂ = ½ * 1000 * (2.5² – 1²) = 2625 Pa.

We know, P_A = P_B                

Or, P₁ + h₁ ρ_w g = P₂ + h₂ ρ_w g + Δh ρ_Hg g

Or, Δh ρ_Hg g = P₁ – P₂ + (h₁ – h₂) ρ_w g

Or, Δh (ρ_Hg – ρ_w) g = 2625

Or, Δh * 12600 * 9.8 = 2625

Or, Δh = 0.0197 m = 1.97 cm.

34. Water leaks out from an open tank through a hole of area 2 mm² 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 m². 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.

Sol:

Given: height of water, h = 80 cm = 0.8 m; cross section area of tank, A = 0.4 m²; cross section area of hole, a = 2 mm² = 2 * 10^-6 m².

(a) Appling Bernoulli’s equation between point A and B, we get

→ P_A + ½ ρ (V_A)² + ρgh = P_B + ½ ρ (V_B)²

P_A = P_B = atmospheric pressure, and V_A = 0 (Neglect)

Or, ρgh = ½ ρ (V_B)²

Or, V_B = √ (2gh) = √ (2 * 10 * 0.8) = 4 m/s

So, the initial speed of water coming out of the hole is 4 m/s.

(b) New height = h/2 = 0.4

∴ V_B = √ (2gh) = √ (2 * 10 * 0.4) = √8

So, the speed of water coming out when half of water has leaked out is √8.

(c) Velocity at the hole, v_B = √ (2gh)

The volume of water leaked out during a time interval dt is

→ Volume, V = av_B * dt

Or, V = √ (2gh) * 2 * 10^-6 dt.

From the continuity equation,

→ Volume decrease in tank = volume of water leaked

Or, A * dh = √ (2gh) * 2 * 10^-6 dt

Or, dh = √ (2gh) * 2 * 10^-6 dt/A

Or, dh = √ (2gh) * 5 * 10^-6 dt.

(d) From the continuity equation,

→ Volume decrease in tank = volume of water leaked

Or, A * dh = √ (2gh) * a * dt

Or, ∫dt = ∫ [(A * dh) / (√ (2gh) * a)]

Or, T = (A/a) * √ (2/g) * [√h – √ (h/2)]

Or, T = 6.5 hours.

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).

Sol:

Let, we assume the height of the hole be h.

Velocity of the jet at the height (h), V = √ [2g (H – h)].

Let we assume that the time taken by the jet to reach the ground be‘t’.

So, h = ½ gt² → t = √ (2h/g)

Range, x = V * t = √ [2g (H – h)] * √ (2h/g)

Or, x = 2 √ (Hh – h²)

For maximum range, dx/dh = 0

Or, dx/dh = d (2 √ (Hh – h²))/dh = 0

Or, H – 2h = 0

Or, h = H/2

So, the height of the hole is H/2.

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