HC
Verma Concepts of Physics Solutions - Part 1, Chapter 13 - Fluid Mechanics:
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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, patm
= 1.01 * 105 N/m2; density of mercury, ρHg =
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, pA = pB
Or, pg = ρHg
*g*Δh + patm
Or, pg = 13600 * 10 * 0.06
+ 1.01 * 105
Or, pg = (0.0816 + 1.01) *
105 = 1.0916 * 105 N/m2.
(b)
The pressure of mercury at the bottom of the U tube is
= patm + ρHg
*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, pA = pB
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:
(a)
Force exerted at the bottom
= Force due to cylindrical water column
+ atm. Force
= Ab * h * ρ * g + patm
* A
= Ab * (h * ρ * g + patm)
= 0.002 * (0.2 * 1000 * 10 + 105)
= 204 N.
(b)
Let the resultant force exerted by the sides of the glass be Fside.
From the free body diagram of water inside
the glass
→ Fside + Fbottom
– pa * A – mg = 0
Or, Fside + 204 – 105
* 0.002 – 0.5 * 10 = 0
Or, Fside = 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, At
= area of bottom of glass, Ab = 20 cm2 = 0.002 m2;
height, h = 20 cm = 0.2 m; Atmospheric pressure, patm = 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
= Ab * h * ρ * g
= 0.002 * 0.2 * 1000 * 10
= 4 N.
(b)
Let the resultant force exerted by the sides of the glass be Fside.
From the free body diagram of water
inside the glass
→ Fside + Fbottom
– mg = 0
Or, Fside + 4 – 0.5 * 10 =
0
Or, Fside = 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, ρw
= 1000 kg/m3; Density of mercury, ρHg = 13600 kg/m3;
height of mercury column, hHg = 76 cm.
Let the height of water column be hw.
As the atmospheric pressure is same
for both the cases.
Therefore, hw * ρw
* g = hHg * ρHg * g
Or, hw * 1000 = 76 * 13600
Or, hw = 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.
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 = pat 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.
