CBSE Test Paper 03
Chapter 10 Mechanical Properties of Fluids
1. Density is defined as 1
a. volume of 1 kg of the material
b. mass per unit volume
c. volume per unit mass
d. volume of 10 kg of the material
2. The coefficient of viscosity for a fluid is defined as the ratio of 1
a. shearing stress to the flow rate
b. shearing stress to the strain rate
c. compressive stress to the strain rate
d. tensile stress to the strain rate
3. A man in a lift is holding a can filled with water but having a hole in its bottom. When
will the drainage of water be maximum? 1
a. when the lift goes down with uniform velocity
b. when the lift is stationary
c. when the lift is going down with acceleration
d. when the lift is going up with acceleration
4. The height of mercury column in a simple barometer is h. As tube is inclined to the
vertical at an angle the length of the mercury column along the length of the tube is
l, then 1
a. l = h/cos
b. l = h sin
c. l = h cos
d. l = h
5. The surface tension of soap solution is 25 x . The excess pressure inside a
soap bubble of diameter 1 cm is 1
a. 15 Pa
b. 20 Pa
c. 5 Pa
d. 10 Pa
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6. Why is it difficult to stop bleeding from a cut in the human body at high altitudes? 1
7. The blood pressure of humans is greater at the feet than at the brain? 1
8. Is viscosity a vector? 1
9. A mercury barometer gives a reading of 74.5 cm. Calculate the atmospheric pressure
at the place. Given that density of mercury = 13.6 2 103 kg m-3.
10. Why is it painful to walk barefooted on a road covered with pebbles having sharp
edges? 2
11. The terminal velocity of a tiny droplet is v. N number of such identical droplets
combine together forming a bigger drop. Find the terminal velocity of the bigger drop.
2
12. Two mercury droplets of radii 0.1 cm. and 0.2 cm. collapse into one single drop. What
amount of energy is released? The surface tension of mercury T = 435.5 × 10–3 N m–1.
3
13. The flow rate of water is 0.58 L/mm from a tap of diameter of 1.30 cm. After some
time, the flow rate is increased to 4 L/min. Determine the nature of the flow for both
the flow rates. The coefficient of viscosity of water is 10-3Pa - s and the density of
water is 103 kg/m3. 3
14. The free surface of oil in a tanker, at rest, is horizontal. If the tanker starts
accelerating the free surface will be titled by an angle θ. If the acceleration is a ms–2,
what will be the slope of the free surface? 3
15. Two narrow bores of diameters 3.0 mm and 6.0 mm are joined together to form a U-
tube open at both ends. If the U-tube contains water, what is the difference in its
levels in the two limbs of the tube? Surface tension of water at the temperature of the
experiment is N m-1. Take the angle of contact to be zero and density of
water to be kg m-3 (g = 9.8 m s-2). 5
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CBSE Test Paper 03
Chapter 10 Mechanical Properties of Fluids
Answer
1. b. mass per unit volume
Explanation: Density is defined as the compactness of substance.
Mathematically,
Density(D)=
2. b. shearing stress to the strain rate
Explanation: The degree to which a fluid resists flow under an applied force,
expressed as the ratio of the shearing stress to the velocity gradient. The
coefficient of viscosity of liquids decreases as temperature increases because
the bonds between molecules are weakened.
3. d. when the lift is going up with acceleration
Explanation: Force acting on the liquid contained in the cylinder in the vertical
directions are
i. upward due to the liquid below it
ii. downward due to the liquid above it
Applying newton's second law
iii. weight mg = Sz g where is density of liquid
Net force = mass acceleration
Thus pressure differece is greatest when lift moves upwards with acceleration a
because for uniform motion will be zero and for lift moving downwards the
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term in bracket would be (g - ). So due to maximum pressure difference
between top and bottom, drainage of water will be maximum in this case.
4. c. l = h cos
Explanation: This will form a right angled triangle with base l and hypertonius
h and angle between base and hypertonius is
5. b. 20 Pa
Explanation: given surface tension S=25 10-3 N/m
diameter= 1 cm
thus radius r = = 0.5 cm
excess pressure inside soap bubble
6. The atmospheric pressure decreases as the altitude increases. Thus, The atmospheric
pressure is low at high altitudes. So the pressure difference between blood pressure
and atmospheric pressure become large at high altitudes. As the pressure difference
is large thus the rate of flow of blood from the cut will be greater. Therefore it is
difficult to stop bleeding from a cut in the body at high altitudes.
7. The height of the blood column in the human body is more at the feet than at the
brain. Since pressure is directly dependent on height of the column, so pressure is
more at feet than at the brain.
8. It is the property of liquid which is equal to the magnitude of dragging force per unit
area between the two layers of liquid whose velocity gradient is unity. So it has no
direction (only magnitude of force) so it is not vector.
9. Here h = 74.5 cm = 0.745 m, density of mercury and
Atmospheric pressure Pa =
or
10. It is painful to walk barefooted on a road covered with pebbles having sharp edges
because they have small area and since Pressure, Area is less i.e. pressure is
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more. It Means out feet exert greater pressure on pebbles and in turn pebbles exert
equal reaction on the feet.
11. The terminal velocity is given by ....(i), where coefficient of
discovery. = density of the liquid and = density of the object. If N drops coalesce,
then
Volume of one big drop = volume of N small droplets
R = N1/3 r
Placing these values in the equation..(i) , we get
, is the terminal velocity of the larger drop.
12. Energy due to surface Tension
By law of conservation of mass, volume of drop
R is the radius of new drop formed by the combination of two smaller drops.
Energy is released due to formation of bigger drop from smaller drops because final
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area will be smaller than former case.
13. Given, diameter,
Coefficient of viscosity of water,
Density of water,
The volume of the water flowing out per second is
r2 = v
Reynold’s number, Re = =
Case I When
Re =
Re , so the flow is steady or streamline
Case II When
=
Re =
Re > 3000, so the flow will be turbulent.
14. In accelerated fluids, fluid in a container can orient itself in a direction due to
acceleration. In that case, pressure at different heights (at the surface) is same (atm).
Then at the same height, pressure is different. Consider the tanker is pulled by force F
which produces acceleration in the truck in forward direction. consider the bit of
fluid just to the left of that. It must be exerting more pressure: it has to accelerate the
bit to its right (the bit in the middle) plus all the stuff to the right of that.
Consider only an small element of mass at P. When tanker is pulled by forward
acceleration ‘a’ then this element of mass also experiences it in forward direction. But
due to inertia of rest it tries to remain in rest. Due to same acceleration it moves in
backward direction as it is free and not rigidly connected to tanker. Hence force
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acting on in horizontally backward direction due to tanker’s
acceleration a, vertically downward due to gravity.
The resolution of component off along an perpendicular to the inclined
surface of oil are resolved. Normal reaction is balanced by component
and when surface is inclined at maximum angle then
So is required slope
is required slope.
So slope depends on the acceleration(a) of the fluid taking container as shown in
figure.
15. Diameter of the first bore,
thus, the radius of the first bore,
Diameter of the second bore,
Hence, the radius of the second bore,
Surface tension of water,
Angle of contact between the bore surface and water,
Density of water,
Acceleration due to gravity, f = 9.8 m/s2
Let h1 and h2 be the heights to which water rises in the first and second tubes
respectively.
These heights are given by the relations:
...(i)
....(ii)
The difference between the levels of water in the two limbs of the tube can be
calculated as:
= 4.97 mm
Hence, the difference between levels of water in the two bores is 4.97 mm.
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