i. What will be the difference in liquid pressure on their bases if A of them is completely filled and B is filled half and kept at the same place.
A will have more pressure on its base and B will have pressure half of A if they are kept in the same place.
ii. What will be the difference in liquid pressure on their bases if both A and B are filled with water completely but one of them is kept in Lumbini and another in Pokhara? Why?
The drum which is kept in Lumbini will experience more pressure. Because the pressure of a liquid is directly proportional to the value of g and the value of g is more at Lumbini than at Pokhara.
iii. What will be the difference in liquid pressure on their base if A is filled with water and B is filled with salty water and kept at Biratnagar in the same position? Why?
B will have more liquid pressure because the density of the salty water is greater than the water.
d. When a body is fully or partially immersed in a liquid, name the forces acting on the body.
When a body is fully or partially immersed in liquid, the force acting on a body are:
i. Weight of body acting vertically downward
ii. The upthrust on the body acting vertically upward.
f. Write definitions yourself.
Differences between:
g. The experiment proves Archimedes' principle.
h. According to the law of floatation the weight of water displaced by the ship must be 5 x 10⁵ N.
K. We can
prove the occurrence of atmospheric pressure by performing different experiments.
One of the experiments is a tin can experiment.
In this experiment,
a tin can filled with less than one-third of the water is taken and is allowed to
heat until the water boils. As the water boils the lid of the can is closed and
is cooled by pouring cold water. After a certain time, the tin can crushes. This
is due to the presence of atmospheric pressure.
QN 3
Differentiate between:
A.
B.
C.
|
Air pump |
Water
pump |
|
1. It is a device used to pump air in
stoves, balloons, balls, bicycles, etc. |
1. It is a device used to pump underground
water. |
|
2. When
the piston is pushed down the air is released. |
2. When
the piston is pulled up (by pushing down the handle) the water is released
from the spout. |
D.
|
Downstroke |
Upstroke |
|
1. When the handle of a water pump is lifted up, the
piston is pushed down. It is a downstroke. |
1. When the handle is pushed down, it lifts the
piston up. It is upstroke. |
|
2. In this
condition, water between the two valves moves above the upper valve. |
2. In this condition, water comes out from the
spout. |
4. Give
reasons.
a. It is easier
to pull a bucket of water from a well until it is inside the water but difficult
when it is out of water.
Water exerts
upthrust on the bucket which reduces weight of the bucket. Outside water there
is very less upthrust of air. Thus, the force required to pull a bucket when
immersed in water is less than that required to pull when it is out of water.
b. A
bucket of water is filled faster in the downstair's tap than in the upstair's
tap.
Water tanks
are kept on the roof of a building. The downstairs taps have more depth. Since
liquid pressure increases with depth, in domestic water supply, the pressure on
the ground floor is higher than the upstairs and hence a bucket of water is
filled faster in the downstair's tap than in the upstair's tap.
c. The
blood pressure in human body is greater at the feet than at the brain.
This is
because the depth of the feet from the heart is more than the depth of the
brain. Since P=dgh (where d-density, g-acceleration due to gravity and
h-depth], it implies that pressure is more in the feet where h is more and it
is less in the brain where h is less.
d.
Deep-sea divers wear diving suits.
Liquid pressure increases with increase in depth of the liquid, so the
large liquid pressure is exerted on the diver. So, to bear this large liquid
pressure, the sea divers wear diving suits.
e. An egg
floats on salty water.
Density of
an egg is higher than the density of pure water but it is less than the density
of saturated salt solution. The salt solution exerts more upthrust on the egg
than the pure water due to its more density. It makes the egg float.
f. An
iron nail sinks in water but a ship made of the same material floats on it.
According to
law of floatation, the object which displaces liquid equal to its own weight
floats in liquid. In case of an iron nail, it has less surface area. Due to
this, the weight of water displaced by an iron nail in less than the weight of
the iron nail. So, it sinks in water. In case of ship, it has very wide surface
area. Due to this the weight of water displaced by the ship is equal to the
weight of the ship. So, the ship floats in water.
g. It is
easier to swim in an ocean than in a river.
Upthrust is
directly proportional to the density of the liquid. Sea water has a greater
density than river water. Thus, upthrust experienced by a swimmer in sea water
is comparatively more than in river water. So, it is much easier to swim in the
sea water.
h. Water cannot be used in a barometer.
Water cannot be used in barometer because it’s density is lower than Mercury . Water’s density is 1000 gram per cubic meter, hence it requires
a barometer whose height is around 11 meters which is practically
impossible.
i. The
washer of an air pump is used with grease/oil.
Washer of an
air pump is used with lubricants because the washer works as a piston as well
as a valve. In flexible condition only, it can expand and contract to work
properly. The smooth lubricant also reduces friction between the piston and the
cylinder.
j. A
water pump cannot lift water more than 10 metre high from the earth's surface.
It is
because a water pump is based on the principle of atmospheric pressure.
Atmospheric pressure cannot push up water more than 10 metre or even higher in
the water pump.
5.
Diagrammatic questions
a. In the
diagram, a body, ABCD is placed in water. Study the diagram and answer
following questions.
I. On
what surface of the body, will pressure be the maximum and why?
The pressure
will be maximum on the surface CD because of more depth.
II. On
what level, will thrust be the maximum?
Explain
why?
Upthrust
will be greater on surface CD because of more depth.
b. Answer the following questions on the
basis of the diagram given below.
i. Is
there any difference in upthrust of water at B and C ?
Upthrust
is greater at C than at B because C has more depth than B.
i. Why is the wall made thicker at the
bottom?
Wall is made thicker in the bottom
because there is more pressure in the bottom. In order to hold the large pressure
the wall is made wider in the bottom.
c. Densities of some of the substances
are given in the table. Answer the following questions on the basis of them.
i. If equal masses of all are taken, which one will have the
largest volume?
Z will
have more volume because density is inversely proportional to volume.
ii. If all have equal
volume, which one will have least mass?
Z will have least mass as density
is directly proportional to mass.
ii.
Among the substances, which one will float on water?
Z will float on water because its
density is less than the density of water.
d. In the diagram, the weight of a stone
in air and in water are shown, where the weight of the beaker is 1N. Study the
diagram and answer the following questions.
i. Find upthrust due to water and weight
of the liquid displaced.
Upthrust due to water = 10 N – 7 N = 3N
Weight of liquid displaced =upthrust = 3N
ii. Write the conclusion of the
experiment.
This
experiment proves that the weight of liquid displaced is equal to the upthrust.
iii. State the principle can be verified
by the experiment.
The principle verified by the experiment
states that, " When a body is partially or wholly immersed in a liquid, it
experiences an upthrust equal to the weight of the liquid displaced by it."
e. Study the diagram and
answer the following questions.
i. What is shown in the
diagram?
A water pump is shown in the
diagram
ii. Label the parts A and E.
The name of part A is valve and
E is cylinder.
iii. Is it downstroke or
upstroke? How do you identify the condition?
It is upstroke because water
is coming out from the spout.
iv. Which part is the spout?
Part D is the spout.
v. Which part blocks the water of the pump from going back into the source of water?
Part A blocks the water of
the pump from going back into the source of water.
F. Numericals.
a. Calculate the pressure exerted by a mercury column of 76 cm high at its bottom. Given that the density of mercury is 13600 kg/m³ and g= 9.8 m/s².
b. The
depth of water in a rectangular tank is 6 m. Find pressure exerted by water at
the bottom of it (g= 10 m/s², density of water= 1000 kg m⁻³).
Given
Depth of
rectangular tank (h) = 6m
Acc due to
gravity (g) = 10 ms⁻²
Density of
Water (d) = 1000 Kg m⁻³
Pressure (P)
=?
We know
Pressure =
dgh
= 1000 x 10 x
6
= 60000
Pascal
= 6 x 10⁴ Pascal
Hence
pressure exerted by water is 6 x 10⁴ Pascal.
c. The depth
of a circular well is 5 m. Water level is below 2 m from the upper face of the
wall. Find the pressure at its base? (g = 9.8 m/s²)
Given,
Depth of circular well = 5m
Depth of circular well having Water in it (h) = 5m – 2m =
3m
Acc due to gravity (g) = 9.8 ms⁻²
Density of
Water (d) = 1000 Kg m⁻³
We Know,
Pressure = dgh
=
1000 x 9.8 x 3
=
29400 Pa
=
2.94 x 10⁴ Pa
Hence Pressure is 2.94 x 10⁴ Pa.
d. The
weight of a piece of stone when fully immersed in water is 18 N and it
displaces 4 N of water. What is the weight of the stone in air?
Given,
Weight
of Stone in Water (Ww) = 18 N
Weight
of Water displaced = 4 N
Weight
of Stone in Air (Wa) =?
We Know,
Upthrust
(U) = Weight of Water displaced
= 4 N
Also,
Upthrust
= Wa - Ww
Or,
4 = Wa – 18
Or,
Wa = 18 + 4
= 22N
Hence
Weight of Stone in air is 22N.
e. Calculate
the pressure exerted by a mercury column of height 75 cm at its bottom. Given
that the density of mercury is 13600 kg/m3 and g= 10 m/s².
Same as que no
a
f. A cube of
wood of volume of 0.2 m³ and density 600 kg/ m3 is placed in a
liquid of density 800 kg/m3. What fraction of the volume of the wood
be immersed in the liquid?
Given
Volume of Wood (V₁) = 0.2 m³
Density of Wood (d₁) = 600 Kg/m³
Fraction of wood in liquid (V₂/V₁) =?
Density of liquid (d₂) = 800 Kg/m³
We Know
According to Law of Floatation,
W₁ = W₂
Or, M₁ x g = M₂ x g
Or, d₁ x V₁ = d₂ x V₂ (Density = M/V)
Or, 600 x V₁ = 800 x V₂
Or,
800
Or,
600 = V₂
800 V₁
V₁ 4
g. A block
of wood of mass 24 kg floats on water. The volume of the wood is 0.032 m³. Find
the volume of the block below the surface of the water and the density of the
wood. (density of water- 1000 kg/m³).
Given
Mass of Wood (M₁) = 24 Kg
Volume of Wood (V₁) = 0.032 m³
Density of Wood (d₁) =?
Density of water (d₂) = 1000 Kg/m³
We Know,
According to Law of Flotation,
W₁ = W₂
Or, M₁ x g = M₂ x g
Or, M₁ = d₂ x V₂
Or, 24 = 1000 x V₂
Or,
24 = V₂
1000
Or, V₂ = 0.024 m³
Also,
Density of Wood =
= 24
0.032
= 750 Kg/m³
Hence the
volume of block under water is 0.024 m³ and density of wood is 750 Kg/m³.
h. A 15 cm long cube made of oak floats in water with 10.5 cm of its depth below the surface and with its sides vertical. What is the density of the cube? (density of water = 1000 kg/m³)
Given,
In cubical
oak
Length (l) =
15cm
Length of
oak inside water (lin) = 10.5 cm
Volume of
oak inside water(V1) =(15 x 15 x 10.5) cm³
= 2362.5 cm³
Density of
water (d₁) =1000 kg/m³= 1 g/cm³
Volume of oak (V2) = (15
cm)³ = 3375 cm³
Density of
oak (d₂) =?
We
know that,
Volume of
water displaced (V1)= mass of water displaced
Density of water (d₁)
Or, 2362.5 = mass of water displaced
1
Or, mass of water
displaced = 2362.5 gm
Or, Mass of oak
= 2362.5 gm
Density
of oak (d₂) = mass of oak
Volume of oak (V2)
= 2362.5
3375
= 0.7 gm/cm3
= 700 kg/m3
Hence
density of oak is 700 kg/m3.
i. Calculate
the mass of displaced water when a piece of 30 cm thick iceberg with surface
area 1000 cm2 floats on water (density of ice 0.9 g/cm³ and density
of water = 1 gm/cm³).
Here,
Surface
area of iceberg (A) =1000 cm2
Thickness
of iceberg (t) = 30 cm
Density
of ice (d1) = 0.9 gm/cm³
Density
of water (d2) = 1 gm/cm³
Now,
Volume
of iceberg (V) = A x t
= 1000 cm2
x 30 cm
= 30000
cm³
Mass
of iceberg (m) = d1 × V
= 0.9 × 30000
=27000 gm
=27 kg
Mass
of water displaced by iceberg = mass of iceberg =27 kg
Hence,
the mass of water displaced is 27 kg.
j.
k. A piece
of stone with its volume 400 cm³ and density 7.8 x 10³ kg /m³ is immersed
totally in water of density 1000 kg/m³. Calculate the weight of the stone in
air and the upthrust of water.
Volume of stone (V1) = 400 cm³
= 400
1000000
= 4 x 10⁻⁴ m³
Density
of stone (d1) = 7.8 x 10³ kg /m³
Density
of water (d2) = 1000 kg/m³
Mass
of stone in air (m₁) = V1 x d1
= 4 x 10⁻⁴ x 7.8 x 10³
= 31.2 x 10⁻⁴⁺³
= 31.2 x 10⁻¹
= 3.12 kg
Acceleration
due to gravity (g) = 10 ms⁻²
We
know that,
Weight
of stone in air (W) = m₁g
=
3.12 x 10
= 31.2 N
Also,
Upthrust
of water (U) = V1 x d2 x g
= 4 x 10⁻⁴ x 1000 x 10
= 4 x 10⁻⁴ x 10⁴
= 4 N
Hence the weight
of stone in air is 31.2N and upthrust of water is 4 N.
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These all answers are very helpful for the students.
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