Ans: Force acting
perpendicularly per unit area is called pressure. Its SI unit is pascal (Pa)
i.e. Nm^{2}
2) Name two factors that
affect pressure.
Ans: The two factors
affecting pressure are;
·
Force
·
Area
Mathematically; Pressure (p) =
3) What relation does
exist between pressure and area?
Ans: Pressure is
inversely proportional to Area (A)
i.e. P µ
4) It is easier to cut
the things by using a sharp knife than by using a blunt one. Why?
Ans: The edge of sharp
knife has less surface area than that of blunt one. hence, it exerts more
pressure on the thing to be cut. Since, pressure is inversely
Proportional to Area . Therefore, it is easier
to cut things by suing a sharp knife
than by using a blunt one.
5) The shoes of football
players are provided with studs. Why?
Ans: Studs reduce the
surface area of contact between shoes and ground. Hence, more pressure is
applied to the ground and players can run safely without slipping on the
ground. Since, pressure µ
6) The foundation of the
building is made wider than the walls. Why?
Ans: The foundation of
the building is made wider than the walls of a building because walls and its
roof exert pressure on the wide area of the foundation and this reduces
pressure over the earths surface and the building is saved from collapsing.
Since, pressure µ
7) Camel can walk on the
desert but not the horse. Why?
Ans: Camels are provided
with flat hoofs. These flat hoofs exerts less pressure on the sand because
pressure µ
. But horse has narrow hoofs
and exerts more pressure on the sand and sinks there. Thus camel can walk on
the desert but not the horse.
8) Tyres of tractors are
wider than the tyres of cars. Why?
Ans: Tractors are used
in muddy areas. They must apply less pressure on the ground in order to run
there. Hence, to decrease the pressure they apply, tyres of tractors are made
wider. However, cars are often used in blacktopped road and that road can
withhold more pressure. Hence, tyres of tractors are wider than that of cars.
9) The backtyres of
trucks are paired. Why?
Ans: Pairing of tyres
increases the area of contact of tyres with the ground and less pressure is
exerted on the ground and to the tyres as well. This increases the life of
tyres and helps the truck to run swiftly on the road without getting collapsed.
Therefore, the backytres of trucks are paired.
10) Write any two
differences between thrust and pressure.
Ans: The differences
between thrust and pressure are as follows;
Thrust (Force) 
Pressure 
a) It is an external
agency that changes or tends to change the state of a matter from motion the
rest and from rest motion. 
It is the force
acting perpendicularly per unit area. 
b) Its SI unit is
Newton (N) 
Its SI unit is
pascal (pa) 
c) Mathematically Force (F)= mass (m)
X acceleration (a) 
Mathematically, Pressure (p)= 
11) Mention three factor on
which liquid pressure depends.
Ans: Liquid pressure
depends on the following factors;
i)
Depth of liquid from the free surface (h)
ii)
Density of liquid (d)
iii)
Acceleration due to gravity (g)
12) The base of dam is mode
wider than the top. Why?
Ans: We know that the
liquid pressure increases with the depth of the liquid. Hence, water in dam
exerts more pressure at the bottom than at the top. To overcome that pressure,
the base of dam is made wider.
13) Deep sea divers need
swim suit. Why?
Ans: Since, liquid
pressure is directly proportional to the depth of the liquid. Sea divers reach
to the deep part of the sea where the liquid pressure is excessively high.
Hence to overcome that pressure, deep sea divers need swim suit.
14) Two identical tanks are
filled with water and mercury respectively. On which tank more pressure is
experienced at the bottom?
Ans: The tank filled
with mercury will experience more pressure at its base.
Since, liquid pressure
directly proportional to the density of the liquid at constant depth and
acceleration due to gracility. Density of mercury is more than that of the
water. Hence tank filled with mercury will experience more pressure at its
base.
15) The bucket at the
bottom floor fills faster than at the top floor from the identical taps. Why?
Ans: The depth of water
column to the bottom floor is more than the depth to the top floor. Hence,
water from the tap of bottom floor comes with greater pressure to fill the bucket
fester than at the top floor. As liquid pressure is directly proportional to
the depth of the liquid i.e. P µ h at constant density and acceleration due to gravity.
16) The blood pressure in
human body is grater at feet than at the brain. Why?
Ans: Since, the depth
of feet from the heart is more than the depth of the brain. As liquid pressure
is directly proportional to the depth of the liquid column, blood exerts more
pressure at feet.
17) A tank filled with
water is brought from himalyan region to Terai region. What change in pressure
is seen at the bottom of the tank? Why?
Ans: The pressure
exerted at the bottom of the tanks will be more in Terai region. The value of
acceleration due to gravity is more in the terai region than that in the himalyan
region. Since liquid pressure is directly proportional to the acceleration due
to gravity at constant acceleration due to gravity deth of liquid so the
pressure exerted by water will be more in Terai region.
18) State Pascal's law.
Ans: Pascal's law states
that 'the external pressure applied on a liquid contained in a closed vessel is
transmitted equally and perpendicularly in all the directions"
19) Name any two device
constructed on the basis of Pascal's law.
Ans: The following
devices are constructed on the basis of Pascal's law;
 Hydraulic press
 Hydraulic crane
 Hydraulic brake
20) On which property of
the liquid Pascal's law is based?
Ans: Pascal's law is
based on the property of the liquid that the "liquid can not be
compressed"
21) What is up thrust?
Write its SI unit?
Ans: The upward force
exerted by the fluid on an object partially or wholly sinking it in it is
called up thrust.
Its SI unit is
Newton (N)
[up thrust is
also called buoyant force]
22) State Archimedes
principle.
Ans: According to
Archimedes principle, "when a body is partially or wholly immersed in a
liquid it experiences an up thrust which is equal to the weight of liquid
displaced by the body"
23) The weight of a body is
less in water than in air. Why?
Ans: Upthrust exerted by
the fluid decreases the weight of an object immersed in it. Because of its
higher density, the upthrust exerted by the air on an object. Hence the weight
of a body is less in water in air.
24) It is easier to swim in
the sea than in the river. Why?
Ans: it is easier to
swim in the sea than in the river because the density of sea water is more than
the density of river water. Hence the upthrust exerted by sea water on an
object is more than the upthrust exerted by the river water.
25) On which factor the
upthrust depends?
Ans: The upthrust
depends upon the following factros;
i)
Volume of the liquid displaced (v)
ii)
Density of liquid (d)
iii)
Acceleration due to gravity (g)
26) A ship is so
constructed that its external dimension is similar to that of an aeroplane will
it fly in air? Why?
Ans: No, it won't fly in
the air.
Because of less
density of air the upthrust exerted by the air is much less than the upthrust
exerted by water. This upthrust of air will not be sufficient to lift up the
ship in the air.
27) How does the plane fly?
Ans: Due to the convex
wings of plan, the air pressure above the wings is less and below the wing is
more. Thus, this results the plane to fly up in the air as it is lifted up.
28) The upthrust exerted by
mercury is more than the upthrust exerted by water why?
Ans: The density of
mercury is more than that of water. Since upthrust is directly proportional to
density i.e. U µ
d. Hence the upthrust exerted by the mercury is more than that of water.
29) What is density? Write
its SI unit?
Ans: Density is defined
as mass per unit volume.
Its SI unit is
kgm^{3}
30) An egg sinks in pure
water but floats on the concentrated salt solution. Why?
Ans: An egg sinks in the
pure water because its' density is greater than that of pure water. However, on
egg floats on the salt solution because its density is less than that of salt
solution.
31) An iron nail sinks in
water but floats on mercury. Why?
Ans: An iron nail sinks
in water because its density is greater than that of water. However an iron
nail floats on the mercury because its density is less than that of mercury.
32) An ice cube flats on
water. Why?
Ans: An ice cube flats
on water because its density is less than that of water and it can displace the
water equal to its weight.
33) What is relative
density? Why it doesn't have any unit?
Ans: Relative density is
defined as the ratio of density of substance to the density of water at 4°c. It
doesn't have any unit, because it is the simple ration of the same physical quantities i.e.
34) The relative density of
mercury is 13.6. What do you mean by it?
Ans: It means that
mercury is 13.6 times denser tan water at 4°c i.e. for the same volume. Weight
of mercury is 13.6 times more than the weight of water.
35) What is specific
gravity? Why doesn't it contain any unit?
Ans: Specific gravity is defined as the ratio of mass of
certain volume of substance to the mass of same volume of water at 4°c.
It doesn't
contain any unit because it is a simple ratio of same physical quantities.
36) At which condition does
a body float in water?
Ans: A body floats in
water in the following two conditions;
·
If the density of the body is less than the
density of water
·
If the body can displace the water equal to its
weight when place in water.
37) State law of
floatation.
Ans: According to the
law of floatation "A body floats in liquid if it can displace the liquid
equal to its weight when place in that liquid"
38) An iron nail sinks in
water but the ship made by iron floats in it. Why?
Ans: An iron nail can't
displace the water equal to its weight, hence gets sink in water. However, the
ship is so constructed that it can displace the water equal to its weight and
can flat in water.
39) Water change in
floatation is seen when a cargo loaded ship travel from sea to the river?
Ans: The density of
river water is less than the density of sea water. Hence more volume of water
is to be displaced on he river compared to the sea in order to maintain the
weight of the water displaced equal to the weight of the ship. So, it sinks
more in river than in sea.
40) What is the difference
in floating of empty ship and the cargoloaded ship?
Ans: In order to float
in water ship must displace the water equal to its weight. Since the weight of
empty ship is less than the weight of cargo loaded ship empty ship has to
displace less than the weight of cargo loaded ship empty ship has to displace
less amount of water and the cargo loaded ship has to displace more amount of
water. hence, cargo loaded ship sinks more in water than the empty ship.
41) What is hydrometer? On
which principle is it based?
Ans: Hydrometer is
device that is used to measure the specific gravity and the density of a
liquid.
Its is based on
the law of floatation.
42) The base of hydrometer
is made heavier. Why?
Ans: The base of
hydrometer is made heavier in order to float vertically in liquid and to sink
partially .
43) The neck region of
hydrometer is mode narrower. Why?
Ans: The neck region of hydrometer is made narrower for its
grater sensitivity.
44) What is lactometer?
Ans: Lactometer is a
device which is use to measure the purity of milk.
Q) An iron nail sinks in
water but the ship made by iron floats in water. Why?
Ans: According to the
law of flotation object floats in liquid if it can displace the liquid equal to
its weight. At iron nail can't displace the water equal to its weight. Hence it
gets sink in water however the ship is so constructed that it can displace the
water equal to its weight and can float in water.
Prove that: P=hdg
Solution:
Let
us consider, liquid of density (d) is kept in a vessel. Let h be the depth of
the liquid column from the free surface.
Since,
Pressure
(p) =
or, P = [\F=m. g: where
m=mass of the body
g= acceleration of body]
or, P =
[\d = ; where
v=volume
or, p
= [ \V=A×h A® Area
h=
height ]
or, p =
dhg. Proved.
Study the given figure and answer the following questions.
1) Which tank will
experience more pressure at the base. Why?
Ans: Tank a will experience more pressure at the
base. Because, the depth of liquid column is more in tank A than in tank B. We
know that the liquid pressure is directly proportional to the depth of liquid
column and is independent to the volume of liquid [\P=hdg]
Q. Study
the given figure and answer the following questions.
i) On what
surface of the body, the pressure is maximum?
 The pressure
is maximum on face CD. Because this face lies at more depth from the free
surface. Since liquid pressure µ depth of liquid from the free surface.
ii) What will
be the resultant of pressure on AD & BC?
·
The resultant of pressure on AD and BC is zero.
As the pressure in these two faces acts equal and in opposite direction.
iii)
What is the effect of pressure difference between two faces AB and CD?
·
The difference in pressure in these two faces
generates up thrust.
Study the given figure and answer the following questions.
i) Which
device is shown in the figure?

"Hydraulic press" is shown in the figure.
ii) On which
principle is it based? State it
·
It is based on Pascal's law which states that
"the external pressure applied on a liquid contained in a closed vessel is
transmitted equally and perpendicularly in all the directions"
iii) What is the use of this
device?
·
This device is used to lift the heavy equipments
(load) by applying small force.
iv) What is the role of the liquid
in the given figure?
·
The role of the liquid in the hydraulic press is
to transmit the pressure from small piston to the large piston.
v) Why the force applied on the
small piston?
 Since pressure is inversely
proportional to the area. Hence, the applied force generates high pressure on
small piston.
Prove that,
U=vdg, where
symbols carry their usual meaning
where;
u
= upthrust
d
= density of liquid
v
= volume of liquid displaced
g
= acceleration due to gravity
Solution;
Let us consider a cylindrical object
is immersed in a liquid of density 'd' contained in a vessel let 'h_{1}'
be the depth of the upper end of cylinder from the free surface and 'h_{2}'
be the depth of the lower end of the body from free surface of the liquid as
shown in the above figure.
Here,
The liquid pressure at upper
surface,
P_{1} = h_{1}.d.g.
...................eqn (i)
[where, g ® acceleration due to
gravity'
The Force at upper
end;
F_{1} = P_{1} × A_{1}
.....................eqn (ii)
[where A_{1} ®Area
of upper surface]
Now,
From eqn (i) and (ii)
F_{1}
= h_{1} dg A_{1} .................eqn (iii)
Similarly, Force on
lower end,
F_{2} = h_{ dg }A_{2}
.................eqn (iv)
Since, Upthrust (U) = F_{2  }F_{1}
= h_{2}dg A_{2}  h_{1
}dg A
= dg (h_{2}A_{2}  h_{1}
A_{1})
Since the object is symmetrical, A_{1}
= A_{2}
U
= dg (h_{2} A  h_{1}
A)
= dgA (h_{2}  h_{1})
But,
h_{2}
 h_{1} = h; which is the height of the object.
or, U = vdg [where
v = A × h is the volume of substance]
Q. Write any three
properties of liquid pressure?
i) The pressure given in liquid is distributed
equally in all the directions.
ii) A liquid
maintain its own level.
iii) The
pressure of liquid is independent of the shape of the container in which it is
kept .
Q. The weight of the
stone is measured in three different media A,B and C study the given table and
answer the following questions.
Medium 
Weight 
A 
20 N 
B 
12 N 
C 
16 N 
i) Which one is air, water and salt solution among A,B and c Why?
 From the table.
Medium A ¾¾® Air
Medium B ¾¾® Salt solution
Medium C ¾¾® Water
Out of the given medium air, water and alt solution the
density of air is minimum and it exerts minimum upthrust. As a result, weight
of a body should be minimum in the air. In the given table weight of a body is
maximum in medium A Hence, "A" must be the air.
Again the upthrust exerted by the salt solution is maximum
and the weight of a body should be minimum in the salt solution. In the given
table, weight of the body is minimum in medium B. Hence "B" must be
the salt solution.
And the rest medium i.e. "C" is the water
ii) What will be water displaced?
Solution,
Weight in air (w_{1}) = 20 N
Weight in water (w_{2}) 16 N
Now,
We know,
Upthrust
= w_{1} w_{2}
=
20 N  16 N
=
4 N
=
weight of water displace.
Q. What
is error in given figure regarding the
level of liquid. Why?
 The level of liquid in column B is less than in column A and
column C. Which is the error in the given figure. Because liquid maintains its
own level.
Q. What are fluids?
 Any thing
that can flow or blow are called fluids.
Q. The upthrust of the
stone is measured in three different media; A, B and C. study the given table
and answer the following question.
Medium 
Upthrust 
A 
40 N 
B 
60 N 
C 
30 N 
i) Which one is air
water and salt solution among the given media? Why?
 From the
table
Medium
B = Salt solution
Medium
C = Air
Medium
A = water
Out of the given medium Air, water and retribution, the
upthrust exerted by salt solution is Maximum. Hence according to the table B is
salt solution. The upthrust exerted by air is minimum. Hence C is air and the
remaining A is water in the given table.
Practice Question
Q) What is hydraulic
garage lift ? On which principle it is bared? State the principle.
Q) Why doesn't the
pressure of atmosphere break windows?
Q) Why does a man
weigh lighter in air than in vacuum?
Q) It is easier to
pull a bucket from well until it is a water. Why?
Q) Why a balloon
filled with hydrogen?
Q) Can a substance
heavier than water float in it? How?
Q) Will a hydrogen
filled balloon raise in moon? Why?
Q) The ships can
carry more cargo in sea water than in river water. Why?
Q) Wooden sleepers
are used below the rails Why?
Q) What is the cause of
upthrust?
Ans: When a substance is
partially or wholly immersed in a liquid the upper and lower surfaces feel
different pressures. This difference in pressure causes a resultant upward
force which is called upthrust.
P_{2 }>_{ } P_{1}; as h_{2} > h_{1}
Q) Why are passengers
advised to remove ink from their pens while going up in the aeroplane?
Ans: The atmospheric
pressure goes on decreasing with increase in altitude. So, the ink level which
was previously at a compressed state will rise up. This may cause overflow of
the inks. So, in order to avoid it, passengers are advised to empty the ink
before lying.
Q) What is the apparent
weight of a floating body? Explain
Ans: The apparent weight
of a floating body is zero Newton.
Since, W
apparent = W true  upthrust
For a floating
body W true = upthrust
\ W
apparent = O Thee
force body will be weightless.
Q) An iron ball sinks in
water but floats in mercury. Why?
Ans: Iron is highly
dense compared to water. When an iron ball is placed in water it can displace
very less weight of water. So, it sinks in water. But iron is less dense compared
to mercury. When placed in mercury the weight of mercury displaced will be very
high and will exceed the weight of iron ball. So iron will flat on mercury.
Q) A fish can raise and
descend in water. how?
Ans: According to law of
floatation, A body floats in liquid if it can displace the liquid to its
weight. When a fish stores more air in it, it can cover more volume and can
displace more water than its own weight and raises u. But when it releases air
the displaced wt. of water will less than the weight of fish themselves and
starts to sink.
Q) A body is thrown into a
deep pond. As it sinks deeper and deeper into water does the upthrust (buoyant
force) increase or decrease?
Ans: Since, upthrust (U)
= Ndg
v® vol.
of liq. displaced
d ®
density ; g ®
acceleration due to gravity.
Whether a body
sinks less or more, the volume of liquid displaced the density and the
acceleration due to gravity are same. Hence, upthrust exerted on body is same.
Pressure Numerical
Q) A
sold object weighs 24N in air and 10 N in water what is the upthrust of the
liquid on the object? What is the upthrust and weight of the liquid displaced?
Q) If pistons A, B
and C of the apparatus
given in the diagram are supposed to
be frictionless, what i the area of
piton B? What force is
exerted in piston C? [Ans: ]
Q) What
is the volume of water
displaced when
a wooden box
of length 50cm,
breadth 40 cm
and thickness
30 cm is immersed in
water? [density
of water
= 1 gm/cm^{3}
and density of wooden box= 0.56 gm/cm^{3}
1) A model weighing
54 kg wears high heels with surface area 20 cm^{2} each. Calculate the
pressure exerted by her on the ground.
Sol,
Weight of model
(w) = mg
=
54×10
=
540 N
Surface area of a heel (a) = 20 cm^{2}
=
20×cm×cm
=
20× m × m
=
2×10^{3} m^{2}
\Total
surface area (A) = 2×2×10^{3} m^{2}
Pressure
exerted (p) = ?
we know that,
P =
=
= 135000N
\The
girl exerts the pressure of 135000 N.
2) Calculate the
pressure at the bottom of the tank of height 1m 75% of which is filled with the
liquid of density 100 kg/m^{2}
Sol,
Height of tank
(H) = 1 m
Now,
depth of
liquid column from free surface is;
h = 75% of h
= ×1 = 0.75 m.
density of
fluid (S) = 1000 kg/m^{3}
acceleration
due to gravity (g) = 10m/s^{2}
Pressure (P) =
?
we have
P = sgh
= 1000×10×0.75
= 7500 Pa.
\Pressure
at the bottom = 7500 Pa.
3) If
cross sectional area of piston a
A is 20 cm^{2}
and effort of 50 N is
applied on A, how much load is needed
on pistonB to balance it?
Sol,
For Piston A For Piston B
Area (A_{1}) = 20 cm^{2}
=2×10^{3}m^{2} Area
(A_{2}) = 60cm^{2} = 6×10^{3}m^{2}
Force (F_{1}) = 50 N Force (F_{2}) =
?
Let
pressure at A be P_{1} and that in B be P_{2}.
Since
pressure is equal at both pistons.
P_{1}=P_{2}
=
=
or =
=
F_{2}
= 25×6 = 150N
\A load of 150 N is req. to
balance the piston.
4) Calculate
the mass of water displaced when a piece of ice having 40 cm length, 20 cm
breadth and 15 cm height is kept or water. The density of ice and water is 0.92
gm/cm^{3} and 1 gm/cm^{3} respectively.
Sol,
(v_{1})
Volume of ice =
l×b×h
=40×20×15
=
12000 cm^{3}
density of ice (S_{1}) = 0.92 gm/cm^{3}
density of water (S_{2}) = 1 gm/cm^{3}
volume of water displaced (v_{2}) = ? let m_{1}, and m_{2}
be masses of ice
and water respectively
Acc. to law of
floatation,
Weight of ice
= weight of water displaced
or, M_{1} × S_{1}
= M_{2}
=
12000×0.92
=
11040 gm
\Mass
of water displaced = 11040 gm,
5) Study
the given figure and answer the
questions.
i) Which
principle is explained by
the given figure.
 It explains
the Archimedes's principle.
ii) What is the
upthrust acting on the body. Why?
 Teh upthrust
acting on the body is 2 N. Because according to Archimedes uptrhust = wt. of
liquid displaced.
iii) What will
be the wt. of object in air here
wt
of object in water = 10 N
wt
of liquid displaced = 2 N
liquid
upthrust = 2 N
Hence, wt in
air = 10N+2N=12N
iv) Calculate
the mass of body in air
Since, w = mg
or 12
= M×10 = m=12 kg
[Mass is measured in air]
6) An ice block
(60cm×40cm×30cm) is floating on water. What portion of ice should be outside
the water is the density of ice is 0.75 gm/cm^{3}
Sol,
Given
Volume of ice
block = (60×40×30)cm^{3}
=
72000 cm^{3}
density of ice
(S_{1}) = 0.75 gm/cm^{3}
density of
water (S_{2}) = 1 gm/cm^{3}
vol. of water
displaced (v_{2}) = ?
Acc. to law of
floatation,
wt. of ice =
wt. of water displaced
or, m_{ice}
× y = m _{water} ×y
or, v_{1}×s_{1}
= v^{2} ×s_{2}
or, = v^{2}
r_{2} =
54000 cm^{3}
Since, vol. of
water displaced is equal to vol. g ice inside water, vol. of ice outside water,
= (72000 
54000) cm^{3}
= 18000 cm^{3}
\
Portion of ice above water =
= past.
7) The ratio of
areas of piston A and B shown in the following diagram is 1:20. Find the load
that can be lifted at the piston B by applying 100 N effort at the piston A.
Sol,
Let A_{1}
and A_{2} be the cross
sectional areas
of the piston
A and B
respectively.
Given,
=
Effort applied
at piston A (F_{1}) = 100N
Load lifted at
piston B (F_{2}) = ?
Using
Pascal's law,
=
or, =
\ F_{2}
= 1000 N
\ 1000
N load can be lifted at piston B by applying 100 N force at piston A.
8) Find the pressure
exerted by the following box. The density of the box is 3000 kg/m^{2}
and g=9.8m/s^{2}
Given,
Density (S) = 3000 kg/m^{2}
Volume (V) = 2m×1.5m×2m
= 6m^{3}
Area of the
bottom (A) = 2 m×1.5m = 3m^{2}
we have,
Mass (M) = Vol. ×density
= 6×300 kg
Again,
Force (F) = mg
= 18000×9.8
= 176400 N
Now,
Pressure (P) = =
Thus, the pressure exerted by the body is 326.67 Pa.
9) The area of the
edge of a knife is 0.3cm^{2}. 75 N force is applied on the head of
knife find the pressure exerted by the knife on the material to be cut.
Given,
Force (F) = 75N
Area (A) =
0.025 cm^{2} = 0.03 m^{2} = 0.00003 m^{2}
1000×100
Now,
Pressure (P)
= = Pa = 2.5×10^{0} Pa.
Thus, pressure exerted by the knife is 2.5×10^{7} Pa.
Pressure
Practice Questions.
Q) A cube of side 0.5
m and density 8000 kg/m^{3} is placed on a table. Find the pressure
exerted by the cube on the table. [g=9.8m/s^{2}] (Ans.
19600 Pa)
Q) A man exert 32666
Pa pressure on the ground. If his feet occupy the area of 150 cm^{2}
find the mass of the man if g=9.8m/s^{2} (Ans: 49.99 kg.)
Q) A boat of mass 200
kg displaces maximum of 3m^{3} water when it is floating. Find the
maximum mass of load the boat can carry safety. the density of water is 1000
kg/m^{3}
(Ans: 2800 kg)
Q) What will be the
pressure exerted at a depth of 6m in a pond where the acceleration due to
gravity is 9.8m/s^{2} (Density of water = 1000 kg/m^{3})
Q) The weight of an
object is 20N in air, the weight of that object in the water is seen 12 N only
then;
i) What is the
value of upthrust on the body?
ii) What is the
weight of the displaced water by it?
Q) There are 2
pistons A and B. The area of A is 100cm^{2} and that of B is 0.9 m^{2}.
If the force of 101.3N is applied at piston A calculate the force at piston B.
Pressure
Solution (Numerical)
Q) A block of wood
floats on a liquid with fourfifth of its volume submerged. If the density of
the wood is 800 kg/m^{3} find the density of the liquid.
Sol,
Let V be the
total volume of the block and let 4/5 V be the volume under the liquid.
\ Vol.
of liquid displaced = 4/5 V
Here,
Fraction of block
inside the liquid (v^{2}/v_{2}) = 4/5
Density of wood (S^{1}) = 800 kg/m^{3}
^{ }Density
of liquid (S_{2}) = ?
Acc. to law of
floatation,
wt of block =
wt. of liquid displaced
m_{1} g
= m_{2} g
or, V_{1} ×S_{1}
= V_{2} × S_{2}
or, =
\ S_{2}
= 1000 kg/m^{3}
\
Density of liquid = 1000 kg/m^{3}.
Q) A balloon filled with air can withstand with maximum of 6.174×10^{5} Pa. Pressure. At what depth of water of density 1000 kg/m^{3} should it be taken so that the balloon just burst? g= 9.8 m^{2}
Sol,
Given,
Max. pressure
(P_{max}) = 6.174×10^{5} Pa
Max. depth (H_{max})
= ?
Density of water (d) = 1000 kg/ m^{3}
^{ }We
know that,
P = dgh
or, 6.174×10^{5}
= 1000×9.8×h
\ h =
=
63m
\ It
can be taken to maximum depth of 63 m where it just brusts.
Q) The density of a
brick is 2.5 gm/cc and its mass 1 kg. Find the upthrust of water on the brick
the density of water is 1 gm/cc.
Given,
Density of brick (d_{1}) = 2.5 gm/cc = 2500 kg/ m^{3}
Density of
water (d_{2}) = 1 gm/cc = 1000 kg/m^{3}
Mass of brick
(m) = 1 kg
Upthrust (u) =
?
Now,
Volume of the brick (v) = = m^{3 }= 0.004 m^{3}
Since,
density of brick is more than the
density of water it sinks in water.
for sinking case,
U = vdg
= 0.0004×1000×9.8
= 3.92 N
\ Upthrust of water on the
brick =
3.92 N.
Q) Egg sinks in pure
water but floats in salty
water why?
 The density
of salty water is more
than that of fresh water. As a result
it provides more upthrust to the egg
and hence, egg floats in salty water
but
sins in fresh water.
Pressure: Theory practice
Q) In
the given diagram three taps A,B and
C having holes of the same diameter are
connected in tank full of water. In
which
tap does more water flow per second
and
why?
Q) It is difficult to
immerse empty plastic bottle into water. Why?
Q) What will be the
atmospheric pressure inside vacuum?
Q) A beiger can sleep
on the bed of nails. Why?
Q) What will happen
is the base of a dam is made narrower than the tap?
Q) Why does a ship
carry more load than an aeroplane?
Q) A stone tied with
a string is completely immersed in a liquid. Draw a diagram to illustrate the
forces acting on the stone.
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