EXERCISE
7.1
i.
ii.
iii.
iv.
v.
vi.
vii.
viii.
ix.
x.
A
B
B
B
A
C
B
C
D
A
7.2
According to kinetic molecular theory the intermolecular distance is
smallest in solids and it is largest in case of gases. And in case of liquid,
the intermolecular distance is in-between solid and gases.
This helps to differentiate between solids, liquids and gases.
7.3
The fourth state of matter is called plasma which is collection of positive
ions at very light temperature.
7.4
WHOLE TOPIC OF DENSITY
7.5
Hydrometer is instrument used to measure density of any liquid.
7.6
DEFINITION, FORMULA AND UNIT OF PRESSURE FROM BOOK(7.3)
7.7
In atmosphere of earth, there are gases water vapors and dust particles.
All of these consist of material particle. Due to force of gravity acting on
these particles the object inside the atmosphere experience pressure in
all around equally.
7.8
By applying pressure, air can be entered into balloon. Due to atmosphere
pressure it is difficult to remove air from a glass bottle.
Because air tries to maintain its pressure equally energy where.
7.9
Barometer is a device to measure the atmospheric pressure of air.
7.10
Water has smaller density. Therefore, the height of water column in the
tube of barometer would be very large about 10m which becomes
inconvenient. Therefore, use of water in the construction of barometer
is not suitable.
7.11
When sucker is pressed on a smooth surface, then air pressure below it
becomes very small (due to air displaced), as compared with air pressure
above it. This help to stick it with smooth surface.
7.12
At very high positions in the atmosphere, the density of air becomes
small. That is why pressure of air becomes small.
7.13
When press of atmosphere falls suddenly then molecular distance (of air)
increases. And work in done against the force of gravitational attraction
between them. It is done by using internal energy air. It is done by using
internal energy of air. Therefore, internal energy of air decreases and
coldness is produced.
7.14
When barometer reading increases suddenly then due to increase of
pressure of air, the molecules are brought closer to one another and
their energy increases. So that, there is rise of temperature of
atmosphere
7.15
STATEMENT FROM PASCAL’S LAW.
7.16
IN APPLICATIONS OF PASCAL’S LAW ON PAGE 153
7.17
The property of matter to retains their shape is called elasticity.
7.18
WHOLE TOPIC OF ARCHIMEDE’S PRINCIPAL(7.6)
7.19
The fluids (liquids) exert force in the upward direction when some object
is immersed into them. This is called upward thrust. If weight of fluid
displaced is equal to weight floating object inside liquid. Then object will
not sink and keep floating. It is called. It is called principle of floating.
7.20
If the submarine is not filled with sea water. Then its weight is less than
upward thrust. So that it floates on surface of sea water. But when it is
filled with water. Then its weight becomes larger as compared with
upward thrust done to water then it sinks into water.
7.21
When piece of stone is placed in the water, then upward thrust acting on
it due to water is much smaller because the weight of water displaced is
less than the weight of piece of stone. Whereas the shape of ship is made
such that the weight of water displaced is much greater than weight of
ship. So that upward thrust because large, as compared with weight of
the ship and it floats over water surface.
7.22
WHOLE TOPIC OF HOOKE’S LAW.
7.23
Take a rubber band and hang it with a hook. Then pointer is attached at
the lower end of it with a scale in front of pointer. Different known
weights are suspended, one by one, at lower end of this rubber band.
The pointer position on the scale are marked for each different known
weight suspended. It is called calibration of scale for weight
measurements. This makes a balance for weight measurement.
NUMERICALS
7.1
Data
V=volume = (40 X 10 X 5) cm3 = 2000 cm3 = 0.002 m3
m=mass = 850 g = 0.850 kg
density = p = ?
Formula
Density=m/V
p = 425 kg m-3
7.2
Data
volume of ice =?’
volume of water = 1 litter
Formula
Vol of ice/vol of water = density of water/density of ice
Vol of ice=( density of water/density of ice) x vol of water
Vol of ice=(1000/920)x1
Volume of ice = 1.09 litter
7.3
(1) An iron sphere of mass 5 kg, the density of iron is 8200 kgm-3.
Data mass = 5 kg
density = 8200 kg m-3
volume = ?
Formula
volume =mass/density
volume=5/8200
Volume = 6.1 X10-4 m3
(ii) 200 g of lead shot having density 11300 kgm-3 .
Data mass = 200g=200/1000=0.2kg
density = 8200 kg m-3
volume = ?
Formula
volume =mass/density
volume =0.2/11300
volume = 1.77 X 1O-5 m3
(iii) A gold bar of mass 0.2kg.density of gold is 19300kgm-3
Data mass = 0.2kg
density = 19300 kg m-3
volume = ?
Formula
volume =mass/density
volume =0.2/19300
volume = 1.04 X 1O-5 m3
7.4
Data
density = p = 1.3 kg m-3
mass = m = ?
volume =(8x5x4)m3
Formula
mass = volume X density
Putting values, we get A
m = (8 x 5 x 4)(1.3)
m = 208 kg
7.5
Data
F =75N
A = 1.5 cm2= 1.5 x10m2
Formula
P = F/A
Putting values
P=75/(1.5 x 10-4)
P =5x105Nm-2
7.6
Data
A =(10x10-3)(10x10-3) m2
F =20N
P =?
Formula
P=F/A
P=20/((10x10-3)(10x10-3))
P=2 x 105Nm-2
7.7
Data
volume = (20 x 10-2) (7.5 x 10-2) (7.5 x 10-2) m3
mass = 1000 g = 1 kg
Area of side = (7.5 x 10-2) (7.5 x l0-2)m2
(i) pressure =?
(ii) A density =?
Formula
P=F/A
P=mg/A
P=1x10/(7.5 x 10-2) (7.5 x l0-2)
P =1778Nm-2
Density=mass/volume
Density=1/(20 x 10-2) (7.5 x 10-2) (7.5 x 10-2)
Density=889kgm-3
7.8
Data
V = volume= (0.05)3 m3
mass=(306 x1O-3)kg
p = density = 2.55 g cm-3 = 2.55 x 103 kg m-3
Volume ofcavity = ?
Volume without cavity = (5 x 5 x 5)10-6 m3
= 125 x 10-6 m3
Volume with cavity=mass / density
Volume with cavity=306 x1O-3/2.55 x 103
Volume with cavity=120 x 10-6m3
Volume of cavity= volume without cavity – volume with cavity
Volume of cavity= ( 125 x10-6 ) – (120 x 10-6)
Volume of cavity=5x 10-3m3
Volume of cavity=5cm3
7.9
Data
In air weight = 18 N
In water weight = 11.4kg
m=1.8kg
Upward thrust = (18.0 — 11.4) N
= 6.6 N
10x(density of water)(volume of water) = upward thrust
Volume =upward thrust/(density of water x 10)
Volume= 6.6/(1000 x 10)
Volume=6.6 x 10-4
Density=mass /volume
Density=1.8/(6.6 x 10-4)
Density = 3.67 x 10-4kgm-3
7.10
Data
density of wood = 0.6 gcm3 = 0.6 X 103 Kg m-3
weight = 3.06 N
mass = 3.06 /10= 0.306 kg
(a) volume = ?
Formula
Volume=mass/density
Volume=0.306/0.6 x 103
volume = = 0.51 X 10-3 m3
volume = 510 cm3
Formula
Weight=10 x vol x density
3.06=10x vol x 0.9 x103
Vol=0.00034m3
Vol=340cm3
7.11
Data
D1 = Diameter = 30 cm =0.3 m
F = w = 20000 N
D2 = 3.0 cm = 0.03 m
F1/π(D1/2)2= F2/π(D2/2)2
F2=F1 D12/D22
F2=20000(0.03)2/0.32
F2=200N
7.12
A=2 x 10-5m2
ΔL=2x10-3m
F=4000N
Lo=2m
Y=?
Y=FLo/A ΔL
Y=(4000x2)/(2x10-5x2x10-3)
Y=2x1011Nm-2