أهم 100 سؤال فى مادة الفيزياء لغات - الجزء الثاني عشر وبدون تحميل

أهم 100 سؤال فى مادة الفيزياء لغات - الجزء الثاني عشر




Important applications


A-mention the applications for each the following:


1) The density measurement as an analytical technique, in the medicinal field and industry.

2) The pressure measurement.

3) Pascal's principle.

4) Archimedes principle.


B-Mention one function only for each of the following:


1) The manometer

2) The mercury barometer

3) pressure gauge

4) u- shaped tube.

5) The hydraulic press.





H- (i)Write the unit used to measure each of the following physical quantities:

a- Pressure at a point inside liquid

b- Standard atmospheric pressure

(ii)Write the unit that equivalent to each of the following then
mention the physical quantities that measured by the following units:

a-Kg/m.. sec. (May 2006)

b-J/m3 (August 2003)

c-N/m2 (May 1994 )


I-STRUCTURED QUESTIONS;-


1-In the opposite figure four barometric tubes;

a-Why is the volume of the vacuum is different from one tube to another?

b-What do you expect if one of the barometers is transferred to the top of high mountain? And why?

C- Mention two reasons for the use of mercury in such barom


2-While a student is measuring the
pressure of a gas enclosed in a container by mercury manometer,
another student advice him to use water instead of mercury.


a-Why it is preferable to use water in this case?

b-Complete the second figure (B) showing the water level in the two sides of the manometer.


Selected problems

1) August 1996

A layer of water of thickness 50cm rests on a layer of mercury of
thickness 20cm. what is the difference in pressure at two points, one
at the interface between water and mercury and other at the bottom of
the mercury layer (the density of water  = 1000kg/m3, the density of
mercury  =13600 kg/m3, the acceleration due to gravity g = 10 m/sec2).

1-P2(at the bottom) = Pa + (1g h1)w + (2g h2 )mer.

P1(at the interface) = Pa + (1g h1)w

P = (2g h2 )mer =13600 × 10 × 0.2 =27200 N/m2





2) A submarine is balanced
horizontally under sea water. The pressure inside it equals atmospheric
pressure at sea level, calculate the force acting on one of its circular
windows of radius 21cm when it is at a depth of 50meters below sea
level. (density of sea water 1030Kg/m3, g = 9.8m/sec2).

2- P = gh P = 1030X9.8X50N/m2 = 504700 N/m2

F = P.A = P. r2



3) A man carries a mercuric
barometer whose reading at the ground floor is 76cm Hg and at the upper
floor is 74.15cm Hg. If the height of the building is 200m, find the
average density of the air between these two floors. (acceleration due
to gravity 9.8m/s2 , density of mercury 13600 Kg/m3).

3-The weight of air column with unit cross- sectional area =the weight of mercury column with unit cross- sectional area

I.e. the decrease in air pressure = the decrease in reading of barometer


airH = Hg(H2-H1).

H; The height of air column.

H2 ,H1; Reading of barometer

 200 = 13600(0.76-0.7415)

air = 1.25Kg/m3


4)A U-shaped tube of uniform cross-sectional area contains water,
its density 1000Kg/m3, oil is poured in one arm of the tube its density
600Kg/m3, the height of the water surface decreases 3cm, calculate the
height of the oil above the separating surface.


4-Since the tube having a uniform cross-sectional area


Therefore, the level of water in one arm is depressed

3 cm as well as raise in the second by 3cm

1gh1 =2gh2 600xh=1000x6  h =10cm





5) May 1998

A mercury manometer is used to measure the pressure of a gas
inside a container. The surface of the mercury inside the free side of
the manometer is lower than that in the other side attached to the
container by 20cm. what is the value of the pressure in “bar” units of
the gas enclosed in the container, given that the atmospheric pressure
when performing such measurements was 105 Pascal, density of mercury =
13600 Kg/m-3 and g = 10m/s-2.

5-(May 1995

6-A force of 200N is acting on the small piston of hydraulic
press of diameter 2cm. If the diameter of its large piston is 24cm and
the acceleration due to gravity is 10m/sec2 (=3.14): find

a-The maximum mass which can lifted with the large piston.

b-The mechanical advantage of the hydraulic press.

c-The pressure on each of large and small pistons.

6- Area(a) = (0.01)2 , Area( A) = (0.12)2

7-May 2007

A body of volume 50 cm3 is placed in water, it displaced 40 cm 3
of water. Is this body immerse or float ? Why ? Calculate the density of
the body if the water density equals 103 kg /M3.


7-The body is float because the volume of the displaced water is less than the total volume of the body.

Fb = (Fg)s w.g.(Vol )1= mg w.g.(Vol )1=s. g (Vol)t

1000×40=s.×50 s =800kg/m3

8-May 1997

A piece of wood of density 800 kg/m3 , floats on water such that the volume of the immersed part is 8 cm3,

Find ,

1)Its mass 2)The volume of the apparent part

Given that the density of water 1000 kg/ m3 and the acceleration due to gravity = 10m/s2


8- a- Fb = (Fg)s w.g.(Vol )1= mg m =1000x8x10-6 =0.008kg

b- The total volume Vol =(0.008/800) =10-5m3  (Vol)2=10-5-8x10-6 =2x10-6m3





9)A submarine of volume (1000m3)
is suspended in water of density 1000Kg/m3 such that its upper surface
is such below water surface. If it is transferred to sea water of
density 1030Kg/m3, calculate:

1) The up thrust force in the two cases.

2) The volume of the apparent part of the submarine in the second case.

3) The volume of the sea water that can be put in the tanks to immerse the submarine under surface.

a) Fb remains the same in the two cases because, the weight is constant.

Fb = rgVo1 = 1000x10x103 = 107N

b)In salty water Fb = sg(Vo1)1

( (Vo1)1 = EMBED Equation.3 = 970.87m3

 The volume of floating part

= 103 – 970.8 = 29.7m3

c)The volume of water equals the volume of floating part from the submarine = 29.1m3.



10)An empty ferry boat used to
transporting petroleum oil, the boat with vertical sides with dimension
(3020)m2, when the boat filled with the oil it sink 20cm in addition to
the original depth. Find the mass of oil that transported by the
ferryboat, density of sea water 1.03g/cm3.,g =9.8 m/s2

[m = 123600 Kg]

10- Additional weight = additional Fb

∴mg= r1g Vol m= 1030´30x20x0.2 = 123600 kg.


11) A metallic cube of length 10cm
of specific weight 2.7 is suspended using a thread. Find the tension in
the thread (in Newton’s) in the following cases:

a) When the cube is suspended in air.

b) When half of the cube is immersed in water.

c) When the cube is totally immersed in water (g = 10m/sec2, w=1000Kg/m3).


11- a) (Ft)1 = (Fg)s \ (Ft)1 = rsgV = 2700´10´(0.1)3 = 27N

b) (Ft)2 + Fb = (Fg)s \ (Ft)2= (Fg)s – Fb = 27-1000´10´(1/2)´(0.1)3=22N

c) (Ft)3+ Fb = (Fg)s \ (Ft)3 = 27 - 1000´10´(0.1)3 = 17N


12)1st s 94

A balloon is filled with hydrogen of density 0.09 Kg/m3
until its volume becomes 14104m3. find the lifting force of the balloon
knowing that density of air 1.29 Kg/m2, mass of balloon and its
(without gas) = 105 Kg.(10m/s2)

[680000N]

13- The lifting force is given by

F = Fb – (Fg)b – (Fg)gas

F = 1.29x10x14x104 –105x10-0.09x10x14x104 = 68x104N