b) Compare FX, FY and FZ 2. - odtugvofizik.com worksheet - Answer... · Physics 10 Pressure & Buoyancy Worksheet : Solid & Liquid Pressure 3 5. 6. 2 5. Three containers K, L, M are

Physics 10 Pressure & Buoyancy Worksheet : Solid & Liquid Pressure

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PHYSICS WORKSHEET Grade - 10

PRESSURE Solid & Liquid Pressures

1. Solid objects X, Y and Z are made of the same substance. The pressures applied by X, Y, Z on the horizontal surface are PX, PY, PZ and the magnitudes of the forces applied by them on the horizontal surface are FX, FY, FZ. (“A” represents the base areas of the objects.)

a) Compare PX, PY and PZ. b) Compare FX, FY and FZ. 2. Solid objects K and L are placed on a horizontal surface in two different arrangements as shown in Figure-1 and Figure-2.

Read the following statements. If the statement is ABSOLUTELY CORRECT print "A", if it is POSSIBLE print "P" or if it is WRONG then print "W". You have to correct the wrong statements by using an appropriate word(s) of phrase(s).

____ The weight of object K is equal to the weight of object L.

____ The pressure applied by the objects in Figure-2 is greater than the pressure applied by the objects in Figure-1.

____ The total force applied by the objects on the horizontal surface in Figure-1 is equal to the total force applied by the objects on the horizontal surface in Figure-2.

3. Three identical cubes are arranged as in three different (I, II and II) positions on a horizontal surface. The pressure applied on the horizontal surface by the cubes in arrangement I is “P”.

What are the pressures applied on the horizontal surface by the cubes in arrangement II and III in terms of “P”? 4. Two cylinders X and Y are placed on a horizontal surface as shown in the figure. The radius of cylinder Y is twice the radius of cylinder X and their heights are equal. The pressure applied by cylinder X on the horizontal surface is “P” and the pressure applied by cylinder Y on the horizontal surface is “2P”.

What would the pressure be, in terms “P”, on the horizontal surface if cylinder X is placed on the top of cylinder Y?

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2.

PZ > PY > PX

FZ > FY > FX P

A

A


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3. 4.

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P= 3GA

PII =3G3A

= GA⇒PII =

P3

PIII =3G2A

⇒PIII =P2

P= GX

πr2⇒GX =Pπr

2

2P = GY

4πr2⇒GY = 8Pπr

2

Pfinal =GX +GY

4πr2= (Pπr

2)+ (8Pπr2)4πr2

Pfinal =9P4


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5. 6.

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5. Three containers K, L, M are placed on a horizontal surface as shown in the figure. The forces applied by the liquids on the bases of containers K, L and M are equal.


____ The liquid pressures at the bottom of the containers are equal.

____ The base areas of the containers are equal.

____ The densities of the liquids in the containers are related as dL>dK>dM.

____ The heights of the liquids in the containers are related as hM>hK>hL.

____ The masses of the liquids in the containers are related as mL>mK>mM.

6. A liquid is poured into a container that is placed on a horizontal surface as shown in the figure. The pressure applied by the liquid on the base of the container is “P”, the force applied by the liquid on the base of the container is “F” and the force applied by the container on the horizontal surface is “Fh”.

How would “P”, “F” and “Fh” change if the container were placed on its surface having base area of “A”? P : ______________________ F : ______________________ Fh : ______________________

7. Containers X and Y are identical. X contains some liquid as in the figure. The liquid pressure on the base of container X is “P”, the force applied by the liquid on the base of container X is “F” and the force applied by the container on the horizontal surface is “Fh”. (Figure is composed of identical squares.)

Read the following statements. If the statement is CORRECT print "C" or if it is WRONG then print "W". You have to correct the wrong statements by using an appropriate word(s) of phrase(s).

____ If the liquid in container X was poured into container Y, the liquid pressure on the base of container Y would be “P/2”.

____ If the liquid in container X were poured into container Y, the force applied by the liquid on the base of container Y would be “F/2”.

____ If the liquid in container X were poured into container Y, the force applied by container Y would be “Fh”.

8. A container is fulfilled with a liquid as shown in the figure. The areas of the lateral surfaces X and Y are "3A" and "A" respectively. The force applied by the liquid on lateral surface X is "F".

What is the force applied by the liquid on the lateral surface Y in terms of "F"?

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P increases

decreases P P

P

A

remains the same


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7. 8.

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2

















FX =F =3h2.d.g.3A = 9hdgA

2

FY =5h2.d.g.A = 5hdgA

2

FY =5F9

C

W

C 3F/2


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9. 10.

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9. Two immiscible liquids X and Y are in equilibrium in a closed container as shown in the figure.


____ If the container was made upside-down, the liquid pressure at the bottom of the container would not change.

____ If the container was made upside-down, the force applied by the liquids at the base of the container would be less.

____ If the container was made upside-down, the pressure applied by the container on the horizontal surface would be greater.

10. The dimensions of a rectangular container are 15 cm, 20 cm and 30 cm. It is completely filled with a liquid. The force applied by the liquid on the lateral surface K is "FK" and the force applied by the liquid on the base of the container is "FB".

What is the ratio of FK to FB?

11. Two containers are filled with two liquids X and Y as in figure. The weight of liquid X is “G1” and the weight of liquid Y is “G2”. The force applied by liquid X on the base of its container is “F1” and the force applied by liquid Y on the base of its container is “F2”. The ratio of F1 to F2 is 1/2.

What is the ratio of G1 to G2? 12. Two immiscible liquids of densities “2d” and “3d” are poured into a container as shown in the figure. The surface areas of lateral surfaces are “A” and “2A”. The force exerted by the liquid on the lateral surface having area of “A” is “F”.

Calculate the forces exerted by the liquid on the lateral surface having the area of “2A” and on the base of the container in terms of “F”?

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C

W

C

increase AB = 15.20 = 300 cm2

AK = 20.30 = 600 cm2

FK = 15.d.g.600

FB = 30.d.g.300

⎫⎬⎪

⎭⎪⇒ FK

FB

= 1


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11. 12.

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AB = 15.20 = 300 cm2

AK = 20.30 = 600 cm2

FK = 15.d.g.600

FB = 30.d.g.300

⎫⎬⎪

⎭⎪⇒ FK

FB

= 1

VY = 3V⇒ VX = 7VAA = A⇒ Ay = A

F1F2= 12= h.dX.g.Ah.dY.g.A

⎫⎬⎭⇒ dXdY= 12

G1

G2

= dX.VX .gdY.VY .g

= 1.7V2.3V

= 76


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13. Four containers X, Y, Z and T are placed on a horizontal surface. Containers X, Y and Z are filled with liquids and container T is empty. The liquid pressures at the bottoms of the containers X, Y and Z are all equal to "P". (The liquids in the container are immiscible.)

If the liquids were poured into container T, what would be the total liquid pressure at the bottom of container T in terms of "P"? 14. The hydraulic system is in equilibrium as shown in the figure. The piston is supposed to be weightless and frictionless. The pressure at point X is “P”. (“A” and “2A” represent the cross-sectional areas of the columns of the combined container.)

What would be the pressure at point X (in terms of P), if the piston were pushed from level K to level L?

15. Two miscible liquids having densities of “d” and “2d” are in equilibrium as in the figure. The liquid pressure at point X is “P”. (“A” and “2A” represent the cross-sectional areas of the columns of the combined container.)

What will be the liquid pressure at point X in terms of “P” after the valve is opened and the final equilibrium had reached? 16. The hydraulic system is in equilibrium as shown in the figure. Pistons are supposed to be weightless and frictionless. A1 and A2 are the base areas of the pistons. The density of liquid is “d”.


____ The liquid pressure at point X is equal to “h.d.g”.

____ The liquid pressures at point X and Y are equal.

____ The total pressure at point X is greater than the liquid pressure at point Y.

P= 4hdXg⇒dX = 3d

P = 2hdYg⇒dY = 6d

P = 3hdzg⇒dZ = 4d

⎫

⎬

⎪⎪⎪

⎭

⎪⎪⎪

⇒P=12hdg

PT = 6hdg( )+ 5h4.4d.g⎛

⎝⎜⎞⎠⎟ + 3hdg( )

PT =14hdg

PT =7P6

hhhh

T

h

h5h/4

6d4d3d


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14.

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PXinitial =P = 2hdg⇒hdg= P2

PXfinal =5h2dg⇒PXfinal =

5P4

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Vinitial = Vfinal ⇒ 4Ah+2Ah=h/.3A

h/ = 2h

dmix =(d.4V)+ (2d.2V)

4V+2V=4d3

P = 2hdg

P/ = 2h.4d3.g

⎫⎬⎪

⎭⎪⇒P/ =

4P3


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16. 17.

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C

C

C

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17. The flow rates of the valves M and N are the same. They are open at the same time and the final equilibrium is reached. The liquid pressures at the points X, Y and Z are PX, PY and PZ. (All the columns of the combined container have the same cross-sectional area. Dotted lines are equally spaced.)

How can PX, PY and PZ be compared? 18. Liquid A of density "2d" and liquid B of density "3d" are in equilibrium in a combined container as shown in the figure. (Atmospheric pressure is ignored.)

What is the pressure at point X in terms of "hdg"?

19. A combined container is in equilibrium when an object of mass "m" is placed on piston K and object of mass "2m" is placed on piston L as shown in the figure. K and L are weightless and frictionless pistons. Their cross-sectional areas are "A" and "3A" respectively. The vertical distance between the levels of the pistons is "h".

If the places of the objects were interchanged, what would be the vertical distance between the levels of the pistons in terms of "h"? 20. Column K of a combined container is filled with a liquid as shown in the figure. The liquid pressure on the bottom of column K is “P” when the valve is closed. (“A” and “2A” represent the cross-sectional areas of the columns K and L of the combined container.)

The valve is opened and the final equilibrium is reached. What will be the liquid pressure in terms of “P” at the bottom of column K?

After the valves are opened

PX = PY < PZ

MN

X YZ


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18.

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PK =PL = 4hdg+9hdg=13hdg

PL =13hdg=7h2.3d.g+PX

PX =5hdg2

L K

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19.

1st condition;mgA

=2mg3A

+hdg⇒ mA

= 3hd

2nd condition;2mg

A=

mg3A

+h/dg⇒ 5m3A

= h/d

h/ = 5h


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20.

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Vinitial = Vfinal ⇒6Ah= 2A+3A( ).h/

h/ = 2h

The valve is opened;

P/ = 2hdg⇒P/ =P3

h

A 2A

horizontal

5h

K L

2h 2h

The valve is closed;

P = 6hdg⇒hdg =P6


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21. 22.

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21. Two immiscible liquids X and Y are poured into a container and their equilibrium condition is given in the figure. The density of liquid X is “3d” and the density of liquid Y is “2d”. The liquid pressure at point L is two times the liquid pressure at point K.

What is the ratio of h1 to h2? 22. Liquid X having density of “d” and weight of “G” is poured into the container as shown in the figure. The liquid pressure at the bottom of the container is “P”. The empty part of the container is filled with liquid Y having weight of “2G”. (Container is composed of identical cubic parts.)

If liquids X and Y are miscible, what will be the liquid pressure at the bottom of the container in terms of “P”?

23. The liquids of densities “2d”, “d” and liquid X are in equilibrium in a U-tube as shown in the figure.

What is the density of liquid X in terms of “d”? 24. Mercury is poured into a U-tube as given in Figure-1. The left arm of the U-tube has a cross-sectional area of 10 cm2, and the right arm has a cross-sectional area of 5 cm2. One hundred grams of water are then poured into the right arm, as shown in Figure-2. (The density of water is 1 g/cm3.)

a) What is the length of the water column in the

right arm of the U-tube in cm? b) If the density of mercury is 13,6 g/cm3, what

distance (h) will the mercury rise in the left arm?

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PK =h1.2d.g

PL = (h1.2d.g)+ (h2.3d.g)

⎫⎬⎪

⎭⎪⇒ PKPL= 12

4h1 = 2h1 +3h2 ⇒h1h2= 32

VX = 2V⇒ VY = V

G=d.2V.g2G=dY.V.g

⎫⎬⎭⇒dY = 4d

dmix =(d.2V)+ (4d.V)

V+2V= 2d

P =hdg

P/ = 2h.2d.g= 4hdg⎫⎬⎭⇒P/ = 4P


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23. 24.

PA =PB(3h.d.g)+ (h.dX.g) = 4h.2d.g

(3d)+ (dX) = 8ddX = 5d

Vwater = 100 cm3

hwater = 1005

= 20 cm

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2x2x

=x

A B

PA = PB

3x.1.g = 20.13,6.g⇒ x = 0,49 cm

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b) Compare FX, FY and FZ 2. - odtugvofizik.com worksheet - Answer... · Physics 10 Pressure & Buoyancy Worksheet : Solid & Liquid Pressure 3 5. 6. 2 5. Three containers K, L, M are