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GRADE 7 PRACTICE

Physics Problems: Pressure, Grade 7

Pressure of solids, liquids, gases and hydraulic devices.

27 problems across 6 sub-topics
Practise in the interactive mode

Here the conditions and analysis of the solution can be read without an account. In the simulator, you need to collect the same analyzes yourself: “Given,” formulas, order of steps and answer.

Problem statements

The answer options are there to check against, and the walkthrough opens under each problem.

27 problems
Position in this set: 1 from 27

Problem #34

Pressure from a solid object and Pascal's law

What concept does the description correspond to: “Gas pressure arises from continuously moving particles striking the container walls.»?

Answer options

  1. \text{Gas pressure}
  2. \text{Pressure from a solid object}
  3. \text{Pascal's law}
  4. \text{Physical phenomenon}
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Given
\text{Description}
\text{Gas pressure arises from continuously moving particles striking the container walls.}
Find

\text{name concepts}

  1. \text{Let's compare approval with definition}

    When volume decreases, impacts can become more frequent, so the gas exerts more force per unit wall area.

  2. \text{Let's formulate conclusion}\text{Concept} = \text{Gas pressure}

Answer\text{Concept} = \text{Gas pressure}

Position in this set: 2 from 27

Problem #35

Pressure from a solid object and Pascal's law

What concept does the description correspond to: “A physical quantity equal to the force perpendicular to a surface divided by that surface's area.»?

Answer options

  1. \text{Pressure from a solid object}
  2. \text{Gas pressure}
  3. \text{Pascal's law}
  4. \text{Physical phenomenon}
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Given
\text{Description}
\text{A physical quantity equal to the force perpendicular to a surface divided by that surface's area.}
Find

\text{name concepts}

  1. \text{Let's compare approval with definition}

    Wide skis reduce pressure on snow; a sharp blade increases pressure for the same applied force.

  2. \text{Let's formulate conclusion}\text{Concept} = \text{Pressure from a solid object}

Answer\text{Concept} = \text{Pressure from a solid object}

Position in this set: 3 from 27

Problem #36

Pressure

The machine presses on the floor with a force of 600 N through a support with an area of 0.03 m². What pressure does it create?

Answer options

  1. 20\,\text{кПа}
  2. 0{,}05\,\text{кПа}
  3. 18\,\text{кПа}
  4. 180\,\text{кПа}
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Given
F
600\,\text{Н}
S
0{,}03\,\text{м}^{2}
Find

p

  1. p=\frac{F}{S}p=\frac{600}{0{,}03}\,\text{Па}p = 20000\,\text{Па}
  2. 1000\,\text{Па}=1\,\text{кПа}p = 20\,\text{кПа}

Answer20\,\text{кПа}

Position in this set: 4 from 27

Problem #37

Hydrostatic pressure and vessels

What concept does the description correspond to: “Pressure in a resting liquid due to the weight of the layers above, increasing with depth.»?

Answer options

  1. \text{Hydrostatic pressure}
  2. \text{Communicating vessels}
  3. \text{Measuring liquid pressure}
  4. \text{Physical phenomenon}
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Given
\text{Description}
\text{Pressure in a resting liquid due to the weight of the layers above, increasing with depth.}
Find

\text{name concepts}

  1. \text{Let's compare approval with definition}

    At the same depth in a homogeneous liquid, the pressure is the same and does not depend on the shape of the vessel.

  2. \text{Let's formulate conclusion}\text{Concept} = \text{Hydrostatic pressure}

Answer\text{Concept} = \text{Hydrostatic pressure}

Position in this set: 5 from 27

Problem #38

Hydrostatic pressure and vessels

What concept does the description correspond to: “Containers connected below their free surfaces so liquid can flow between them.»?

Answer options

  1. \text{Communicating vessels}
  2. \text{Hydrostatic pressure}
  3. \text{Measuring liquid pressure}
  4. \text{Physical phenomenon}
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Given
\text{Description}
\text{Containers connected below their free surfaces so liquid can flow between them.}
Find

\text{name concepts}

  1. \text{Let's compare approval with definition}

    The same liquid settles at equal heights if the pressure above both surfaces is equal.

  2. \text{Let's formulate conclusion}\text{Concept} = \text{Communicating vessels}

Answer\text{Concept} = \text{Communicating vessels}

Position in this set: 6 from 27

Problem #39

Hydrostatics

Find the pressure of the water column at a depth of 2 m. The density of water is 1000 kg/m³, the acceleration of gravity is 9.8 m/s².

Answer options

  1. 19{,}6\,\text{кПа}
  2. 4{,}9\,\text{кПа}
  3. 9{,}8\,\text{кПа}
  4. 196\,\text{кПа}
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Given
\rho
1000\,\text{кг/м}^{3}
g
9{,}8\,\text{м/с}^{2}
h
2\,\text{м}
Find

p

  1. p=\rho ghp=1000\cdot9{,}8\cdot2\,\text{Па}p = 19600\,\text{Па}
  2. p=19{,}6\,\text{кПа}

Answer19{,}6\,\text{кПа}

Position in this set: 7 from 27

Problem #40

Atmospheric pressure and hydraulics

What concept does the description correspond to: “The pressure that atmospheric air exerts on objects due to its weight and the motion of its molecules.»?

Answer options

  1. \text{Atmospheric pressure}
  2. \text{Mercury-column experiment}
  3. \text{Hydraulic machine}
  4. \text{Physical phenomenon}
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Given
\text{Description}
\text{The pressure that atmospheric air exerts on objects due to its weight and the motion of its molecules.}
Find

\text{name concepts}

  1. \text{Let's compare approval with definition}

    With increasing altitude, less air remains above the observer, so a barometer's reading usually decreases.

  2. \text{Let's formulate conclusion}\text{Concept} = \text{Atmospheric pressure}

Answer\text{Concept} = \text{Atmospheric pressure}

Position in this set: 8 from 27

Problem #41

Atmospheric pressure and hydraulics

What concept does the description correspond to: “The balance between a liquid column and atmospheric pressure allows atmospheric pressure to be measured from the column's height.»?

Answer options

  1. \text{Mercury-column experiment}
  2. \text{Atmospheric pressure}
  3. \text{Hydraulic machine}
  4. \text{Physical phenomenon}
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Given
\text{Description}
\text{The balance between a liquid column and atmospheric pressure allows atmospheric pressure to be measured from the column's height.}
Find

\text{name concepts}

  1. \text{Let's compare approval with definition}

    A barometer uses this principle directly, or indicates pressure through the deformation of a sealed metal capsule.

  2. \text{Let's formulate conclusion}\text{Concept} = \text{Mercury-column experiment}

Answer\text{Concept} = \text{Mercury-column experiment}

Position in this set: 9 from 27

Problem #42

Hydraulics

A force of 120 N acts on a piston with an area of 0.003 m². The area of the second piston is 0.06 m². Find the force on the second piston without taking into account losses.

Answer options

  1. 2400\,\text{Н}
  2. 6\,\text{Н}
  3. 120\,\text{Н}
  4. 600\,\text{Н}
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Given
F_1
120\,\text{Н}
S_1
0{,}003\,\text{м}^{2}
S_2
0{,}06\,\text{м}^{2}
Find

F_2

  1. \frac{F_1}{S_1}=\frac{F_2}{S_2}
  2. F_2=F_1\frac{S_2}{S_1}F_2=120\cdot\frac{0{,}06}{0{,}003}\,\text{Н}F_2 = 2400\,\text{Н}

Answer2400\,\text{Н}

Position in this set: 10 from 27

Problem #124

Pressure from a solid object and Pascal's law

Perpendicular pressing force 120 N distributed by area 0{,}02 m². Find pressure.

Answer options

  1. 6000\,\text{Па}
  2. 12000\,\text{Па}
  3. 3000\,\text{Па}
  4. 18000\,\text{Па}
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Given
F
120\,\text{Н}
S
0{,}02\,\text{м}^{2}
Find

p

  1. p=\frac{F}{S}p=\frac{120}{0{,}02}p=6000\,\text{Па}

    Pressure shows force per unit area.

Answer6000\,\text{Па}

Position in this set: 11 from 27

Problem #125

Pressure from a solid object and Pascal's law

Perpendicular pressing force 240 N distributed by area 0{,}04 m². Find pressure.

Answer options

  1. 12000\,\text{Па}
  2. 6000\,\text{Па}
  3. 3000\,\text{Па}
  4. 18000\,\text{Па}
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Given
F
240\,\text{Н}
S
0{,}04\,\text{м}^{2}
Find

p

  1. p=\frac{F}{S}p=\frac{240}{0{,}04}p=6000\,\text{Па}

    Pressure shows force per unit area.

Answer6000\,\text{Па}

Position in this set: 12 from 27

Problem #126

Pressure from a solid object and Pascal's law

Body presses on support with by force 200 N, creating pressure 10 kPa. Find Contact area B m².

Answer options

  1. 0{,}04\,\text{м}^{2}
  2. 0{,}02\,\text{м}^{2}
  3. 0{,}01\,\text{м}^{2}
  4. 0{,}06\,\text{м}^{2}
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Given
F
200\,\text{Н}
p
10\,\text{кПа}
Find

S

  1. P=1000pP=10 \cdot 1000P=10000\,\text{Па}

    Convert kilopascals to pascals.

  2. S=\frac{F}{P}S=\frac{200}{10000}S=0{,}02\,\text{м}^{2}

    We express the area from the definition of pressure.

Answer0{,}02\,\text{м}^{2}

Position in this set: 13 from 27

Problem #127

Pressure from a solid object and Pascal's law

Body presses on support with by force 360 N, creating pressure 12 kPa. Find Contact area B m².

Answer options

  1. 0{,}06\,\text{м}^{2}
  2. 0{,}015\,\text{м}^{2}
  3. 0{,}03\,\text{м}^{2}
  4. 0{,}09\,\text{м}^{2}
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Given
F
360\,\text{Н}
p
12\,\text{кПа}
Find

S

  1. P=1000pP=12 \cdot 1000P=12000\,\text{Па}

    Convert kilopascals to pascals.

  2. S=\frac{F}{P}S=\frac{360}{12000}S=0{,}03\,\text{м}^{2}

    We express the area from the definition of pressure.

Answer0{,}03\,\text{м}^{2}

Position in this set: 14 from 27

Problem #128

Pressure from a solid object and Pascal's law

block of mass 6 kg lies on edges 20 cm on 10 cm. Find pressure on table at g = 10 N/kg.

Answer options

  1. 6000\,\text{Па}
  2. 1500\,\text{Па}
  3. 3000\,\text{Па}
  4. 9000\,\text{Па}
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Given
m
6\,\text{кг}
a
20\,\text{см}
b
10\,\text{см}
g
10\,\text{Н/кг}
Find

p

  1. S=\frac{ab}{10000}S=\frac{20 \cdot 10}{10000}S=0{,}02\,\text{м}^{2}

    There are ten thousand square centimeters in a square meter.

  2. F=mgF=6 \cdot 10F=60\,\text{Н}

    At rest, the pressure force is equal to the weight.

  3. p=\frac{F}{S}p=\frac{60}{0{,}02}p=3000\,\text{Па}

    We divide the pressure force by the area of the face.

Answer3000\,\text{Па}

Position in this set: 15 from 27

Problem #129

Pressure from a solid object and Pascal's law

block of mass 9 kg lies on edges 30 cm on 10 cm. Find pressure on table at g = 10 N/kg.

Answer options

  1. 6000\,\text{Па}
  2. 1500\,\text{Па}
  3. 9000\,\text{Па}
  4. 3000\,\text{Па}
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Given
m
9\,\text{кг}
a
30\,\text{см}
b
10\,\text{см}
g
10\,\text{Н/кг}
Find

p

  1. S=\frac{ab}{10000}S=\frac{30 \cdot 10}{10000}S=0{,}03\,\text{м}^{2}

    There are ten thousand square centimeters in a square meter.

  2. F=mgF=9 \cdot 10F=90\,\text{Н}

    At rest, the pressure force is equal to the weight.

  3. p=\frac{F}{S}p=\frac{90}{0{,}03}p=3000\,\text{Па}

    We divide the pressure force by the area of the face.

Answer3000\,\text{Па}

Position in this set: 16 from 27

Problem #130

Hydrostatic pressure and vessels

Liquid density 1000 kg/m³, depth 0{,}5 m. Find the excess pressure at this depth at g = 10 N/kg.

Answer options

  1. 10000\,\text{Па}
  2. 5000\,\text{Па}
  3. 2500\,\text{Па}
  4. 15000\,\text{Па}
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Given
r
1000\,\text{кг/м}^{3}
h
0{,}5\,\text{м}
g
10\,\text{Н/кг}
Find

p

  1. p=\rho ghp=1000 \cdot 10 \cdot 0{,}5p=5000\,\text{Па}

    We calculate only the pressure of the liquid without atmospheric pressure.

Answer5000\,\text{Па}

Position in this set: 17 from 27

Problem #131

Hydrostatic pressure and vessels

Liquid density 800 kg/m³, depth 0{,}8 m. Find the excess pressure at this depth at g = 10 N/kg.

Answer options

  1. 12800\,\text{Па}
  2. 3200\,\text{Па}
  3. 6400\,\text{Па}
  4. 19200\,\text{Па}
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Given
r
800\,\text{кг/м}^{3}
h
0{,}8\,\text{м}
g
10\,\text{Н/кг}
Find

p

  1. p=\rho ghp=800 \cdot 10 \cdot 0{,}8p=6400\,\text{Па}

    We calculate only the pressure of the liquid without atmospheric pressure.

Answer6400\,\text{Па}

Position in this set: 18 from 27

Problem #132

Hydrostatic pressure and vessels

The excess water pressure is 24 kPa. Density of water 1000 kg/m³, g = 10 N/kg. Find the depth.

Answer options

  1. 4{,}8\,\text{м}
  2. 1{,}2\,\text{м}
  3. 2{,}4\,\text{м}
  4. 7{,}2\,\text{м}
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Given
p
24\,\text{кПа}
r
1000\,\text{кг/м}^{3}
g
10\,\text{Н/кг}
Find

h

  1. P=1000pP=24 \cdot 1000P=24000\,\text{Па}

    We reduce pressure to SI.

  2. h=\frac{P}{\rho g}h=\frac{24000}{(1000 \cdot 10)}h=2{,}4\,\text{м}

    We express the depth from the law of hydrostatics.

Answer2{,}4\,\text{м}

Position in this set: 19 from 27

Problem #133

Hydrostatic pressure and vessels

The excess water pressure is 35 kPa. Density of water 1000 kg/m³, g = 10 N/kg. Find the depth.

Answer options

  1. 7\,\text{м}
  2. 1{,}75\,\text{м}
  3. 10{,}5\,\text{м}
  4. 3{,}5\,\text{м}
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Given
p
35\,\text{кПа}
r
1000\,\text{кг/м}^{3}
g
10\,\text{Н/кг}
Find

h

  1. P=1000pP=35 \cdot 1000P=35000\,\text{Па}

    We reduce pressure to SI.

  2. h=\frac{P}{\rho g}h=\frac{35000}{(1000 \cdot 10)}h=3{,}5\,\text{м}

    We express the depth from the law of hydrostatics.

Answer3{,}5\,\text{м}

Position in this set: 20 from 27

Problem #134

Hydrostatic pressure and vessels

In a vessel above water height 0{,}3 m is the oil height 0{,}2 m. Densities of water and oil 1000 and 800 kg/m³. Find the excess pressure at the bottom; g = 10 N/kg.

Answer options

  1. 9200\,\text{Па}
  2. 2300\,\text{Па}
  3. 13800\,\text{Па}
  4. 4600\,\text{Па}
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Given
h_1
0{,}3\,\text{м}
h_2
0{,}2\,\text{м}
r_1
1000\,\text{кг/м}^{3}
r_2
800\,\text{кг/м}^{3}
g
10\,\text{Н/кг}
Find

p

  1. p_1=\rho_1 gh_1p_1=1000 \cdot 10 \cdot 0{,}3p_1=3000\,\text{Па}

    We consider the contribution of the water layer.

  2. p_2=\rho_2 gh_2p_2=800 \cdot 10 \cdot 0{,}2p_2=1600\,\text{Па}

    We calculate the contribution of oil.

  3. p=p_1+p_2p=3000 + 1600p=4600\,\text{Па}

    The pressures of the layers add up.

Answer4600\,\text{Па}

Position in this set: 21 from 27

Problem #135

Hydrostatic pressure and vessels

In a vessel above water height 0{,}5 m is the oil height 0{,}4 m. Densities of water and oil 1000 and 800 kg/m³. Find the excess pressure at the bottom; g = 10 N/kg.

Answer options

  1. 8200\,\text{Па}
  2. 16400\,\text{Па}
  3. 4100\,\text{Па}
  4. 24600\,\text{Па}
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Given
h_1
0{,}5\,\text{м}
h_2
0{,}4\,\text{м}
r_1
1000\,\text{кг/м}^{3}
r_2
800\,\text{кг/м}^{3}
g
10\,\text{Н/кг}
Find

p

  1. p_1=\rho_1 gh_1p_1=1000 \cdot 10 \cdot 0{,}5p_1=5000\,\text{Па}

    We consider the contribution of the water layer.

  2. p_2=\rho_2 gh_2p_2=800 \cdot 10 \cdot 0{,}4p_2=3200\,\text{Па}

    We calculate the contribution of oil.

  3. p=p_1+p_2p=5000 + 3200p=8200\,\text{Па}

    The pressures of the layers add up.

Answer8200\,\text{Па}

Position in this set: 22 from 27

Problem #136

Atmospheric pressure and hydraulics

Drop pressure by parties piston 2000 Pa, area piston 0{,}02 m². Find resulting force pressure.

Answer options

  1. 80\,\text{Н}
  2. 20\,\text{Н}
  3. 40\,\text{Н}
  4. 120\,\text{Н}
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Given
p
2000\,\text{Па}
S
0{,}02\,\text{м}^{2}
Find

F

  1. F=pSF=2000 \cdot 0{,}02F=40\,\text{Н}

    We use differential pressure rather than pressure on one side only.

Answer40\,\text{Н}

Position in this set: 23 from 27

Problem #137

Atmospheric pressure and hydraulics

Drop pressure by parties piston 3000 Pa, area piston 0{,}04 m². Find resulting force pressure.

Answer options

  1. 240\,\text{Н}
  2. 60\,\text{Н}
  3. 360\,\text{Н}
  4. 120\,\text{Н}
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Given
p
3000\,\text{Па}
S
0{,}04\,\text{м}^{2}
Find

F

  1. F=pSF=3000 \cdot 0{,}04F=120\,\text{Н}

    We use differential pressure rather than pressure on one side only.

Answer120\,\text{Н}

Position in this set: 24 from 27

Problem #138

Atmospheric pressure and hydraulics

area small and big pistons 4 and 100 cm². On small piston valid strength 20 N. Find force big piston. Friction and difference heights not take into account.

Answer options

  1. 1000\,\text{Н}
  2. 250\,\text{Н}
  3. 1500\,\text{Н}
  4. 500\,\text{Н}
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Given
S_1
4\,\text{см}^{2}
S_2
100\,\text{см}^{2}
F_1
20\,\text{Н}
Find

F_2

  1. k=\frac{S_2}{S_1}k=\frac{100}{4}k=25\,\text{раз}

    In terms of area, equal units are reduced.

  2. F_2=F_1 kF_2=20 \cdot 25F_2=500\,\text{Н}

    According to Pascal's law, the pressures under the pistons are equal.

Answer500\,\text{Н}

Position in this set: 25 from 27

Problem #139

Atmospheric pressure and hydraulics

area small and big pistons 5 and 150 cm². On small piston valid strength 30 N. Find force big piston. Friction and difference heights not take into account.

Answer options

  1. 900\,\text{Н}
  2. 1800\,\text{Н}
  3. 450\,\text{Н}
  4. 2700\,\text{Н}
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Given
S_1
5\,\text{см}^{2}
S_2
150\,\text{см}^{2}
F_1
30\,\text{Н}
Find

F_2

  1. k=\frac{S_2}{S_1}k=\frac{150}{5}k=30\,\text{раз}

    In terms of area, equal units are reduced.

  2. F_2=F_1 kF_2=30 \cdot 30F_2=900\,\text{Н}

    According to Pascal's law, the pressures under the pistons are equal.

Answer900\,\text{Н}

Position in this set: 26 from 27

Problem #140

Atmospheric pressure and hydraulics

big piston area 200 cm² holds Load weight 400 kg. Small piston area 5 cm². Which force necessary attach K small piston? g = 10 N/kg; friction and mass pistons neglect.

Answer options

  1. 100\,\text{Н}
  2. 200\,\text{Н}
  3. 50\,\text{Н}
  4. 300\,\text{Н}
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Given
S_2
200\,\text{см}^{2}
S_1
5\,\text{см}^{2}
m
400\,\text{кг}
g
10\,\text{Н/кг}
Find

F_1

  1. F_2=mgF_2=400 \cdot 10F_2=4000\,\text{Н}

    The large piston must balance the weight of the load.

  2. k=\frac{S_2}{S_1}k=\frac{200}{5}k=40\,\text{раз}

    We calculate the gain in strength.

  3. F_1=\frac{F_2}{k}F_1=\frac{4000}{40}F_1=100\,\text{Н}

    A small piston requires k times less force.

Answer100\,\text{Н}

Position in this set: 27 from 27

Problem #141

Atmospheric pressure and hydraulics

big piston area 300 cm² holds Load weight 600 kg. Small piston area 6 cm². Which force necessary attach K small piston? g = 10 N/kg; friction and mass pistons neglect.

Answer options

  1. 240\,\text{Н}
  2. 120\,\text{Н}
  3. 60\,\text{Н}
  4. 360\,\text{Н}
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Given
S_2
300\,\text{см}^{2}
S_1
6\,\text{см}^{2}
m
600\,\text{кг}
g
10\,\text{Н/кг}
Find

F_1

  1. F_2=mgF_2=600 \cdot 10F_2=6000\,\text{Н}

    The large piston must balance the weight of the load.

  2. k=\frac{S_2}{S_1}k=\frac{300}{6}k=50\,\text{раз}

    We calculate the gain in strength.

  3. F_1=\frac{F_2}{k}F_1=\frac{6000}{50}F_1=120\,\text{Н}

    A small piston requires k times less force.

Answer120\,\text{Н}

The theory behind these problems

Work through the topic if a problem will not give in.

Pressure from a solid object and Pascal's lawContact area, gas pressure and the transmission of pressure through liquids.Hydrostatic pressure and vesselsHow pressure depends on depth, and equilibrium in communicating vessels.Atmospheric pressure and hydraulicsBarometers, manometers, pumps and the force gain of a hydraulic press.

Formulas by topic

Look up symbols and units of measurement in the reference.

Height and depthh · мPressurep · ПаDensityρ · кг/м³