Oxford Revise AQA GCSE Physics | Chapter P10 answers

P10: Pressure in liquids and gases

Answers

Extra information

Mark

AO / Specification reference

01.1

\(\rm{pressure} = \frac{\rm{force}}{\rm{area}}\) \(\rm{accept\ P} = \frac{\rm{F}}{\rm{A}}\)

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AO1

4.5.5.1.1

01.2

\(
\begin{array}{lll}
\rm{pressure}& = & \frac{2000}{0.02}\\
& = & 100\ 000\\
\rm{Pa\ or\ N/m^2}
\end{array}
\)

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AO1

AO2

4.5.5.1.1

01.3

at right angles/perpendicular to the wall

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AO1

4.5.5.1.1

01.4

the direction of the force is the same

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AO2

4.5.5.1.1

02.1

the force of the Earth/weight/gravity

the force of the water/upthrust

do not accept ‘acceleration due to gravity’ or ‘g’

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AO1

AO2

4.5.5.1.2

02.2

the upthrust

balances/is equal to the weight

accept ‘no resultant force’

do not accept ‘no forces acting’

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AO2

4.5.5.1.2

02.3

any three from:

  • (as the child adds sand) the boat floats deeper in the water/more of the boat is submerged in the water
  • the pressure exerted by the water on the boat increases with depth (so) the upthrust increases (as the area is the same, force \(\propto\) pressure)
  • the boat still floats because the larger weight is balanced by the larger upthrust

or words to that effect

one point per correct answer up to a maximum of three points

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AO1

AO2

4.5.5.1.2

03.1

X, Y, W

one mark for X before Y

one mark for Y before W

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AO3

4.5.5.1.2

03.2

the weight of Y is greater

Y moves/floats further into the liquid before the pressure is big enough

for the upthrust to balance the weight

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AO3

4.5.5.1.2

03.3

\(\rm{pressure} = \frac{\rm{force}}{\rm{area}}\) \(\rm{accept\ P} = \frac{\rm{F}}{\rm{A}}\)

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AO1

03.4

\(
\begin{array}{lr}
\rm{surface\ area} = 0.01 \times 0.01 = 1 \times 10^{-4}\ \rm{m}^2\\
\rm{pressure} = \frac{0.015}{1 \times 10^{-4}}\\
= 150\ \rm{N/m}^2\ \rm{\textbf{or}\ Pa}
\end{array}
\)

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AO1

AO2

04.1

centre of mass

accept centre of gravity

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AO2

4.5.1.3

04.2

appropriate diagram with scale given

clear identification of vertical component

2.6 N

allow 2.2 – 3.0 N

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AO2

4.5.1.4

04.3

2.6 N

allow answers using range 2.2 – 3.0 N from last question

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AO2

4.5.1.4

04.4

appropriate diagram with scale given

clear identification of resultant of two forces using a parallelogram

3.1 N

allow 2.7 – 3.5 N

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AO2

4.5.1.4

05.1

independent – volume of water in the bottle

dependent – distance travelled by jet

accept ‘height of water column’

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AO2

4.5.5.1.2

05.2

a graph that shows increasing volume produces increasing distance

as the volume/height decreases, the pressure exerted by the column decreases

so the force on the water is less

so the jet travels a smaller distance

allow reverse argument

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AO2

4.5.5.1.2

05.3

the ruler does not start in the right place

move the ruler so that zero is next to the edge of the bottle where the hole is.

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AO3

06.1

\(
\begin{array}{lr}
\rm{pressure\ increase} = \rm{depth\ \times\ density\ \times gravitational\ field\ strength}\\
= 10 \times 1 \times 10^3 \times 9.8.\\
= 98\ 000\ \rm{Pa}\\
= \frac{98\;000}{1000}\ \rm{kPa} = 98\ \rm{kPa}\ (\rm{approx.} 100\ \rm{kPa})\\
\end{array}
\)

accept P = h × ρ × g

must explicitly convert to kPa

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AO2

4.5.5.1.2

06.2

\(
\begin{array}{lr}
\rm{water\ pressure} = 405\ \rm{kPa} – 101\ \rm{kPa} = 304\ \rm{kPa}\\
\rm{using\ diver’s\ rule}, \frac{304}{100} = 3.040\\
3.04 \times 10 = 30.4\ \rm{m}
\end{array}
\)

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AO2

4.5.5

07.1

downwards force – interaction of the ball with the Earth

upwards force – interaction of the ball with the ground

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AO2

4.5.1.2

07.2

work = force distance

allow W = F × d

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AO1

4.5.2

07.3

\(
\begin{array}{lr}
19 = \rm{friction}\ 4.6\\
\rm{friction} = \frac{19}{4.6}\\
= 4.13\\
= 4.1\ \rm{N}\ (\rm{to\ two\ significant\ figures})
\end{array}
\)

answer to two significant figures

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AO2

4.5.2

07.4

on ball, arrow downwards labelled weight

arrow horizontally to left labelled air resistance

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AO2

4.5.1.2

4.5.1.4

07.5

the force of the ground on the ball/reaction force

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AO2

4.5.1.2

4.5.1.4

08.1

air molecules collide with a surface

and produce a force

\(\rm{pressure} = \frac{\rm{force}}{\rm{surface\ area}}\)

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AO1

AO2

4.5.5.2

08.2

the atmospheric pressure increases

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AO1

4.5.5.2

08.3

no

there are more air molecules (and more weight) above (the phone)

so the pressure is greater

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AO1

AO2

4.5.5.2

09

sketch graph showing smooth curve decreasing pressure with height

approximately halving every 5 km

labelled axes

at higher altitudes there is less weight of air above that point and so less pressure is exerted.

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AO1

AO2

10.1

weight = mass gravitational field strength

accept W = mg

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AO1

4.5.1.3

10.2

80 × 9.8

= 784 N

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AO2

4.5.1.3

10.3

\(
\begin{array}{lr}
\rm{force\ on\ each\ spring} = \frac{784}{4}\\
= 196\ \rm{N}\\
\end{array}
\)

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AO2

4.5.3

10.4

\(
\begin{array}{lr}
\rm{force} = \rm{spring\ constant\ extension}\\
3.4\ \rm{cm} = 0.034\ \rm{m}\\
196 = \rm{spring\ constant}\ 0.034\\
\rm{spring\ constant} = \frac{196}{0.034}\\
= 5765\ \rm{N/kg}
\end{array}
\)

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AO1

AO2

4.5.3

10.5

the same

the spring has not deformed elastically

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AO2

4.5.3

11.1

pressure = height × density × gravitational field strength

= 1 × 103 × 9.8

= 9800 Pa

accept P = h × ρ × g

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AO2

4.5.5.1.2

11.2

the pressure gauge reads the pressure due to the column of water and the column of air above it )atmospheic pressure)

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AO2

AO2

4.5.5.1.2

11.3

pressure is proportional to density

if the salt content is higher, the density is higher

if the density is higher, the pressure will be higher

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AO3

4.5.5.1.2

12.1

any three from:

  • (if the bag is sealed at the bottom of the mountain) the gas inside the bag will be at atmospheric pressure
  • as he goes up the mountain the atmospheric pressure decreases /will be less than the pressure of the gas inside the bag (so the volume increases)
  • pressure is inversely proportional to volume
  • (so) if the volume increases by a factor of three, the pressure has decreased by a factor of three

one mark for each correct answer up to a maximum of three marks

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AO2

4.5.5.1.2

12.2

\(
\begin{array}{lr}
\frac{100}{3}\\
= 33\ \rm{kPa}
\end{array}
\)

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AO2

12.3

if the density decreases the pressure that you calculate at a given height is actually less than that predicted assuming density is constant

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AO3

4.5.5.1.2

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