Pressure Worked ExamplesIGCSE Physics 0625

Pressure is force per unit area, so divide each weight by its area.

Work out each pressure.

C exerts the greatest pressure even though C is the lightest, because the contact area is the smallest. D is the tempting wrong answer, since D has the greatest weight.

(a) The pressure is greater at Y, because Y is deeper. The pressure beneath the surface of a liquid increases with depth — there is a greater height of water above Y pressing down.

(b) Near the bottom of the dam the water is deepest, so the pressure there is greatest. Greater pressure means a greater force on each square metre of the dam wall, so the lower part of the wall must be stronger (thicker) to withstand it.

(c) At the same depth, a denser liquid exerts a greater pressure, so the pressure at Y increases.

The force on the floor is the wardrobe's weight.

Find the total area of all four feet and convert to square metres ($1\text{ cm}^2 = 1 \times 10^{-4}\text{ m}^2$).

Use $p = \frac{F}{A}$.

(a) Convert the head area to m²: $1\text{ cm}^2 = 1 \times 10^{-4}\text{ m}^2$.

(b) The same force acts at the point. Convert: $1\text{ mm}^2 = 1 \times 10^{-6}\text{ m}^2$, so $0.050\text{ mm}^2 = 5.0 \times 10^{-8}\text{ m}^2$.

(c) The force is the same at both ends, but the point has a very much smaller area, so the pressure there is very much greater ($2000$ times greater). The high pressure at the point is enough to push into the board; the low pressure spread over the wide head is not enough to hurt the thumb.

Start from the equation you recall.

Divide both sides by $\rho g$ to make $\Delta h$ the subject.

So the answer is B. A quick check: a greater pressure increase means a greater depth, so $\Delta p$ must be on the top — this rules out C. D puts $g$ on the wrong side.

(a) The change in depth is $32 - 5.0 = 27\text{ m}$. Use the equation for the change in pressure beneath the surface of a liquid.

Evaluate.

(b) Rearrange $p = \frac{F}{A}$ to $F = pA$, using the unrounded pressure change.

When the student sucks, air is removed from the straw, so the pressure inside the straw falls below atmospheric pressure.

The atmosphere still presses on the surface of the juice in the glass. There is now a pressure difference, so there is a resultant force pushing juice from the higher pressure (the glass) to the lower pressure (inside the straw). The answer is C.

A has the pressure difference the wrong way round — that would push juice down. B is the common misconception: a gas at low pressure cannot "pull"; the juice is pushed from outside. D is irrelevant — the density does not change.

(a) Calculate each pressure, using weight $= mg$ as the force.

Divide the car's pressure by the tractor's, as the question asks.

Check with ratios: the tractor's weight is $3.0$ times the car's, but its area is $15$ times the car's. Pressure is proportional to force and inversely proportional to area, so the car's pressure is $\frac{15}{3.0} = 5.0$ times greater.

(b) The car is more likely to sink. Although it is lighter, its weight acts on a much smaller area, so it exerts five times the pressure on the mud.

(a) Count the squares: whole squares plus the part squares that are more than half inside.

Each square has an area of $2\text{ cm} \times 2\text{ cm} = 4\text{ cm}^2$.

(b) She stands on two feet, so double the area and convert to m².

The force is her weight.

Use $p = \frac{F}{A}$.

(c) Use paper with smaller squares (for example $1\text{ cm}$ or $1\text{ mm}$ squares), so that fewer squares are only partly inside the outline. (Alternatively: draw round the part of the sole that actually touches the floor, since the heel and toe gap may not be in contact.)

(a) At zero depth the sensor is at the surface, so it measures only the atmospheric pressure. Read the intercept. Note that the vertical axis starts at $90\text{ kPa}$, not zero.

(b) Go up from $2.5\text{ m}$ to the line and across.

(c) The pressure due to the liquid increases by $\rho g$ for every metre of depth, so the gradient of the line is $\rho g$. Use a large triangle.

Set the gradient equal to $\rho g$ and solve.

(a) Each layer adds its own $\rho g h$. Work out the pressure due to the oil first.

Then the pressure due to the water layer.

Add them.

(b) Add the atmospheric pressure acting on the surface.

(c) Two straight-line sections. From the surface to $0.40\text{ m}$ the pressure rises steadily from zero with gradient $850 \times 9.8 = 8330\text{ Pa/m}$. Below $0.40\text{ m}$ it continues to rise in a straight line, but steeper (gradient $9800\text{ Pa/m}$), because water is denser than oil.

(a) Find the volume, then the mass, then the weight.

Weight:

(b) The weight is the same on every face, so the greatest pressure is on the smallest face ($0.10 \times 0.060 = 0.0060\text{ m}^2$), standing on its end.

The least pressure is on the largest face ($0.20 \times 0.10 = 0.020\text{ m}^2$), lying flat.

(c) For a block of base area $A$ and height $h$: $W = mg = \rho V g = \rho (Ah) g$. Divide by the base area.

Check with the brick on its end ($h = 0.20\text{ m}$): $2000 \times 9.8 \times 0.20 = 3920\text{ Pa}$, as in (b). For the pillar: