Answers · Pressure

Pressure — return questions key

Check meaning, not exact wording. These are teaching points, not official Cambridge marks.

1. PRR1 · Return

Model answer. (a) Total area = 4 × 2.0 cm² = 8.0 cm² = 8.0 ÷ 10 000 = 0.00080 m². (b) F = pA = 2 500 000 Pa × 0.00080 m² = 2000 N. (c) Yes, the wheels dent the floor. The piano, at 2500 N, is heavier than the 2000 N the floor can carry on this area: its pressure is 2500 N ÷ 0.00080 m² = 3 125 000 Pa, more than 2 500 000 Pa. (d) Stand each wheel in a wide, flat cup, or the piano on a wide board. The weight is the same, but it is spread over a much larger area, so the pressure on the floor is much smaller.

A full answer includes

(a) Method: I multiplied 2.0 cm² by 4 to get 8.0 cm².
(a) Method: I divided 8.0 cm² by 10 000.
(a) Answer: I wrote 0.00080 m².
(a) I did not divide by 100, which gives 0.080 m².
(b) Method: I multiplied 2 500 000 Pa by 0.00080 m².
(b) Answer: I wrote 2000 N.
(c) I wrote that the wheels dent the floor.
(c) I said 2500 N is more than 2000 N (or that 2500 N ÷ 0.00080 m² = 3 125 000 Pa, more than 2 500 000 Pa).
(c) I did not compare the weight on one wheel with 2000 N.
(d) I suggested something wide and flat under each wheel, such as a cup, a mat or a board.
(d) I said the same weight is then spread over a larger area.
(d) I said a larger area gives a smaller pressure on the floor.

Also accept

(a) 0.00080 m² (b) 2000 N (c) yes: 2500 N is more than 2000 N (d) a wide cup, mat or board under the wheels: a larger area gives a smaller pressure

Watch for

“The total area is 8.0 ÷ 100 = 0.080 m².” — There are 100 × 100 = 10 000 cm² in 1 m², so divide by 10 000.
“Each wheel carries only 2500 ÷ 4 = 625 N, less than 2000 N, so the floor holds.” — 2000 N is the largest weight on all four wheels together, so compare it with the whole weight, 2500 N.
“A mat under the wheels makes the piano lighter.” — The weight stays 2500 N; the mat only spreads it over a larger area.

2. PRR2 · Return

Model answer. (a) Less dense. Y is deeper than X, so if L were water, the pressure at Y would be greater. The pressure is only the same, so each metre of L must add less pressure than water does: L is less dense than water. (b) ∆p = 30.4 kPa = 30 400 Pa. ∆h = ∆p ÷ (ρg) = 30 400 Pa ÷ (1240 kg/m³ × 9.8 N/kg) = 30 400 ÷ 12 152 = 2.5016… m ≈ 2.50 m.

A full answer includes

(a) I wrote that liquid L is less dense than water.
(a) I said Y is deeper than X.
(a) I said that in water, the greater depth at Y would give a greater pressure.
(a) I said only a less dense liquid gives no more pressure at the greater depth.
(a) I did not write that L is denser because Y is deeper.
(b) Method: I changed 30.4 kPa into 30 400 Pa.
(b) Method: I divided 30 400 Pa by (1240 kg/m³ × 9.8 N/kg).
(b) Answer: I wrote 2.50 m.
(b) I did not leave the pressure in kPa, which gives 0.00250 m.

Also accept

(a) less dense: Y is deeper but has no more pressure (b) 2.50 m

Watch for

“Y is deeper, so liquid L must be denser.” — In water, Y’s greater depth would give a greater pressure. To give only the same pressure, L must be less dense.
“∆h = 30.4 ÷ (1240 × 9.8) = 0.00250 m.” — Change kPa into Pa first: 30.4 kPa = 30 400 Pa.