A mercury barometer reads 0.76 m. What is atmospheric pressure? (ρ of mercury = 13 600 kg m−3, g = 9.81 m s−2)
- A 1.01 kPa
- B 10.3 kPa
- C 76 kPa
- D 101 kPa
Show answer
Answer: D
Free Forces, density & pressure questions for AS Level Physics (9702), part of Mechanics & materials. Each question has the answer and the marking points an examiner would look for.
Generate a full Forces, Density & Pressure worksheet →All Mechanics & materials questionsA mercury barometer reads 0.76 m. What is atmospheric pressure? (ρ of mercury = 13 600 kg m−3, g = 9.81 m s−2)
Answer: D
Which condition is needed for an object to be in equilibrium?
Answer: B
Which statement describes centre of gravity?
Answer: D
A block floats in water with part of its volume submerged.
Explain the origin of the upthrust on the block.
State Archimedes’ principle.
The block has density 750 kg m−3. Calculate the fraction of its volume below the surface of water (density 1000 kg m−3).
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | pressure greater at bottom than top pressure in a fluid increases with depth B1 so pressure on the bottom surface is greater than on the top surface B1 | 2 |
| (b) | upthrust = weight of fluid displaced upthrust equals the weight of fluid displaced B1 | 1 |
| (c) | 0.75 weight = upthrust: ρbV = ρwVsub C1 0.75 A1 | 2 |
A solid cube of side 0.10 m and density 700 kg m−3 floats in water of density 1000 kg m−3.
Define density.
Calculate the weight of the cube.
Explain the origin of the upthrust on the cube.
Use your answer to (b) to determine the depth of the cube below the water surface.
Calculate the extra downward force needed to push the cube so that it is just completely submerged.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | mass per unit volume mass per unit volume B1 | 1 |
| (b) | 6.87 N m = 700 × 0.103 C1 W = 6.87 N A1 | 2 |
| (c) | pressure increases with depth; force on bottom face greater than on top pressure in a fluid increases with depth B1 so the upward force on the lower face is greater than the downward force on the upper face B1 | 2 |
| (d) | 0.0700 m upthrust = weight C1 1000 × 0.102 × d × 9.81 = 6.87 C1 d = 0.0700 m A1 | 3 |
| (e) | 2.94 N full upthrust = 9.81 N C1 extra force = 2.94 N A1 | 2 |
A uniform beam of weight 40 N and length 1.2 m is hinged at a wall. It is held horizontal by a cable attached to its far end, making 30° with the beam.
State the principle of moments.
Calculate the tension in the cable.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | sum clockwise = sum anticlockwise about any point for equilibrium, sum of clockwise moments about a point = sum of anticlockwise moments about the same point B2 | 2 |
| (b) | 40 N moments about hinge: T sin 30° × 1.2 = 40 × 0.6 C1 correct substitution C1 40 N A1 | 3 |
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