3 free questions from this chapter (the full set has 30). Marking: Multi-correct: +4 all correct, +1 per correct option if none wrong, −2 if any wrong · Numerical: +4 / 0 · Matching: +3 / −1.
Question 1Level 2 · ExamSingle correct
A U-tube of uniform cross-section stands vertically with both limbs open to the atmosphere. It contains water (density 1000 kg m−3), and the water surfaces in the two limbs are initially level; call this the original level. Oil of density 800 kg m−3, which does not mix with water, is poured slowly into the right limb until the oil column is 10.0 cm long. No liquid leaves the tube, the oil stays entirely in the right limb, and the system is in equilibrium. Neglect surface-tension effects. The height of the free surface of the oil above the original level is ______. Exactly one option is correct.
A2.0 cm
B4.0 cm
C6.0 cm
D10.0 cm
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Answer: C
Let the oil-water interface in the right limb fall by x; then the water surface in the left limb rises by x (uniform tube).
Pressures at the level of the interface are equal (same liquid, water, connects the two points): ρog(0.10)=ρwg(2x), so 2x=0.10×800/1000=0.080 m and x=4.0 cm.
The oil top is at −x+10.0=+6.0 cm relative to the original level.
(The oil top is 2.0 cm above the left water surface; the left water surface rises 4.0 cm; these are the distractors.)
Difficulty: HardQuestion 2Level 3 · AdvancedSingle correct
An open rectangular tank, 2.0 m long (along the x-axis), 1.0 m wide and 2.0 m tall, contains water (density 1000 kg m−3) to a depth of 1.0 m when at rest. The tank is then given a constant acceleration with components 2.5 m s−2 along +x (forward, horizontal) and 2.0 m s−2 vertically upward. After the water settles into a steady state relative to the tank (no spilling), consider the gauge pressure at the bottom of the tank at the rear wall (the wall facing −x). Take g=10 m s−2. This gauge pressure is ______. Exactly one option is correct.
A14.5 kPa
B15.0 kPa
C12.5 kPa
D9.5 kPa
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Answer: A
In the tank frame the effective gravity has components ax=2.5 backward and g+ay=12 m s−2 downward. The free surface is perpendicular to it, so its slope is tanθ=ax/(g+ay)=2.5/12, rising toward the rear.
The volume is unchanged, so the surface still passes through depth 1.0 m at the middle; the depth at the rear wall is h=1.0+1.0×122.5=1.2083 m (less than 2.0 m, so no spilling).
Gauge pressure at a point below the free surface along the vertical: p=ρ(g+ay)h=1000×12×1.2083=14500 Pa.
Equivalently p=ρ(g+ay)h0+ρax(L/2)=12000+2500.
Distractors: ignoring the vertical acceleration gives 12.5 kPa; using slope ax/g with g+ay gives 15.0 kPa; the front wall gives 9.5 kPa.
Difficulty: Very hardQuestion 3Level 2 · ExamSingle correct
A block of ice of mass 500 g (density 917 kg m−3) has a stone of mass 50 g (density 2500 kg m−3) frozen inside it. The block floats in water (density 1000 kg m−3) in a vertical-walled beaker of inner cross-sectional area 100 cm2. The ice melts completely; the stone sinks to the bottom. Neglect evaporation and any temperature change of the water's density. As a result of the melting, the water level in the beaker ______. Exactly one option is correct.
Arises by 3.0 mm
Bstays unchanged
Cfalls by 5.0 mm
Dfalls by 3.0 mm
Show answer and solution
Answer: D
Initially the floating block displaces water of weight equal to its total weight: V0=(M+m)/ρw.
After melting, the ice becomes water of volume M/ρw and the sunken stone displaces only its own volume m/ρs: V1=M/ρw+m/ρs.
Δh=AV1−V0=−Am(ρw1−ρs1)=−0.0100.050(10−3−4×10−4)=−3.0×10−3 m.
The level falls by 3.0 mm. (The block does float: its mean density is about 973 kg m−3.) Using m/(Aρw) alone gives the distractor 5.0 mm.
Difficulty: Hard