Physics worksheet page titled "Topic 6 Pressure" featuring eight multiple-choice questions covering barometers, manometers, and pressure calculations.

Let's go through each question in Topic 6: Pressure and solve them step by step with explanations.

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Question 1:

> The diagrams show a simple mercury barometer. Which one shows the distance to be measured to find atmospheric pressure?

Answer: B

Explanation:
In a mercury barometer, atmospheric pressure supports a column of mercury. The height of the mercury column is measured from the surface of the mercury in the reservoir to the top of the mercury column inside the tube. This vertical distance (from the reservoir level to the top of the column) represents the atmospheric pressure.

- A: Incorrect — measures from the bottom of the tube.
- B: Correct — measures from the surface of mercury in the reservoir to the top of the mercury column.
- C: Incorrect — measures only part of the column.
- D: Incorrect — measures too low.
- E: Incorrect — includes the entire length of the tube, not just the column.

✔ So, B is correct.

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Question 2:

> Which of the following does NOT cause the height of the mercury column to vary?

Options:
A. changes in atmospheric pressure
B. changes in temperature of the mercury
C. changes in the value of *g*
D. evaporation of mercury from the reservoir
E. leakage of air into the tube

Answer: D

Explanation:
The height of the mercury column depends on:
- Atmospheric pressure (A): Yes, directly affects it.
- Temperature (B): Mercury expands/contracts with temperature → changes height.
- Gravity (*g*) (C): Since pressure = ρgh, changing *g* changes height.
- Air leakage (E): If air enters the tube, it reduces the vacuum → lowers the column.

But evaporation of mercury (D): Mercury vapor is negligible at room temperature and doesn’t significantly affect the column height because:
- The amount evaporated is tiny.
- It doesn't change the pressure balance in the tube.
- The system remains closed enough that vapor pressure is constant.

So, D is the correct answer — it does not significantly affect the height.

✔ Answer: D

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Question 3:

> In one minute, a diver breathes 1 litre of air at an atmospheric pressure of 100 kPa. To breathe in the same mass of air in one minute, how much air would he need to breathe when the total pressure on him under water is 300 kPa?

Answer: C (1 litre)

Wait! Let’s think carefully.

This is about Boyle’s Law: At constant temperature, pressure × volume = constant for a fixed mass of gas.

We are told:
- At 100 kPa, diver inhales 1 litre → this is the volume at surface pressure.
- Underwater, pressure = 300 kPa.
- But he needs to inhale the same mass of air in the same time.

So, since pressure has increased, the volume of air available per breath decreases if we assume same pressure in lungs (i.e., same as ambient).

But here's the key: To get the same number of molecules (mass), you must take in more volume at higher pressure.

But wait — actually, the amount of air (mass) inhaled is proportional to pressure × volume.

So:

At surface:
P₁ = 100 kPa, V₁ = 1 L → PV = 100 × 1 = 100 kPa·L

Underwater:
P₂ = 300 kPa, let V₂ = ?
For same mass: P₁V₁ = P₂V₂
→ 100 × 1 = 300 × V₂
→ V₂ = 100 / 300 = 1/3 L

So, he needs to breathe 1/3 litre of air at 300 kPa to get the same mass as 1 litre at 100 kPa.

Wait — but the question says: “how much air would he need to breathe... to breathe in the same mass of air”

So he must inhale more volume at lower pressure? No — he's underwater, so the air is compressed.

Actually, the air he inhales is already at 300 kPa, so to get the same number of moles, he must inhale less volume.

But the question asks: "how much air would he need to breathe" — meaning what volume of air at ambient pressure?

Yes — so at 300 kPa, he needs to inhale 1/3 litre to get the same mass as 1 litre at 100 kPa.

So answer should be A: ⅓ litre

Let’s check:

- Mass ∝ P × V
- So at 300 kPa, to get same mass as 100 kPa × 1 L, he needs V such that:
300 × V = 100 × 1 → V = 1/3 L

✔ So Answer: A (⅓ litre)

But let's double-check options:

A) ⅓ litre
B) ½ litre
C) 1 litre
D) 2 litres
E) 3 litres

So yes, A is correct.

Wait — but some might think: "He needs to breathe more air" — but no, because the air is denser under pressure.

You need less volume to get the same mass.

✔ Answer: A

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Question 4:

> In which of the following examples is the greatest pressure exerted?

A) barefooted person on beach
B) brick resting on ground
C) book on table
D) elephant standing on ground
E) knife cutting meat

Answer: E

Explanation:
Pressure = Force / Area

- Person: moderate force, large area → moderate pressure
- Brick: small force, moderate area → low pressure
- Book: small force, small area → still low
- Elephant: huge force, but large feet → moderate pressure
- Knife: very small area, high force concentrated → very high pressure

Even though the force may be small, the area is extremely small, so pressure is very high.

✔ So, E: knife cutting meat exerts the greatest pressure.

Answer: E

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Question 5:

> A tank 3 m long, 1 m wide, 0.5 m deep is filled with oil weighing 12,000 N. What is the pressure on the base due to the oil?

Given:
- Weight of oil = 12,000 N
- Base area = 3 m × 1 m = 3 m²
- Pressure = Force / Area = 12,000 N / 3 m² = 4,000 Pa

✔ Answer: A (4000 Pa)

Note: You could also use density and depth, but weight is given directly.

Answer: A

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Question 6:

> An Eskimo of mass 40 kg stands on snow-shoes with total contact area 0.5 m². Gravitational force = 10 N/kg.

So, weight = 40 × 10 = 400 N

Pressure = Force / Area = 400 N / 0.5 m² = 800 N/m²

✔ Answer: E (800 N/m²)

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Question 7:

> A mass of air occupies 12 m³ at normal pressure. If pressure is increased to 4 times without changing temperature, what is new volume?

Boyle’s Law: P₁V₁ = P₂V₂

P₂ = 4P₁ → V₂ = V₁ / 4 = 12 / 4 = 3 m³

✔ Answer: A (3 m³)

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Question 8:

> Diagram shows a U-tube manometer measuring gas pressure.

From diagram:
- Left side (gas side): 17 cm above reference
- Right side: 15 cm above reference
- Difference = 17 cm - 15 cm = 2 cm
- But there's a 5 cm difference shown in the tube?

Wait — look at the diagram:

It shows:
- Gas side: liquid level is 5 cm below the other side?
No — read carefully:

There’s a horizontal line at the bottom. From that:
- On left: 5 cm from bottom to liquid level
- On right: 15 cm from bottom to liquid level
- And the gas side is connected to the left arm.

Wait — better interpretation:

The difference in levels is shown as:
- The right side is higher than the left by (15 cm - 5 cm) = 10 cm?

Wait — no: both arms are open at the top? No — gas supply is on left.

Standard setup: gas pressure pushes down on left side → liquid level goes down on left, up on right.

So:
- Left arm (gas side): liquid level is lower
- Right arm: liquid level is higher

Difference = 15 cm – 5 cm = 10 cm

So the pressure of the gas = atmospheric pressure + pressure due to 10 cm water column.

But the question asks: What is the pressure of the gas?

But it doesn't give atmospheric pressure. However, the diagram likely implies that the difference is 10 cm, and we're to assume standard conditions or just report the excess.

But looking at the labels:
- Left side: 5 cm from bottom
- Right side: 15 cm from bottom
- So the difference is 10 cm of water.

Thus, gas pressure = atmospheric pressure + 10 cm H₂O

But since the question likely wants the gauge pressure or the difference, and options aren't given (in image), but in original paper, probably asks for the pressure in terms of cm water.

But since no options listed in your image, perhaps just state:

✔ The gas pressure is greater than atmospheric pressure by 10 cm of water.

So the excess pressure is 10 cm H₂O.

If asked for absolute pressure, we’d need atmospheric pressure, but typically in such questions, they want the difference.

So final answer: 10 cm of water pressure above atmospheric

But since no multiple choice shown in image, maybe skip.

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## ✔ Final Answers Summary:

| Q | Answer | Explanation |
|---|--------|-----------|
| 1 | B | Height from reservoir surface to top of mercury column |
| 2 | D | Evaporation doesn’t affect column height significantly |
| 3 | A | Boyle’s Law: P₁V₁ = P₂V₂ → V₂ = (100×1)/300 = 1/3 L |
| 4 | E | Knife has smallest area → highest pressure |
| 5 | A | Pressure = 12,000 N / 3 m² = 4,000 Pa |
| 6 | E | Pressure = 400 N / 0.5 m² = 800 N/m² |
| 7 | A | Volume = 12 / 4 = 3 m³ |
| 8 | 10 cm water | Difference in levels = 15 cm – 5 cm = 10 cm |

Let me know if you'd like a diagram explanation for Q8 or any clarification!