This page from an O-Level Physics topical paper presents a series of problems on pressure, including manometer readings and Boyle's Law calculations.

Let's solve each of the problems shown in your image one by one, with clear explanations.

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Question 9: Manometer connected to a gas supply

Diagram: A U-tube manometer is connected to a gas supply. The left side is open to air (atmospheric pressure), and the right side is connected to the gas. The water level on the left is higher than the right by 4 cm, and the total height difference between the two arms is 8 cm.

> Given:
> - Height difference in water column = 8 cm
> - The gas side has lower water level → so gas pressure is greater than atmospheric pressure
> - The difference in levels is 8 cm of water

> Question: What is the pressure of the gas?

#### ✔ Solution:
In a manometer:
- If the liquid level is lower on the side connected to the gas, then the gas pressure is higher than atmospheric pressure.
- The excess pressure is equal to the height difference of the fluid column.

Here, the height difference is 8 cm of water, and since the gas side is lower, the gas pressure is 8 cm of water greater than atmospheric pressure.

> Answer: A) 8 cm of water more than atmospheric pressure

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Question 10: J-shaped tube with trapped air

Given:
- Initial volume of trapped air = 3.0 cm³
- Initially, mercury levels are equal → pressure inside = atmospheric pressure = 76 cm Hg
- More mercury is added until the levels differ by 76 cm (diagram 2)
- Atmospheric pressure remains 76 cm Hg
- Temperature constant

We need to find the new volume $ V $ of trapped air.

#### Step 1: Use Boyle’s Law
Since temperature is constant:

$$
P_1 V_1 = P_2 V_2
$$

- $ P_1 = 76 $ cm Hg (initial pressure = atmospheric)
- $ V_1 = 3.0 $ cm³
- $ P_2 = ? $

Now, in diagram 2, the mercury level on the open side is 76 cm higher than the closed side. This means the pressure exerted by the trapped air must support this extra column.

So, pressure of trapped air = atmospheric pressure + 76 cm Hg = $ 76 + 76 = 152 $ cm Hg

Thus,
- $ P_2 = 152 $ cm Hg

Now apply Boyle’s Law:

$$
76 \times 3.0 = 152 \times V_2
$$

$$
V_2 = \frac{76 \times 3.0}{152} = \frac{228}{152} = 1.5 \text{ cm}^3
$$

> Answer: D) 1.5 cm³

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Question 11: Which column exerts greatest pressure on base?

We are given four columns of liquids with different densities and heights.

Pressure at the base of a liquid column:

$$
P = \rho g h
$$

But since $ g $ is constant, we can compare $ \rho h $ (density × height)

| Container | Liquid | Density (g/cm³) | Height (cm) | $ \rho h $ |
|----------|------------|------------------|-------------|--------------|
| A | Water | 1.0 | 25 | 1.0 × 25 = 25 |
| B | Mercury | 13.6 | 25 | 13.6 × 25 = 340 |
| C | Paraffin | 0.8 | 50 | 0.8 × 50 = 40 |
| D | Water | 1.0 | 50 | 1.0 × 50 = 50 |

Clearly, B has the highest $ \rho h $ → highest pressure.

Even though C and D have greater height, mercury's high density dominates.

> Answer: B

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Question 12: Gas compressed at constant temperature

Given:
- Initial volume $ V_1 = 2.0 \, \text{m}^3 $
- Initial pressure $ P_1 = 240 \, \text{kPa} $
- Final volume $ V_2 = 0.5 \, \text{m}^3 $
- Temperature constant → use Boyle’s Law

$$
P_1 V_1 = P_2 V_2
\Rightarrow P_2 = \frac{P_1 V_1}{V_2}
= \frac{240 \times 2.0}{0.5} = \frac{480}{0.5} = 960 \, \text{kPa}
$$

> Answer: D) 960 kPa

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Question 13: Balloon under bell jar with vacuum pump

Setup:
- Partially inflated balloon inside a bell jar
- Vacuum pump removes air from the jar → reduces pressure outside the balloon
- The balloon expands (volume increases)

What happens?

- Pressure inside the bell jar decreases (due to vacuum pump)
- The balloon expands → its internal pressure must be higher than the surrounding pressure
- But the pressure inside the balloon is determined by the elastic tension and the amount of air. As it expands, the internal pressure decreases slightly (since the same number of molecules occupy more volume), but the key point is that the external pressure drops faster, causing expansion.

However, the question asks about pressure changes:

- In the bell jar: pressure decreases (vacuum created)
- In the balloon: as it expands, the internal pressure decreases because the air inside expands (but not necessarily to zero — just less than initial)

But actually, the balloon expands because the external pressure drops. The pressure inside the balloon will also decrease (as volume increases, assuming no more air enters), but the net effect is that the pressure difference causes expansion.

But let's think carefully:

- The balloon is sealed → no air enters or leaves
- So when external pressure drops, the balloon expands → internal pressure decreases due to expansion
- However, the pressure inside the balloon is still greater than the outside pressure (that’s why it expands), but both pressures decrease

But the change:
- Bell jar pressure → decreases
- Balloon pressure → decreases (because volume increases, and temperature is constant → pressure drops)

Wait — but the balloon expands because the external pressure drops. The internal pressure must drop too, but less than the external pressure?

Actually, no — the internal pressure does decrease due to expansion.

For example, if you have a fixed amount of gas, and it expands, pressure drops.

So both:
- Pressure in bell jar → decreases
- Pressure in balloon → decreases

But the balloon expands because the outside pressure drops faster.

But the question is: what is the change in pressure?

> Answer: A) decrease, decrease

But wait — check options:

A) decrease, decrease
B) decrease, increase
C) increase, decrease
D) increase, increase

But if the balloon expands, and it's sealed, then internal pressure must decrease (by Boyle's law). So cannot increase.

And external pressure definitely decreases.

So A seems correct.

But some might argue that the balloon expands due to internal pressure being higher — but that's only initially. After expansion, the internal pressure is lower than before.

So the change is: both pressures decrease.

> ✔ Answer: A) decrease, decrease

But wait — let's double-check.

Suppose:
- Initial: balloon pressure = 100 kPa, atmospheric = 100 kPa → balloon doesn't expand
- Then pump removes air → atmospheric pressure drops to 80 kPa
- But balloon is elastic — it expands until internal pressure equals external pressure? No — actually, the balloon expands because the external pressure drops, and the internal pressure is now greater than external → so it expands.

But once expanded, the internal pressure drops due to larger volume.

So:
- Internal pressure decreases (from 100 to say 90 kPa)
- External pressure decreases (from 100 to 80 kPa)

So both decrease, but the balloon expands because internal pressure is still higher than external.

So yes, both pressures decrease.

> ✔ Answer: A

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Question 14: Mercury barometer carried to top of mountain

> "A simple mercury barometer is carried to the top of a mountain, where the atmospheric pressure is less than at its base."

What happens to the mercury column?

- Atmospheric pressure supports the mercury column
- At higher altitude, atmospheric pressure is lower
- So the height of mercury column decreases

But the question is not multiple choice here, but likely implied.

But since it's listed without options, probably just for context.

But if asked: what happens to the mercury column?

→ It becomes shorter

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Summary of Answers:

| Question | Answer | Explanation |
|--------|--------|-----------|
| 9 | A | Gas pressure is 8 cm water above atmospheric (right side lower) |
| 10 | D | Boyle’s Law: $ P_1 V_1 = P_2 V_2 $, $ P_2 = 152 $ cm Hg → $ V_2 = 1.5 $ cm³ |
| 11 | B | Mercury has high density → even with same height, much higher pressure |
| 12 | D | $ P_2 = \frac{240 \times 2.0}{0.5} = 960 $ kPa |
| 13 | A | Both pressures decrease: external due to vacuum, internal due to expansion |
| 14 | N/A | Mercury column height decreases due to lower atmospheric pressure |

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✔ All solved! Let me know if you'd like diagrams or further explanation.