Let's break down these two physics problems step by step, focusing on the principles of gas laws and buoyancy. Understanding the behavior of gases under different conditions is key to solving these types of questions.
Problem 1: Change in Length of Air Column
We start with a glass tube containing a column of air that is 40 cm long, sealed at one end, and has a mercury column of 8 cm above it. The initial temperature is 27 degrees Celsius, and we need to find out how the length of the air column changes when the tube is inclined at 60 degrees and the temperature is raised by 30 degrees Celsius.
Step 1: Initial Conditions
First, let's establish the initial conditions:
- Length of air column (L1) = 40 cm
- Length of mercury column (h) = 8 cm
- Initial temperature (T1) = 27 °C = 300 K (converting to Kelvin)
- Final temperature (T2) = 27 °C + 30 °C = 57 °C = 330 K
- Atmospheric pressure (P0) = 76 cm of Hg
Step 2: Applying Gas Laws
According to Boyle's Law, the pressure and volume of a gas are inversely related at constant temperature. When the tube is inclined, the effective length of the air column will change due to the change in pressure exerted by the mercury column and the change in temperature.
The pressure exerted by the mercury column can be calculated as:
P1 = P0 - (h * density of Hg * g)
Where density of Hg is approximately 13.6 g/cm³ and g is the acceleration due to gravity (9.81 m/s²). However, since we are interested in the change in length, we can simplify our calculations by focusing on the ratios of the initial and final states.
Step 3: Volume and Temperature Relationship
Using Charles's Law, which states that V1/T1 = V2/T2, we can express the change in volume due to the temperature increase:
V1 = A * L1
V2 = A * L2
Thus, we have:
A * L1 / T1 = A * L2 / T2
Canceling A gives us:
L1 / T1 = L2 / T2
Substituting the values:
40 cm / 300 K = L2 / 330 K
Solving for L2 gives:
L2 = (40 cm * 330 K) / 300 K = 44 cm
Step 4: Change in Length
The change in length of the air column is:
ΔL = L2 - L1 = 44 cm - 40 cm = 4 cm
However, since the tube is inclined at 60 degrees, we need to consider the vertical component of the air column:
Effective length = L2 * sin(60°) = 44 cm * (√3/2) ≈ 38.2 cm
Thus, the increase in length of the air column when inclined is approximately 1.5 cm.
Problem 2: Air Bubble Rising in Water
Now, let's tackle the second problem regarding the air bubble that doubles in radius as it rises from the bottom of a lake to the surface.
Understanding the Situation
When the bubble rises, it experiences a decrease in pressure, which allows it to expand. The atmospheric pressure at the surface is equivalent to the pressure due to a 10 m column of water.
Step 1: Pressure at Depth
The pressure at a certain depth in a fluid can be calculated using the formula:
P = P0 + ρgh
Where:
- P0 = atmospheric pressure (10 m of water)
- ρ = density of water (approximately 1000 kg/m³)
- g = acceleration due to gravity (9.81 m/s²)
- h = depth of the lake
Step 2: Volume Relationship
When the bubble doubles in radius, the volume increases by a factor of 8 (since volume V = (4/3)πr³). According to Boyle's Law, P1V1 = P2V2. As the bubble rises, the pressure decreases, allowing the volume to increase.
Step 3: Solving for Depth
Let’s assume the pressure at the bottom of the lake is P1 and at the surface is P2. The pressure at the surface is atmospheric pressure (P2 = 10 m of water). The pressure at depth h is:
P1 = P2 + ρgh
Given that the bubble doubles in size, we can set up the equation:
P1 * V1 = P2 * V2
Substituting the volume change gives us:
P1 * V = P2 * 8V
Thus, P1 = 8P2. Since P2 = 10 m, we have:
P1 = 80 m of water.
Step 4: Finding Depth
Now, we can set up the equation:
80 m = 10 m + ρgh
Solving for h gives us:
h = (80 m - 10 m) / (ρg) = 70 m.
Therefore, the depth of the lake is 70 m, which corresponds to option b.
Both problems illustrate the fascinating interplay between pressure, volume, and temperature in gases and fluids. Understanding these principles helps us predict how systems behave under different conditions.
