Practical Skills and Data Analysis

This problem involves Fizeau's historic experiment to measure the speed of light using a rotating cogwheel.

Understanding the experimental setup:

The experiment consists of:
- A light source that sends a beam toward a mirror 7 km away
- A rotating cogwheel with 720 teeth placed in the path of the light
- Light passes through the gaps between teeth, reflects off the mirror, and returns
a) Why extinction occurs at a particular rotation rate:

For extinction to happen, the timing must be just right: 1. Light passes through a gap in the cogwheel 2. Light travels to the mirror (7 km away) and back (7 km return) 3. Total distance traveled = 14 km 4. During this travel time, the cogwheel rotates 5. When the light returns, a tooth (not a gap) must be in the path

This blocks the returning light, causing the observed extinction (reduction in intensity).

b) Why there are multiple extinction rates:

The key insight is that the cogwheel doesn't need to rotate by exactly one tooth position. It could rotate by:
- 1 tooth + 1 gap (minimum for extinction)
- 2 teeth + 2 gaps
- 3 teeth + 3 gaps
- And so on...
Each complete "tooth + gap" pair takes the same amount of time to pass, so multiple rotation speeds will cause extinction.

c) Calculating the teeth crossing rate: At 283 rps: Number of teeth per second = $283 \times 720 = 203,760$ teeth/second

At this speed, if $n + \frac{1}{2}$ teeth cross the beam, extinction occurs.

At 313 rps: Number of teeth per second = $313 \times 720 = 225,360$ teeth/second How many teeth cross at 313 rps?

Since we're at the next extinction speed, one additional tooth has crossed: $$\text{Teeth crossing} = n + \frac{1}{2} + 1 = n + \frac{3}{2}$$

Difference in crossing rate: $$225,360 - 203,760 = 21,600 \text{ extra teeth per second}$$ Time for one extra tooth to cross: $$t = \frac{1}{21,600} = 4.63 \times 10^{-5} \text{ s}$$

This time interval equals the round-trip travel time of the light!

d) Calculating the speed of light:

The light travels a total distance of 14 km (7 km each way) in the time calculated above.

$$c = \frac{\text{distance}}{\text{time}} = \frac{14,000 \text{ m}}{4.63 \times 10^{-5} \text{ s}}$$

$$c = 3.0 \times 10^8 \text{ m/s}$$

Final answer: The speed of light is approximately $3.0 \times 10^8$ m/s, which matches the modern accepted value!

This problem involves temperature measurement with thermistors vs mercury thermometers.

Understanding the two types of thermometers: Mercury thermometer:
- Linear relationship between volume and temperature
- Mercury expands uniformly with temperature
- Scale is linear: equal divisions represent equal temperature changes
Thermistor thermometer:
- Resistance decreases with temperature (NTC thermistor)
- Non-linear relationship: resistance changes rapidly at high T, slowly at low T
- Need to calibrate against known reference points
Given data:
- At $0^{\circ}$C: $R = 1020 \Omega$
- At $100^{\circ}$C: $R = 154 \Omega$
a) Resistance vs. temperature graph: Actual thermistor behavior:

The relationship is exponential: $$R(T) = R_0 e^{\beta/T}$$

Where $T$ is absolute temperature (Kelvin) and $\beta$ is a material constant.

Graph characteristics:
- Y-axis: Resistance ($\Omega$)
- X-axis: Temperature ($^{\circ}$C)
- Curve starts at $(0, 1020)$ and ends at $(100, 154)$
- Shape: Exponential decay - concave downward
- Steepest at low temperatures
- Flattens out at high temperatures
b) Linear thermistor scale: The approximation:

Instead of the actual exponential curve, we draw a straight line between the two calibration points:
- Line connects $(0, 1020)$ to $(100, 154)$
- This creates a "linearized" temperature scale
Equation of line: Using two-point form: $$R - 1020 = \frac{154 - 1020}{100 - 0}(T - 0)$$

$$R(T) = 1020 - 8.66T$$

Where $T$ is in $^{\circ}$C and $R$ is in $\Omega$.

c) Resistance change per degree:

From the linear scale equation: $$R(T) = 1020 - 8.66T$$

The coefficient of $T$ tells us the rate of change:

$$\frac{\Delta R}{\Delta T} = -8.66 \text{ } \Omega/^{\circ}\text{C}$$

The negative sign means resistance decreases as temperature increases.

d) Measuring body temperature: Given:
- Measured resistance: $R_{body} = 292 \Omega$
Change from $0^{\circ}$C reference: $$\Delta R = R_{initial} - R_{measured} = 1020 - 292$$

$$\Delta R = 728 \text{ } \Omega$$

e) Calculated body temperature: Using linear scale: $$\Delta R = 8.66 \times \Delta T$$

$$728 = 8.66 \times T$$

$$T = \frac{728}{8.66} = 84.0^{\circ}\text{C}$$

This is clearly wrong!

Normal human body temperature is about $37^{\circ}$C (or $98.6^{\circ}$F). A reading of $84^{\circ}$C would be fatal!

Why the error?

The linear scale is a poor approximation because: 1. Thermistor response is highly non-linear 2. Most resistance change happens at low temperatures 3. At body temperature range ($30-40^{\circ}$C), the curve is flatter 4. Linear extrapolation overestimates temperature significantly

Correct approach:

For accurate readings, you'd need to:
- Use the actual exponential calibration curve
- Or calibrate specifically in the body temperature range
- Or use a lookup table/chart
This demonstrates the importance of understanding sensor non-linearity in practical measurements!

a) Diagram should show Earth orbiting Sun counter-clockwise, with Jupiter outside. Point A is where Earth moves away from Jupiter (maximum delay), Point B is where Earth moves toward Jupiter (minimum delay).

b) Speed of Earth: $v = 2\pi \times 1.5 \times 10^{11} / (365 \times 24 \times 3600) = 29900$ m/s

Distance traveled in 42h 28m 42s: $d = 29900 \times (42 \times 3600 + 28 \times 60 + 42) = 4.57 \times 10^{9}$ m

Light travels this extra distance in 15 s: $c = 4.57 \times 10^{9} / 15 = 3.0 \times 10^{8}$ m/s

c) Graph shows sine/cosine curve with period 1 year, with A marked at maximum (Earth moving away from Jupiter) and B at minimum (Earth moving toward Jupiter).

a) Straight line from (0,0) to ($t$, $v_{f}$). Average speed marked at half height: $v_{av}(t) = \frac{v_{f}}{2}$

b) $v_{av}(t)$ is the area under the line (= distance travelled = $\frac{1}{2} v_{f} t$) divided by $t$ to give $\frac{v_{f}}{2}$.

c) Graph of $v$ against $h$ (height): $v = \sqrt{2gh}$, so $v \propto \sqrt{h}$ (curve with decreasing gradient)

d) (i) $v_{av}(s)$ is the area under the $v$-$h$ curve divided by $h$.
(ii) Area = $\int_{0}^{h} \sqrt{2gh} \, dh = \frac{2}{3} \sqrt{2g} h^{\frac{3}{2}}$. So $v_{av}(s) = \frac{\frac{2}{3} \sqrt{2g} h^{\frac{3}{2}}}{h} = \frac{2}{3} \sqrt{2gh} = \frac{2}{3} v_{f}$.

This problem involves Newton's Law of Cooling and experimental calorimetry.

Understanding the cooling curve:

When heating an object and then letting it cool:
- Temperature rises during heating phase
- Temperature exponentially decays toward room temperature during cooling
- The curve depends on power input, heat loss, and object properties
a) What affects $k$:

The constant $k$ in Newton's cooling law: $$\frac{dT}{dt} = -k(T - T_0)$$

depends on:
- Surface area: Larger area = faster heat transfer
- Material properties: Thermal conductivity, emissivity
- Surface characteristics: Color, texture
- Insulation: Any lagging or insulation around the object
NOT dependent on:
- Temperature of object
- Color of object (for same surface properties)
b) Why maximum occurs after heating stops:

Temperature sensor is placed a distance from heater. - Heat takes time to conduct through sample
- Temperature at sensor lags behind temperature near heater
- When heating stops, heater continues to warm sample briefly
- Maximum $T$ occurs when heat from heater stops arriving
This is a common issue in calorimetry called "thermal lag."

c) Area $A_2$ represents thermal energy lost:

Graph shows temperature vs. time. The area under the curve above room temperature represents $\int(T - T_0)dt$, which is proportional to thermal energy lost.

Since $\frac{dQ}{dt} = k(T - T_0)$: $$Q_{lost} = \int k(T - T_0)dt \propto A_2$$

d) Why temperature ratio equals area ratio:

Both heating and cooling follow Newton's law with the same $k$. - Heating: energy in = energy out + energy stored
- Cooling: all stored energy is lost
Geometric similarity of the heating/cooling curves means the ratio of temperature changes equals the ratio of areas.

e) Calculating specific heat capacity:

From energy balance: $$P\Delta t = mc\Delta T + Q_{lost}$$

Where $\Delta T$ is temperature rise above room temperature at peak.

$$c = \frac{P\Delta t - Q_{lost}}{m\Delta T}$$

Measure from graph: power $P$, time $\Delta t$, areas $A_1, A_2$, temperature rise $\Delta T$ and decay $\Delta T_3$.

This problem involves estimating molecular size from oil film spreading.

Given:
- Oil volume: $V = 0.1$ mm$^{3}$ $= 0.1 \times 10^{-9}$ m$^{3}$
- Patch area: $A = 1000$ cm$^{2}$ $= 1000 \times 10^{-4}$ m$^{2}$ $= 0.1$ m$^{2}$
Understanding the phenomenon:

Oil molecules are amphiphilic - they have:
- Hydrophilic (water-loving) head
- Hydrophobic (water-fearing) tail
When oil touches water:
- The hydrophilic heads stick to the water surface
- The hydrophobic tails point upward
- The oil spreads into a monolayer (one molecule thick)
Film thickness calculation:

Assuming the oil spreads into a uniform film: $$V = A \times h$$

where $h$ is the film thickness (molecular length).

$$h = \frac{V}{A} = \frac{0.1 \times 10^{-9}}{0.1}$$

$$h = 10^{-9} \text{ m} = 1 \text{ nm}$$

Physical significance:

This is a classic experiment for determining molecular size! The result of 1 nm is typical for oil molecules, confirming that matter is made of discrete particles with dimensions on the nanometer scale.

Answer: $h = 10^{-9}$ m $= 1$ nm

This problem involves geometry and horizon calculations.

Part (a) - Horizon distance: i) Diagram: Shows Earth's curved surface with observer at height $h$, line of sight tangent to surface at horizon point. ii) Horizon distance formula:

From right triangle geometry: $$(R + h)^{2} = R^{2} + d^{2}$$

$$R^{2} + 2Rh + h^{2} = R^{2} + d^{2}$$

For $h \ll R$: $d^{2} \approx 2Rh$

$$d \approx \sqrt{2Rh}$$

Angle subtended: $$\theta = \frac{d}{R} = \sqrt{\frac{2h}{R}}$$

Part (b) - Angle difference:

For lion ($h_{l} = 0.8$ m): $$\theta_{l} = \sqrt{\frac{2 \times 0.8}{6.37 \times 10^{6}}} = 5.0 \times 10^{-4} \text{ rad}$$

For giraffe ($h_{g} = 3.5$ m): $$\theta_{g} = \sqrt{\frac{2 \times 3.5}{6.37 \times 10^{6}}} = 1.05 \times 10^{-3} \text{ rad}$$

$$\Delta \theta = \theta_{g} - \theta_{l} = 5.5 \times 10^{-4} \text{ rad}$$

Part (c) - Time difference:

Earth rotates $2\pi$ radians in 24 hours: $$\text{Time} = \frac{\Delta \theta}{2\pi} \times 24 \times 3600 \text{ seconds}$$

$$t = \frac{5.5 \times 10^{-4}}{2\pi} \times 86400 \approx 7.5 \text{ seconds}$$

Giraffe sees sunset at 8:00:08 pm (about 7.5 seconds AFTER the lion)

Physical insight:

Being taller lets you see farther over the horizon! The giraffe sees the sun set later because it can "see around" the curve of the Earth slightly more than the lion can.

This problem involves estimation and molecular statistics.

Understanding the mixing assumption:

Water molecules that Caesar drank have been thoroughly mixed through Earth's oceans over 2000+ years. We can assume uniform distribution.

Part (a) - Step-by-step derivation: i) Molecules Caesar consumed:

Volume per day: $V_{day}$ Days in lifetime: $t$ Total volume: $V_{total} = t \times V_{day}$

Moles of water: $n = \frac{\rho V_{total}}{M}$ Molecules: $N_{Caesar} = n \times N_{A} = \frac{\rho t V_{day}}{M} \times N_{A}$

ii) Molecules in oceans:

Ocean volume: $V_{ocean} = f \times 4\pi R_{E}^{2} \times d$

Molecules: $N_{ocean} = \frac{\rho V_{ocean}}{M} \times N_{A}$

iii) Molecules in mug: $$N_{mug} = \frac{\rho V_{mug}}{M} \times N_{A}$$ Part (b) - Caesar molecules in your mug:

Fraction of ocean molecules that Caesar drank: $$f_{Caesar} = \frac{N_{Caesar}}{N_{ocean}} = \frac{t V_{day}}{f \cdot 4\pi R_{E}^{2} d}$$

Expected number in your mug: $$N = N_{mug} \times f_{Caesar}$$

$$N = \frac{\rho V_{mug}}{M} N_{A} \times \frac{t V_{day}}{f \cdot 4\pi R_{E}^{2} d}$$

$$N = V_{mug} \left(\frac{\rho N_{A}}{M}\right) \frac{t V_{day}}{f \cdot 4\pi R_{E}^{2} d}$$

Numerical estimate:

Using typical values: $N \approx 10^{8}$ molecules

Physical insight:

You probably drink about 100 million water molecules that Caesar drank! This shows how incredibly many molecules are in even small amounts of matter, and how thoroughly mixed Earth's water has become over time.

This problem involves rocket dynamics and atmospheric drag.

Part (a) - Event at 50 km:

The speed reaches a MAXIMUM at about 50 km altitude, then decreases. This is when:
- Rocket motor SWITCHES OFF
- Engine stops providing thrust
- Rocket continues upward due to momentum
- Gravity now slows it down
Part (b) - Acceleration calculation:

Using data from graph (55 km onwards):
- $u = 1000$ m/s at 55 km
- $v = 600$ m/s at 88 km
- $s = 33000$ m
Using $v^{2} = u^{2} + 2as$: $$600^{2} = 1000^{2} + 2a \times 33000$$

$$360000 = 1000000 + 66000a$$

$$-640000 = 66000a$$

$$a = -9.7 \text{ m/s}^{2}$$

Comment:

This is essentially free-fall acceleration ($g = 9.8 \text{ m/s}^{2}$ downward), confirming the rocket is coasting with minimal thrust or air resistance at this altitude.

Part (c) - Max Q position:

Max Q (maximum dynamic pressure) occurs when: $$q = \frac{1}{2}\rho v^{2}$$

is maximum.

This depends on:
- Velocity $v$ (increases with time)
- Air density $\rho$ (decreases with height)
Max Q occurs at the balance point where velocity is high but air density is still significant.

Looking at the graph:
- Velocity peaks around 50 km
- Air density decreases exponentially with height
Answer: Max Q occurs around 30-50 km altitude

This is typically the most stressful part of the launch for the vehicle structure.