University Physics I · unit UPI-15
Optional: Fluids
Connect the ideas, make your working visible, and test the result. Use the hints when you need a starting point; open the full solution after your attempt.
Extra challenging questions
A selection from this pathway. Open a topic below for its complete practice and challenge set.
Density & Pressure: infer an input with uncertainty
Density ρ is the mass per unit volume of a material, and pressure P is the force acting perpendicular to a surface per unit area of that surface: a scalar, measured in pascals, with no direction of its own. Use the supplied topic model Y_out = rho*g*t/100000*cosd(theta), where Y_out is PRESSURE P = ρgt cos θ (kPa). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sind/cosd/tand use degrees; g = 9.81 m s⁻². theta is Ramp tilt θ, measured in °. Keep rho (Slab density ρ) = 2400 kg m⁻³; t (Slab thickness t) = 20 cm fixed. For this exercise use only 4.78125 ° ≤ theta ≤ 40.2188 ° on this continuous branch. A grid of test settings is an input restriction for this exercise, not by itself a claim of physical quantisation. This calculation retains the topic model's stated assumptions; it does not extend them to a different physical setup. A reading gives Y_out = 4.2177961 kPa, with a bounded reading uncertainty ±0.071193 kPa. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown theta by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00354375 °, using a numerical bracket if required. Find the full input interval consistent with the bounded reading uncertainty and compare it with a symmetric input-error assumption. Do not assume that the supplied calibration points prove monotonicity outside this branch.
PRESSURE P = ρgt cos θ against theta
Predictions from the supplied Density & Pressure model, with all other inputs fixed as stated in the question. 5 calibration points support a trend comparison; they are not measured data or a straight-line law.
Vertical axis (y): PRESSURE P = ρgt cos θ / kPa
Horizontal axis (x): theta: Ramp tilt θ / °
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| theta: Ramp tilt θ / ° | PRESSURE P = ρgt cos θ / kPa |
|---|---|
| 4.78125 | 4.692414267851309 |
| 13.640625 | 4.575983733671411 |
| 22.5 | 4.350363942689147 |
| 31.359375 | 4.020938492996441 |
| 40.21875 | 3.595567928274655 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
The model's endpoint predictions bracket the reading. Solve Y_out(theta) = Y_measured and then solve again at Y_measured − u and Y_measured + u; a nonlinear inverse can give unequal input deviations.
Reveal the worked solution
- Endpoint predictions are Y_out(4.78125) = 4.69241 kPa and Y_out(40.2188) = 3.59557 kPa. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives theta_line = 4.78125 + (4.2178 − 4.69241)(40.2188 − 4.78125)/(3.59557 − 4.69241) = 20.1155 °. This estimates the inverse of the endpoint chord, rather than assuming that chord is the physical model.
- Use the actual equation, not the chord, and bracket its root. Evaluating the midpoint and retaining the sign-changing half-interval gives theta = 26.3981 °; substituting back gives Y_out = 4.2177961 kPa.
- The interpolation error in the input is |theta_line − theta| = 6.28266 °. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 4.1466031 kPa and 4.2889891 kPa. Sort the roots even if the response decreases: 24.3775 ° ≤ theta ≤ 28.2843 °.
- The lower and upper input deviations are 2.02057 ° and 1.88618 °. They differ because the inverse response is nonlinear; a symmetric error bar would only be a local approximation. This interval includes reading uncertainty only, not model error or uncertainty in fixed inputs.
Answer theta ≈ 26.3981 °; compatible interval [24.3775, 28.2843] °. Endpoint interpolation gives 20.1155 °.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Hydrostatic Pressure & Manometers: infer an input with uncertainty
Hydrostatic pressure is the pressure within a fluid at rest; it rises with depth at the rate ρg, because each layer must hold up the weight of the fluid above it. Use the supplied topic model Y_out = g*(rho*min(h,t)/1000 + max(h-t,0)), where Y_out is GAUGE AT DEPTH h (kPa). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; min(a,b) selects the smaller value; max(a,b) selects the larger value; g = 9.81 m s⁻². h is Probe depth h, measured in m. Keep t (Oil layer thickness t) = 1.5 m; rho (Oil density) = 800 kg m⁻³ fixed. For this exercise use only 0.425 m ≤ h ≤ 3.575 m on this continuous branch. A grid of test settings is an input restriction for this exercise, not by itself a claim of physical quantisation. This calculation retains the topic model's stated assumptions; it does not extend them to a different physical setup. A reading gives Y_out = 20.076165 kPa, with a bounded reading uncertainty ±1.8077 kPa. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown h by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000315 m, using a numerical bracket if required. Find the full input interval consistent with the bounded reading uncertainty and compare it with a symmetric input-error assumption. Do not assume that the supplied calibration points prove monotonicity outside this branch.
GAUGE AT DEPTH h against h
Predictions from the supplied Hydrostatic Pressure & Manometers model, with all other inputs fixed as stated in the question. 5 calibration points support a trend comparison; they are not measured data or a straight-line law.
Vertical axis (y): GAUGE AT DEPTH h / kPa
Horizontal axis (x): h: Probe depth h / m
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| h: Probe depth h / m | GAUGE AT DEPTH h / kPa |
|---|---|
| 0.425 | 3.3354000000000004 |
| 1.2125 | 9.515699999999999 |
| 2 | 16.677 |
| 2.7874999999999996 | 24.402375 |
| 3.5749999999999997 | 32.12775 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
The model's endpoint predictions bracket the reading. Solve Y_out(h) = Y_measured and then solve again at Y_measured − u and Y_measured + u; a nonlinear inverse can give unequal input deviations.
Reveal the worked solution
- Endpoint predictions are Y_out(0.425) = 3.3354 kPa and Y_out(3.575) = 32.1277 kPa. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives h_line = 0.425 + (20.0762 − 3.3354)(3.575 − 0.425)/(32.1277 − 3.3354) = 2.25651 m. This estimates the inverse of the endpoint chord, rather than assuming that chord is the physical model.
- Use the actual equation, not the chord, and bracket its root. Evaluating the midpoint and retaining the sign-changing half-interval gives h = 2.3465 m; substituting back gives Y_out = 20.076165 kPa.
- The interpolation error in the input is |h_line − h| = 0.0899923 m. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 18.268465 kPa and 21.883865 kPa. Sort the roots even if the response decreases: 2.16223 m ≤ h ≤ 2.53077 m.
- The lower and upper input deviations are 0.184271 m and 0.184271 m. They agree on this branch; this does not establish linearity for changes in other inputs or outside the interval. This interval includes reading uncertainty only, not model error or uncertainty in fixed inputs.
Answer h ≈ 2.3465 m; compatible interval [2.16223, 2.53077] m. Endpoint interpolation gives 2.25651 m.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Atmospheric Pressure & Measurement: infer an input with uncertainty
Atmospheric pressure is the force per unit area exerted by the weight of the air above a surface; barometers measure it directly, and manometers and most gauges measure other pressures relative to it. Use the supplied topic model Y_out = 101.325*exp(-34.22*z/tk), where Y_out is ISOTHERMAL p (kPa). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; exp(x) means eˣ. z is Altitude z, measured in km. Keep tk (Air temperature T) = 288 K fixed. For this exercise use only 1.275 km ≤ z ≤ 10.725 km on this continuous branch. A grid of test settings is an input restriction for this exercise, not by itself a claim of physical quantisation. This calculation retains the topic model's stated assumptions; it does not extend them to a different physical setup. A reading gives Y_out = 43.899533 kPa, with a bounded reading uncertainty ±2.3351 kPa. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown z by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000945 km, using a numerical bracket if required. Find the full input interval consistent with the bounded reading uncertainty and compare it with a symmetric input-error assumption. Do not assume that the supplied calibration points prove monotonicity outside this branch.
ISOTHERMAL p against z
Predictions from the supplied Atmospheric Pressure & Measurement model, with all other inputs fixed as stated in the question. 5 calibration points support a trend comparison; they are not measured data or a straight-line law.
Vertical axis (y): ISOTHERMAL p / kPa
Horizontal axis (x): z: Altitude z / km
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| z: Altitude z / km | ISOTHERMAL p / kPa |
|---|---|
| 1.275 | 87.08097046684439 |
| 3.6374999999999997 | 65.76760838547007 |
| 6 | 49.670763767974044 |
| 8.362499999999999 | 37.51367631970869 |
| 10.725 | 28.33207714690376 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
The model's endpoint predictions bracket the reading. Solve Y_out(z) = Y_measured and then solve again at Y_measured − u and Y_measured + u; a nonlinear inverse can give unequal input deviations.
Reveal the worked solution
- Endpoint predictions are Y_out(1.275) = 87.081 kPa and Y_out(10.725) = 28.3321 kPa. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives z_line = 1.275 + (43.8995 − 87.081)(10.725 − 1.275)/(28.3321 − 87.081) = 8.22091 km. This estimates the inverse of the endpoint chord, rather than assuming that chord is the physical model.
- Use the actual equation, not the chord, and bracket its root. Evaluating the midpoint and retaining the sign-changing half-interval gives z = 7.0395 km; substituting back gives Y_out = 43.899533 kPa.
- The interpolation error in the input is |z_line − z| = 1.18141 km. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 41.564433 kPa and 46.234633 kPa. Sort the roots even if the response decreases: 6.60333 km ≤ z ≤ 7.49952 km.
- The lower and upper input deviations are 0.43617 km and 0.460016 km. They differ because the inverse response is nonlinear; a symmetric error bar would only be a local approximation. This interval includes reading uncertainty only, not model error or uncertainty in fixed inputs.
Answer z ≈ 7.0395 km; compatible interval [6.60333, 7.49952] km. Endpoint interpolation gives 8.22091 km.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Pascal's Principle & Hydraulics: infer an input with uncertainty
Pascal's principle: a pressure increase applied to an enclosed fluid at rest reaches every point of that fluid, and every wall confining it, at full strength. Use the supplied topic model Y_out = s/r^2, where Y_out is OUTPUT RISE d₂ (m). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. r is Piston diameter ratio d₂/d₁, measured in ×. Keep s (Input piston stroke d₁) = 0.35 m fixed. For this exercise use only 1.61688 × ≤ r ≤ 2.48313 × on this continuous branch. A grid of test settings is an input restriction for this exercise, not by itself a claim of physical quantisation. This calculation retains the topic model's stated assumptions; it does not extend them to a different physical setup. A reading gives Y_out = 0.076049619 m, with a bounded reading uncertainty ±0.0028929 m. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown r by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000086625 ×, using a numerical bracket if required. Find the full input interval consistent with the bounded reading uncertainty and compare it with a symmetric input-error assumption. Do not assume that the supplied calibration points prove monotonicity outside this branch.
OUTPUT RISE d₂ against r
Predictions from the supplied Pascal's Principle & Hydraulics model, with all other inputs fixed as stated in the question. 5 calibration points support a trend comparison; they are not measured data or a straight-line law.
Vertical axis (y): OUTPUT RISE d₂ / m
Horizontal axis (x): r: Piston diameter ratio d₂/d₁ / ×
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| r: Piston diameter ratio d₂/d₁ / × | OUTPUT RISE d₂ / m |
|---|---|
| 1.616875 | 0.13387983000249976 |
| 1.8334375 | 0.10412039920528013 |
| 2.0500000000000003 | 0.08328375966686492 |
| 2.2665625 | 0.0681290990920828 |
| 2.4831250000000002 | 0.05676372397650918 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
The model's endpoint predictions bracket the reading. Solve Y_out(r) = Y_measured and then solve again at Y_measured − u and Y_measured + u; a nonlinear inverse can give unequal input deviations.
Reveal the worked solution
- Endpoint predictions are Y_out(1.61688) = 0.13388 m and Y_out(2.48313) = 0.0567637 m. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives r_line = 1.61688 + (0.0760496 − 0.13388)(2.48313 − 1.61688)/(0.0567637 − 0.13388) = 2.26649 ×. This estimates the inverse of the endpoint chord, rather than assuming that chord is the physical model.
- Use the actual equation, not the chord, and bracket its root. Evaluating the midpoint and retaining the sign-changing half-interval gives r = 2.14529 ×; substituting back gives Y_out = 0.076049619 m.
- The interpolation error in the input is |r_line − r| = 0.121198 ×. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.073156719 m and 0.078942519 m. Sort the roots even if the response decreases: 2.10561 × ≤ r ≤ 2.18729 ×.
- The lower and upper input deviations are 0.0396746 × and 0.0420053 ×. They differ because the inverse response is nonlinear; a symmetric error bar would only be a local approximation. This interval includes reading uncertainty only, not model error or uncertainty in fixed inputs.
Answer r ≈ 2.14529 ×; compatible interval [2.10561, 2.18729] ×. Endpoint interpolation gives 2.26649 ×.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.