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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.

Extra challenging · 01

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.

Explore the question data

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

PRESSURE P = ρgt cos θ against thetaPredictions 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. Horizontal axis: theta: Ramp tilt θ / °. Vertical axis: PRESSURE P = ρgt cos θ / kPa. Exact values are available in the data table below.3.5523.8484.1444.444.7363.36412.9322.532.0741.64xy

Horizontal axis (x): theta: Ramp tilt θ / °

Axes show the supplied calibration interval; the origin may lie outside the plot.

Reading 1theta: Ramp tilt θ / °: 4.781PRESSURE P = ρgt cos θ / kPa: 4.692
Read the exact data as a table
PRESSURE P = ρgt cos θ against theta — supplied values
theta: Ramp tilt θ / °PRESSURE P = ρgt cos θ / kPa
4.781254.692414267851309
13.6406254.575983733671411
22.54.350363942689147
31.3593754.020938492996441
40.218753.595567928274655

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
  1. 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.
  2. 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.
  3. 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.
  4. The interpolation error in the input is |theta_line − theta| = 6.28266 °. This is larger than the requested numerical tolerance.
  5. 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 °.
  6. 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.

Extra challenging · 02

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.

Explore the question data

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

GAUGE AT DEPTH h against hPredictions 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. Horizontal axis: h: Probe depth h / m. Vertical axis: GAUGE AT DEPTH h / kPa. Exact values are available in the data table below.2.1849.95817.7325.5133.280.2991.14922.853.701xy

Horizontal axis (x): h: Probe depth h / m

Axes show the supplied calibration interval; the origin may lie outside the plot.

Reading 1h: Probe depth h / m: 0.425GAUGE AT DEPTH h / kPa: 3.335
Read the exact data as a table
GAUGE AT DEPTH h against h — supplied values
h: Probe depth h / mGAUGE AT DEPTH h / kPa
0.4253.3354000000000004
1.21259.515699999999999
216.677
2.787499999999999624.402375
3.574999999999999732.12775

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
  1. 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.
  2. 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.
  3. 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.
  4. The interpolation error in the input is |h_line − h| = 0.0899923 m. This is larger than the requested numerical tolerance.
  5. 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.
  6. 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.

Extra challenging · 03

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.

Explore the question data

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

ISOTHERMAL p against zPredictions 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. Horizontal axis: z: Altitude z / km. Vertical axis: ISOTHERMAL p / kPa. Exact values are available in the data table below.25.9841.8457.7173.5789.430.8973.44868.55111.1xy

Horizontal axis (x): z: Altitude z / km

Axes show the supplied calibration interval; the origin may lie outside the plot.

Reading 1z: Altitude z / km: 1.275ISOTHERMAL p / kPa: 87.08
Read the exact data as a table
ISOTHERMAL p against z — supplied values
z: Altitude z / kmISOTHERMAL p / kPa
1.27587.08097046684439
3.637499999999999765.76760838547007
649.670763767974044
8.36249999999999937.51367631970869
10.72528.33207714690376

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
  1. 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.
  2. 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.
  3. 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.
  4. The interpolation error in the input is |z_line − z| = 1.18141 km. This is larger than the requested numerical tolerance.
  5. 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.
  6. 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.

Extra challenging · 04

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.

Explore the question data

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

OUTPUT RISE d₂ against rPredictions 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. Horizontal axis: r: Piston diameter ratio d₂/d₁ / ×. Vertical axis: OUTPUT RISE d₂ / m. Exact values are available in the data table below.0.053680.07450.095320.11610.1371.5821.8162.052.2842.518xy

Horizontal axis (x): r: Piston diameter ratio d₂/d₁ / ×

Axes show the supplied calibration interval; the origin may lie outside the plot.

Reading 1r: Piston diameter ratio d₂/d₁ / ×: 1.617OUTPUT RISE d₂ / m: 0.1339
Read the exact data as a table
OUTPUT RISE d₂ against r — supplied values
r: Piston diameter ratio d₂/d₁ / ×OUTPUT RISE d₂ / m
1.6168750.13387983000249976
1.83343750.10412039920528013
2.05000000000000030.08328375966686492
2.26656250.0681290990920828
2.48312500000000020.05676372397650918

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
  1. 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.
  2. 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.
  3. 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.
  4. The interpolation error in the input is |r_line − r| = 0.121198 ×. This is larger than the requested numerical tolerance.
  5. 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 ×.
  6. 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.

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