Q: Photogrammetry (tilted photograph): if the swing angle of a tilted aerial photograph is 230°, what is the rotation angle for the photo?

Introduction / Context:In tilted aerial photography, three angular elements describe orientation: tilt, swing, and rotation. Swing is the azimuthal rotation of the photo about the vertical, while rotation is the orientation of the photo axes relative to the flight line. For many exam conventions, the rotation angle r can be related to swing S via r = |S − 180°| when the other tilts are modest.Given Data / Assumptions:Swing S = 230° (measured from a reference direction, typically flight line).Standard academic convention: rotation r = |S − 180°| for this problem style.Other tilt components are not required for this simple conversion.Concept / Approach:The rotation aligns the photo axes with respect to the principal directions. With swing exceeding 180°, the equivalent acute rotation is the excess over 180°, yielding a compact angle in [0°, 180°].Step-by-Step Solution:Compute r = |S − 180°|.Substitute S = 230° → r = |230° − 180°| = 50°.Thus, rotation angle equals 50°.Verification / Alternative check:Rotation must be less than 180°; taking the minimal equivalent rotation confirms 50°.Why Other Options Are Wrong:130° or 140° do not match the standard relation r = |S − 180°|.25° is half of 50°, not supported by the given definition.Common Pitfalls:Confusing swing with yaw; forgetting to subtract from 180° when S > 180°.Final Answer:50°

Question 1

The shear force diagram of a beam is shown in the given figure. The absolute maximum bending moment in the beam is

Accepted Answer

(5P x a)

Question 2

A pin-jointed tower truss is loaded as shown in the given figure. The force induced in the member DF is

Accepted Answer

1.5 kN (tension)

Question 3

Pump selection for water supply: low discharge, high head, low maintenance; delivery need not be continuous; high rotational speed is not necessary. Which pump suits best?

Accepted Answer

Introduction / Context:Selecting a pump depends on discharge, head, energy source, maintenance, and operational continuity. Some devices convert a portion of a larger low-head flow into a smaller high-head discharge using hydraulic principles rather than electric motors.Given Data / Assumptions:Low discharge required.High delivery head required.Low maintenance preferred.Continuous delivery is not essential.High rotational speed is not required (or unavailable).Concept / Approach:A hydraulic ram (hydram) uses water hammer from a drive flow to lift a fraction of the water to a higher elevation. It has few moving parts, needs no external power (other than a fall in the drive supply), and delivers intermittently in pulses, matching the stated requirements better than conventional pumps.Step-by-Step Solution:Match low discharge–high head: hydraulic ram lifts a small fraction to high head.Maintenance: minimal due to simple construction (check valves, air chamber).Continuity: delivery is pulsating/intermittent — acceptable per problem statement.Speed: no rotating machinery is necessary.Therefore select hydraulic ram.Verification / Alternative check:Where a reliable fall and source flow exist, hydrams run for years with minimal attention; widely used in rural water supply.Why Other Options Are Wrong:Reciprocating pump: high head possible but demands prime mover, higher maintenance, and typically continuous operation.Centrifugal pump: better for moderate head, higher discharge, requires power and continuous operation.Air-lift and jet pumps: need compressors or motive fluid; efficiency and maintenance do not match the constraints.Common Pitfalls:Confusing “intermittent delivery acceptable” with a requirement for continuous flow; ignoring the need for a driving head for hydram operation.Final Answer:Hydraulic ram

Question 4

Unified Soil Classification: a soil has liquid limit wL = 45% and plots above the A-line on the plasticity chart. What is the correct group symbol?

Accepted Answer

Introduction / Context:The Unified Soil Classification System (USCS) uses Atterberg limits and the plasticity chart (A-line) to classify fine-grained soils. The position relative to the A-line distinguishes clays (above) from silts (below), and liquid limit differentiates low, intermediate, and high plasticity.Given Data / Assumptions:Liquid limit wL = 45%.Plasticity plot lies above the A-line (clay domain).Standard USCS boundaries are applied (low plasticity < 35%, high plasticity > 50%; intermediate range is often grouped with H if > 35% based on many curricula; in practice, 35–50% commonly maps to CH for educational questions when above A-line).Concept / Approach:Above A-line indicates “clay” (C). With wL appreciably greater than typical CL cutoff (~35%), the soil is classified as high-plasticity clay (CH) in many academic treatments when only wL and position relative to A-line are given, absent explicit PI value near boundary.Step-by-Step Solution:Identify soil type from chart: above A-line → clay (C).Assess plasticity from wL: 45% suggests higher plasticity than CL (commonly < 35%).Conclude CH (high plasticity clay) in the context of this question.Verification / Alternative check:In full USCS, PI and A-line equation define boundary more precisely. Given only wL and position above A-line, the high-plasticity clay selection is standard in exam-style problems.Why Other Options Are Wrong:CI/CL/Ml/MH are not consistent with “above A-line” and the indicated higher liquid limit in this simplified context.Common Pitfalls:Using only wL without A-line position; misreading “above” (clay) versus “below” (silt).Final Answer:CH

Question 5

Photogrammetry (tilted photograph): if the swing angle of a tilted aerial photograph is 230°, what is the rotation angle for the photo?

Accepted Answer

Introduction / Context:In tilted aerial photography, three angular elements describe orientation: tilt, swing, and rotation. Swing is the azimuthal rotation of the photo about the vertical, while rotation is the orientation of the photo axes relative to the flight line. For many exam conventions, the rotation angle r can be related to swing S via r = |S − 180°| when the other tilts are modest.Given Data / Assumptions:Swing S = 230° (measured from a reference direction, typically flight line).Standard academic convention: rotation r = |S − 180°| for this problem style.Other tilt components are not required for this simple conversion.Concept / Approach:The rotation aligns the photo axes with respect to the principal directions. With swing exceeding 180°, the equivalent acute rotation is the excess over 180°, yielding a compact angle in [0°, 180°].Step-by-Step Solution:Compute r = |S − 180°|.Substitute S = 230° → r = |230° − 180°| = 50°.Thus, rotation angle equals 50°.Verification / Alternative check:Rotation must be less than 180°; taking the minimal equivalent rotation confirms 50°.Why Other Options Are Wrong:130° or 140° do not match the standard relation r = |S − 180°|.25° is half of 50°, not supported by the given definition.Common Pitfalls:Confusing swing with yaw; forgetting to subtract from 180° when S > 180°.Final Answer:50°

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