सर्भेक्षण – Surveying MCQ
सर्भेक्षण – सिभिल इन्जिनियरिङ
Surveying · Civil Engineering · १,५६५ प्रश्न
१,५६५ प्रश्न उपलब्ध। मोड छान्नुहोस्:
सर्भेक्षण: प्रश्न र उत्तर
सही लागेको विकल्पमा थिच्नुहोस्। सही भए हरियो, गलत भए रातो देखिन्छ; ३ सेकेन्डपछि सही उत्तर र व्याख्या देखिन्छ।
सही: ० / ०
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In carrier-phase GNSS positioning, the 'integer ambiguity' refers to:
सही उत्तर: घ) The unknown whole number of carrier wavelengths between the satellite and receiver at the start of tracking
The receiver measures only the fractional phase; the initial whole number of cycles must be resolved to obtain precise range.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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In GNSS carrier-phase surveying, a 'cycle slip' is:
सही उत्तर: ग) A sudden jump by an integer number of cycles in the accumulated carrier phase, due to temporary loss of lock
When the receiver briefly loses the signal (e.g. due to obstruction), the integer cycle count jumps and must be detected and repaired.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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The 'swath width' of a satellite imaging sensor is:
सही उत्तर: ख) The width of the strip of ground imaged in a single pass
Swath width is the across-track width of the ground strip covered as the satellite moves along its orbit.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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'Selective Availability', the intentional degradation of the GPS civilian signal, was switched off by the United States in:
सही उत्तर: क) 2000
Selective Availability was discontinued in May 2000, improving civilian GPS accuracy greatly.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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A vertical photograph is taken with a camera of focal length 150 mm from 3000 m above mean sea level over ground at an elevation of 500 m. The photo scale for that ground is:
सही उत्तर: घ) 1 : 16,667
Scale = f / (H − h) = 0.15 / (3000 − 500) = 1/16,667.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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Vertical photographs of format 23 cm × 23 cm are taken at a scale of 1:10,000 with 60% forward overlap. The air base (ground distance between successive exposures) is:
सही उत्तर: ग) 0.92 km
Ground coverage = 0.23 × 10,000 = 2.3 km; air base = (1 − 0.6) × 2.3 = 0.92 km.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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For a 20 m length of road embankment, the two end cross-sections have areas of 20 m² and 44 m², and the mid-section has an area of 30 m². The volume by the prismoidal formula is about:
सही उत्तर: ख) 613 m³
V = L/6 (A1 + 4Am + A2) = 20/6 × (20 + 120 + 44) = 3.333 × 184 ≈ 613 m³.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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A road curve of radius 400 m is designed for a speed of 20 m/s. Ignoring friction, the superelevation e = v²/(gR) required is about:
सही उत्तर: क) 0.10 (about 1 in 10)
e = 20² / (9.81 × 400) = 400 / 3924 ≈ 0.102, i.e. about 1 in 10.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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Jäderin's apparatus was used in triangulation for:
सही उत्तर: घ) Measuring base lines with suspended wires or tapes in catenary
Jäderin's method measures base lines with long invar wires or tapes suspended under tension between tripods.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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In a closed traverse, the length and bearing of one line could not be measured. The sums of latitudes and departures of the other lines are +30.00 m and −40.00 m. The missing line has:
सही उत्तर: ग) Length 50.00 m, bearing S 53°08' E
The missing line must close the traverse: latitude −30, departure +40, so length = √(30² + 40²) = 50 m and bearing = S tan⁻¹(40/30) E = S 53°08' E.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
यी अभ्यास प्रश्न हुन्, आधिकारिक प्रश्न होइनन्। पाठ्यक्रम, विज्ञापन र नतिजाका लागि सम्बन्धित निकायको आधिकारिक वेबसाइट हेर्नुहोस्। उत्तरमा गल्ती भेटे प्रश्नमुनिको "टिप्पणी / गल्ती रिपोर्ट" मा लेख्नुहोस्।
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