(a) For a project consisting of several activities, the allotted time and the dependencies of the activities are presented below:

| Activity | Duration (days) | Predecessor |
| P | 5 | – |
| Q | 4 | – |
| R | 6 | Q |
| S | 5 | P |
| T | 7 | P |
| U

Question

(a) For a project consisting of several activities, the allotted time and the dependencies of the activities are presented below:

| Activity | Duration (days) | Predecessor |
| P | 5 | – |
| Q | 4 | – |
| R | 6 | Q |
| S | 5 | P |
| T | 7 | P |
| U | 4 | T, R |

(i) Prepare a network and mark critical path in it.
(ii) Calculate Float, Earliest start, Earliest finish, Latest start and Latest finish times. (20 marks)

(b) Trains of different speeds are to be run on a 2° curve on a broad gauge. The average speed of trains to be run on the track is 80 kmph. Calculate the value of equilibrium cant. Also calculate the maximum permissible speed on the track allowing the maximum cant deficiency. (15 marks)

(c)

(i)

An aircraft flew at the altitude of 5000 m above the mean sea level. Two consecutive photographs were taken with the camera of focal length 300 mm on the flat ground having elevation of 2000 m above mean sea level. The longitudinal overlap is 65% and photograph print size is 300 mm × 300 mm. Calculate the scale of the photograph and distance between the two consecutive exposure stations. 5 marks

(ii)

What is spectral reflectance curve ? Explain its significance in Remote Sensing. 10 marks

(ii) What is spectral reflectance curve ? Explain its significance in Remote Sensing. (10 marks)

UPSC Answer Check · Accepted Answer

(a)(i)
Using the Critical Path Method (CPM), the precedence links are: P precedes S and T; Q precedes R; T and R precede U. In activity-on-node form the network is:
P(5) → S(5), P(5) → T(7)
Q(4) → R(6)
T(7) → U(4), R(6) → U(4)

Forward pass:
P: ES = 0, EF = 5
Q: ES = 0, EF = 4
S: ES = 5, EF = 10
T: ES = 5, EF = 12
R: ES = 4, EF = 10
U: ES = max(12, 10) = 12, EF = 16

Hence project duration = 16 days.

Backward pass, taking terminal activities S and U to finish at 16 days:
U: LF = 16, LS = 12
T: LF = 12, LS = 5
R: LF = 12, LS = 6
S: LF = 16, LS = 11
P: LF = min(LS of S, LS of T) = min(11, 5) = 5, LS = 0
Q: LF = LS of R = 6, LS = 2

Critical activities have Float = 0: P, T and U.
Critical path: P → T → U, duration = 5 + 7 + 4 = 16 days.

(a)(ii)
Using total float as Float = LS − ES = LF − EF:
P: ES = 0, EF = 5, LS = 0, LF = 5, Float = 0
Q: ES = 0, EF = 4, LS = 2, LF = 6, Float = 2
R: ES = 4, EF = 10, LS = 6, LF = 12, Float = 2
S: ES = 5, EF = 10, LS = 11, LF = 16, Float = 6
T: ES = 5, EF = 12, LS = 5, LF = 12, Float = 0
U: ES = 12, EF = 16, LS = 12, LF = 16, Float = 0

Final times: P(0,5,0,5,0); Q(0,4,2,6,2); R(4,10,6,12,2); S(5,10,11,16,6); T(5,12,5,12,0); U(12,16,12,16,0).

(b)
For broad gauge, G = 1.676 m = 167.6 cm. For a 2° curve, using the chord definition for broad gauge:
R = 1720/D = 1720/2 = 860 m.

Equilibrium cant formula:
e = G V²/(127 R), where e is in cm, G in cm, V in kmph, R in m.

Given average speed V = 80 kmph:
e = 167.6 × 80²/(127 × 860)
e = 167.6 × 6400/109220
e = 9.82 cm.

Equilibrium cant = 9.82 cm.

For maximum permissible speed, allow maximum cant deficiency for broad gauge, taken as 7.5 cm. Then:
e + cant deficiency = G Vmax²/(127 R)
Vmax² = (9.82 + 7.5) × 127 × 860/167.6
Vmax² = 17.32 × 109220/167.6 = 11286.94
Vmax = √11286.94 = 106.24 kmph.

Maximum permissible speed ≈ 106.24 kmph ≈ 106 kmph.
This is valid for the given average-speed cant and the standard maximum cant deficiency of 7.5 cm.

(c)(i)
Focal length f = 300 mm = 0.300 m.
Flying height above mean sea level = 5000 m.
Ground elevation = 2000 m.
Flying height above ground = 5000 − 2000 = 3000 m.

Scale of photograph = f/(H − h) = 0.300/3000 = 1/10000.
Scale = 1:10,000.

Print size along flight = 300 mm = 0.300 m.
Ground coverage along flight = 0.300 × 10000 = 3000 m.
Longitudinal overlap = 65%, so new ground covered between exposures = (1 − 0.65) × 3000 = 0.35 × 3000 = 1050 m.

Distance between consecutive exposure stations = 1050 m.

(c)(ii)
A spectral reflectance curve is a graph showing the reflectance, usually as a percentage, of an object or surface as a function of wavelength of electromagnetic radiation. It is also called a spectral signature. It shows how much incident energy is reflected at different wavelengths and where absorption occurs.

Its significance in remote sensing is that different materials have different spectral reflectance curves. For example, healthy vegetation reflects strongly in the near-infrared and green regions but absorbs red and blue; water reflects little in the near-infrared; dry soil shows increasing reflectance with wavelength. These characteristic curves help in identifying and classifying land cover, crops, forests, water bodies, soils, minerals and urban surfaces. They also guide selection of suitable spectral bands and sensors for a particular application, help in developing vegetation indices such as NDVI, and enable monitoring of crop health, moisture stress, pollution and environmental change. Thus, spectral reflectance curves form the physical basis of image interpretation and digital classification in remote sensing.

(a) For a project consisting of several activities, the allotted time and the dependencies of the activities are presented…

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More from Civil Engineering 2023 Paper II

क्रिया	अवधि (दिन)	पूर्ववर्ती
P	5	–
Q	4	–
R	6	Q
S	5	P
T	7	P
U	4	T, R