Solution :-

True Scale = 1 : 4000

Incorrect Scale = 1 : 5000

%Error in the length measured,

= [(True Scale - Incorrect Scale)/Incorrect Scale] × 100%

= [(1/4000 - 1/5000)/(1/5000)] × 100%

= [{(5 - 4)/20000}/(1/5000)] × 100%

= [(1/20000)/(1/5000)] × 100%

= (5000/20000) × 100%

= 0.25 × 100%

= 25%

Solution :-

Relief displacement (d),

d = rh/H

d = Relief displacement = 8.8 mm = 0.88 cm

r = Radial distance of the image of the object top = 10 cm

h = Height of the object above datum = 88 m

Flying height of the camera (H),

= rh/d

= 10 × 88/0.88

= 1000 m

Explanation :-

The vertices of an astronomical triangle :

(i) Zenith (Z)

• The point directly above the observer on the celestial sphere.

(ii) Pole (P)

• The point on the celestial sphere where the Earth's axis of rotation intersects.

(iii) Heavenly body (S)

• The position of the observed star, planet, or celestial object.

These three points form a triangle on the celestial sphere, that are used for calculations such as altitude, azimuth, and hour angle in spherical astronomy.

Altitude (h),

• The angle between the observed celestial object and the observer’s horizon.

Azimuth (A)

• The angle measured clockwise from the north point (or from the south point) of the observer’s horizon to the point where the vertical circle passing through the celestial body intersects the horizon.

Hour Angle (H)

• The angle between the observer’s meridian and the hour circle passing through the celestial body.

Solution :-

Contour interval (d) = 110 - 105 = 5

Volume of impounded water using trapezoidal formula (V),

= d [{(A1 + An)/2} + A2 + A3 + ..... + A(n-1)]

= 5 [{(100 + 900)/2} + 150 + 350 + 500]

= 5 (500 + 150 + 350 + 500)

= 5 (1500)

= 7500 ha-m

Solution :-

Whole Circle Bearing (WCB) = 327° 24' (4th Quadrant)

Quadrant Bearing (QB) = N (360° - 327° 24') W = N 32° 36' W

Solution :-

Net length covered by each photograph, l= 1.5 km

Length of the strip, L = 18 km

Number of photographs required, N

= L/l + 1

= 18/1.5 + 1

= 12 + 1

= 13

Solution :-

Maximum super elevation (e) = 400 mm = 0.4 m
Rate of change of super elevation (N) = 1 in 400

Length of transition curve (L),

= e × N

= 0.4 × 400

= 160 m

Solution :-

Length of "N" Vernier divisions = Length of "N-1" Main Scale divisions

So,

Length of "10" Vernier divisions = Length of "9" Main Scale divisions

Explanation :-

Direct method of contouring :

• The most accurate method.
• Not suitable for hilly areas because it involves physically marking points on the ground and taking precise level readings, which becomes challenging and inefficient in hilly areas.
• It is suitable for small and detailed surveys requiring precision.
• e.g. Levelling method

Indirect method of contouring :

• In this method, contour lines are interpolated between points of known elevation, rather than marking each contour point directly on the ground.
• Less accurate compared to the direct method.
• Suitable for hilly terrains and large-scale surveys.
• Faster and more efficient for large areas.
• e.g. Grid(Square) Method, Cross-section Method, Tacheometric Method

Explanation :-

Transit Rule :

• This method is used when angular measurements are more precise than linear measurements.
• It distributes the total error in latitudes and departures in proportion to the latitudes and departures of the sides.

Bowditch Rule :

• Bowditch rule is used when both angular and linear measurements are equally precise.
• It distributes errors in proportion to the lengths of the traverse lines.