Architecture and Planning Miscellaneous-topic

Architecture and Planning Miscellaneous-topic

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Architecture and Planning Miscellaneous

Architecture and Planning Miscellaneous-topic

Direction: The scale of a contour map is 1:10,000 and the contour interval is 5 m. Distance between two given points on the map is 2 cm and the elevation difference between the two given points is 10 m.

  1. The actual distance between the two given points in m would be








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    2 cm in map = 2 × 10000 cm in actual = 200 m

    Correct Option: C

    2 cm in map = 2 × 10000 cm in actual = 200 m

  1. In the given project network diagram, the total slack for job A in days would be _______.








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    For activity F to start, activities (B + C) & (A + E) must been finished.
    As activities (B + C) takes 6 days to finish to start the activity F.
    Activities (A + E) takes 5 days to finish to start the activity F.
    Which means that activity A can be delayed by 1 day without affecting the start of activity F.
    The delay by one day is actually slack.

    Correct Option: B

    For activity F to start, activities (B + C) & (A + E) must been finished.
    As activities (B + C) takes 6 days to finish to start the activity F.
    Activities (A + E) takes 5 days to finish to start the activity F.
    Which means that activity A can be delayed by 1 day without affecting the start of activity F.
    The delay by one day is actually slack.

  1. 50 Hectare of residential sector has 65% buildable area. The FAR of the buildable area is 1.5. Within the residential sector, 60% of dwelling units are of area 100 m2 each and 40% of the dwelling units are of area 80 m2 each. The gross residential density, in dwelling units per Hectare, would be _______.








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    Let, N = Total no. of Dwelling units.
    FAR = 1.5 (given)
    So, Total buildable area = 50 Hectare × 65% × 1.5 FAR = (50 × 0.65 × 1.5) Hectare = (50 × 0.65 × 1.5 × 10000) m2
    As per question, 0.6N × 100 m2 + 0.4N × 80 m2 = total buildable area
    So, 0.6N × 100 m2 + 0.4N × 80 m2 = (50 × 0.65 × 1.5 × 10000) m2
    By solving above equation, we get N = 5300

    The gross residential density, in dwelling units per Hectare =
    N
    		 = 106
    50

    Correct Option: D

    Let, N = Total no. of Dwelling units.
    FAR = 1.5 (given)
    So, Total buildable area = 50 Hectare × 65% × 1.5 FAR = (50 × 0.65 × 1.5) Hectare = (50 × 0.65 × 1.5 × 10000) m2
    As per question, 0.6N × 100 m2 + 0.4N × 80 m2 = total buildable area
    So, 0.6N × 100 m2 + 0.4N × 80 m2 = (50 × 0.65 × 1.5 × 10000) m2
    By solving above equation, we get N = 5300

    The gross residential density, in dwelling units per Hectare =
    N
    		 = 106
    50

  1. Flux emitted from a 1cd light source in all directions, in lumens, would be _______.








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    Flux emitted from a 1cd light source in all directions, in lumens, would be 4π.

    Correct Option: B

    Flux emitted from a 1cd light source in all directions, in lumens, would be 4π.

  1. A tank of internal dimension 3 m × 5 m × 4 m (Length × Breadth × Height) has 200 mm thick brick wall on all sides. Volume of brickwork in m3 would be _______.








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    Internal dimension is 3 m × 5 m × 4 m
    External dimension is 3.2 m x 5.2 m × 6.3 m
    Required volume = (3.2 m × 5.2 m × 6.3 m) – (3 m × 5 m × 4 m) = 44.832 m3

    Correct Option: A

    Internal dimension is 3 m × 5 m × 4 m
    External dimension is 3.2 m x 5.2 m × 6.3 m
    Required volume = (3.2 m × 5.2 m × 6.3 m) – (3 m × 5 m × 4 m) = 44.832 m3