Architecture and Planning Miscellaneous-topic

Architecture and Planning Miscellaneous-topic

Easy Questions
Moderate Questions
Difficult Questions

Architecture and Planning Miscellaneous

Architecture and Planning Miscellaneous-topic

Direction: A residential sector planned over an area of 100 hectares has been divided into various plots, each having one dwelling unit with an aver age household size of 5 persons. Remaining area is devoted for schools, roads, parks, shops etc.

Plot size Number
500 sqm 500
300 sqm 500
200 sqm 1000

  1. The gross density of the residential sector in persons per hectare would be








  1. View Hint View Answer Discuss in Forum

    Total Population = Total No. of Units × Average Household Size
    = (500 + 500 + 1000) × 5 = 10,000 persons
    Gross Density = Total Population / Total Land Area = 10,000 persons / 100 hectares = 100 pph
    Plot Size does not matter as Net Density has not been asked.

    Correct Option: A

    Total Population = Total No. of Units × Average Household Size
    = (500 + 500 + 1000) × 5 = 10,000 persons
    Gross Density = Total Population / Total Land Area = 10,000 persons / 100 hectares = 100 pph
    Plot Size does not matter as Net Density has not been asked.

Direction: An auditorium having volume of 4500 cum and total absorption of all acoustic materials is 480 m2 sabine

  1. To reduce reverberation time by 0.5 second. additional absorption (m2 sabine) required would be








  1. View Hint View Answer Discuss in Forum

    To reduce t by 0.5 sec.
    New value of t will be 1 sec.

    t =
    0.161 V
    Sa

    1 =
    0.161 × 4500
    		 = 1.5 sec
    Sa

    Sa = 724.5
    Increase in Sa = 724.5 – 480 = 244.5 ≈ 240 m2 sabine

    Correct Option: C

    To reduce t by 0.5 sec.
    New value of t will be 1 sec.

    t =
    0.161 V
    Sa

    1 =
    0.161 × 4500
    		 = 1.5 sec
    Sa

    Sa = 724.5
    Increase in Sa = 724.5 – 480 = 244.5 ≈ 240 m2 sabine

  1. The critical activities of the project are








  1. View Hint View Answer Discuss in Forum

    Reverberation time =
    0.161 V
    Sa

    Sa(Sabine) = 480
    t =
    0.151 × 4500
    		 = 1.5 sec
    480

    Correct Option: B

    Reverberation time =
    0.161 V
    Sa

    Sa(Sabine) = 480
    t =
    0.151 × 4500
    		 = 1.5 sec
    480

Direction: A building site has a plot of 500 sqm
Maximum allowable height - G+ 7
Area to be utilized for paved access roads- 10%
Maximum ground coverage- 40%
Runoff coefficient for paved surface- 0.9
Maximum allowable FAR - 2.0
Runoff coefficient for unpaved surface- 0.3

  1. If it rains for 30 min. with an intensity of 10 cm/hr, minimum volume of rain water that can be collected will be








  1. View Hint View Answer Discuss in Forum

    Rainwater in half hour = 10 × (1 / 2) = 5 cm or 0.05m
    % paved area = 25% + 10% (roads) = 35% or 0.35 %
    unpaved area = 100 – 35 = 65% or 0.65
    Thus, Rainwater collected = Rainwater collected in paved area + Rainwater collected in unpaved area
    = (0.35 × 500 × 0.05 × 0.9) + (0.65 × 500 × 0.05 ×0.3) = 12.75 cum

    Correct Option: A

    Rainwater in half hour = 10 × (1 / 2) = 5 cm or 0.05m
    % paved area = 25% + 10% (roads) = 35% or 0.35 %
    unpaved area = 100 – 35 = 65% or 0.65
    Thus, Rainwater collected = Rainwater collected in paved area + Rainwater collected in unpaved area
    = (0.35 × 500 × 0.05 × 0.9) + (0.65 × 500 × 0.05 ×0.3) = 12.75 cum

  1. If maximum allowable FAR is utilized, the minimum ground coverage would be








  1. View Hint View Answer Discuss in Forum

    Plot area = 500 sq m
    If Maximum FAR = 2 = 1000 sq m
    Minimum Ground coverage = Max FAR/Max No. of Floors
    = 1000 / 8( = 125 sq m)
    % of Ground Coverage = [1000 ÷ (8 × 500)] × 100 = 25

    Correct Option: B

    Plot area = 500 sq m
    If Maximum FAR = 2 = 1000 sq m
    Minimum Ground coverage = Max FAR/Max No. of Floors
    = 1000 / 8( = 125 sq m)
    % of Ground Coverage = [1000 ÷ (8 × 500)] × 100 = 25