Architecture and Planning Miscellaneous-topic
- Development authorities in India are established under the provision of
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73rd Constitutional Amendment Act: Municipal Act 74th Constitutional Amendment Act: Panchayat Act
Correct Option: C
73rd Constitutional Amendment Act: Municipal Act 74th Constitutional Amendment Act: Panchayat Act
- Toothing is a construction technique used in
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Toothing masonry refers to the process of leaving alternating openings (teeth) for an adjoining block or brick wall to be started from. This allows the adjoining wall t o be st art ed wi t hout havi ng t o adjust or cut br ick. The toothing process is also sometimes used when a window or door opening is to be cut into an existing masonry wall. For t his process, the opening is cut and removed, a lintel installed, alternating bricks removed, and half/partial bricks are cut to be inserted back into the teeth to make the jamb flush for installing the window or door.
Correct Option: D
Toothing masonry refers to the process of leaving alternating openings (teeth) for an adjoining block or brick wall to be started from. This allows the adjoining wall t o be st art ed wi t hout havi ng t o adjust or cut br ick. The toothing process is also sometimes used when a window or door opening is to be cut into an existing masonry wall. For t his process, the opening is cut and removed, a lintel installed, alternating bricks removed, and half/partial bricks are cut to be inserted back into the teeth to make the jamb flush for installing the window or door.
- One litre of acrylic paint can cover 16 sqm of wall area for the first coat and 24 sqm for the second coat. The walls of a lecture hall measuring 12m × 8m × 4m (L × B × H) need to be painted with two coats of this paint. The hall has total glazed fenestration area of 12 sqm. The number of 4 litre paint containers required will be __________
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4 to 4
Total area to be painted = (40 × 4 – 12) =148 m²
Paint to be used in first coat = 148/16 = 9.25 liters
Paint to be used in second coat = 148/24 = 6.16 liters
Total paint required = 9.25 + 6.16 = 15.41 liters
No. of 4 liters containers to be used = 15.41/4
= 3.85 cont ainer, which is approximately 4 containers.Correct Option: A
4 to 4
Total area to be painted = (40 × 4 – 12) =148 m²
Paint to be used in first coat = 148/16 = 9.25 liters
Paint to be used in second coat = 148/24 = 6.16 liters
Total paint required = 9.25 + 6.16 = 15.41 liters
No. of 4 liters containers to be used = 15.41/4
= 3.85 cont ainer, which is approximately 4 containers.
- In 2001, the population and work force participation rate of a town were 30,000 and 30 percent respectively. The work force participation rate in the year 2011 increased to 34 percent. If the decadal population growth rate was 6 percent, the increase in the number of working people in the town in 2011 was __________
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1812 to 1812
Working population (2001) = 9000
In 2011, total population was = 30,000 + 6% of 30,000 = 31,800
Now, 34% of 31,800 is the work force = 10812
Therefore increase in work force = 10812 – 9000 = 1812Correct Option: B
1812 to 1812
Working population (2001) = 9000
In 2011, total population was = 30,000 + 6% of 30,000 = 31,800
Now, 34% of 31,800 is the work force = 10812
Therefore increase in work force = 10812 – 9000 = 1812
- For a room wit h dimensions 4m × 3m × 3m (L × B × H), the details of i ndoor acoustical treatment are as follows.
The reverberation time in seconds at 1000 Hz is _______
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0.4 to 0.5
t = 0.16 × 
V 
= 0.16 (4 × 3 × 3) = 0.37 Sec A 15.44
Calculation of ‘A' Wall area = perimeter of wall × height = [2(4 + 3)] × [4] = 56 m² = P
Ceiling area = 4 × 3 = 12 m² = Q
Floor area = 4 × 3 = 12 m² = R
So, A = P[(0.4 × 30%) + (0.1 × 70%)] + Q[(0.6 × 40%) + (0.1 × 60%)] + R (0.1 × 100%)
= P[(0.4 × 0.3) + (0.1 × 0.7)] + Q[(0.6 × 0.4) + (0.1 × 0.6)] + R[ (0.1 × 1)]
= 56[(0.4 × 0.3) + (0.1 × 0.7)] + 12[(0.6 × 0.4) + (0.1 × 0.6)] + 12 (0.1 × 1)
= 56(0.12 + 0.07) + 12(0.24 + 0.06) + 12 × 0.1
= 56 × 0.19 + 12 × 0.3 + 1.2
= 10.64 + 3.6 + 1.2
= 15.44 secondsCorrect Option: A
0.4 to 0.5
t = 0.16 × V = 0.16 (4 × 3 × 3) = 0.37 Sec A 15.44
Calculation of ‘A' Wall area = perimeter of wall × height = [2(4 + 3)] × [4] = 56 m² = P
Ceiling area = 4 × 3 = 12 m² = Q
Floor area = 4 × 3 = 12 m² = R
So, A = P[(0.4 × 30%) + (0.1 × 70%)] + Q[(0.6 × 40%) + (0.1 × 60%)] + R (0.1 × 100%)
= P[(0.4 × 0.3) + (0.1 × 0.7)] + Q[(0.6 × 0.4) + (0.1 × 0.6)] + R[ (0.1 × 1)]
= 56[(0.4 × 0.3) + (0.1 × 0.7)] + 12[(0.6 × 0.4) + (0.1 × 0.6)] + 12 (0.1 × 1)
= 56(0.12 + 0.07) + 12(0.24 + 0.06) + 12 × 0.1
= 56 × 0.19 + 12 × 0.3 + 1.2
= 10.64 + 3.6 + 1.2
= 15.44 seconds
