GATE Architecture and Planning Mock Test - 1 - GATE Architecture and Planning MCQ

In one hour, the minute hand makes one revolution, and the second hand goes round 60 times. This means that, in one hour, the second hand passes over the minute hand 60 - 1 = 59 times. The minute-hand and second-hand cross each other 30 times between 9:15:00 AM and 9:45:00 AM.

Planting a seed is analogous to ‘raising a child’

If x^a = y^b = z^c and y² = zx, let x^a = y^b = z^c = k.

Since y² = xz

From the exponents;

It is mentioned that p and q are positive integers. The least positive integer is 1. The second inequality holds true only if both p and q are 1. So, the value of (p + q) is 2 

Ventilation shaft (NBC 2016-vol 1)
For ventilating the spaces for water-closets and bath rooms, if not opening on to front, side, rear or interior open spaces, these shall open on to the ventilation shaft, the size of which shall not be less than the values given as,

For overall planning approach following density ranges are suggested in URDPFI Guidelines, Volume 1: 

By observation, there should be continuous link from Entry to ‘A’ and then to ‘B’. Link from A to B can be seen only in Option B and the remaining options can be directly eliminated.

National Monuments Authority has been established under the provisions of ‘The Ancient Monuments and Archaeological Sites and Remains (Amendment and Validation) Act, 2010'. The minimum prohibited area in respect of centrally protected monuments and protected areas is 100m, in all directions beginning at the limit of the protected area of the centrally protected monument or site. The minimum limit of regulated area of centrally protected monuments and protected areas is 200m which will begin from the limit of the prohibited area. These limits area extendable as per the classification of the centrally protected monuments and protected areas on the recommendation of the National Monuments Authority. 

The hydraulic mean depth or hydraulic mean radius is the ratio of wetted area to the wetted perimeter.

y_m = A/P

For circular section:

r = Radius of the circular section

Area of flow through circular section = π r²

Wetted perimeter for circular section = 2πr 

y_m = A/P = πr²/2πr = r / 2 =d/4

Calculation: Given,

d = 450 mm; ym=d4

y_m = 450/4=112.5

Hence the hydraulic radius for the circular pipe running full is 112.5 mm

The projected area of the solar cell, which is perpendicular to the sunlight, can be calculated as follows:

• Area = 3 m² x cos(60°) = 1.5 m²

Given that the efficiency of the cell is 18% and the intensity of full sunlight is 1000 W/m², the output power can be determined:

• Output Power = 0.18 x 1000 W/m² x 1.5 m² = 270 W

Therefore, the power output of the solar cell is 270 Watts.

Indirect project Cost:

Indirect project cost are those expenditures which cannot be clearly allocated to the individual activities of the project, but are assessed as a whole.

The indirect cost rises with an increase in duration. 

Direct Project Cost: 

These include labour cost, material cost, equipment cost, etc.

Direct cost decreases with an increase in duration.

Cost slope:

The cost slope is the slope of the direct cost curve, approximated as straight line. It is defined as follows:

cost slope=Crash cost−Normal cost / (Normal time−Crash time)

C_s = C_c − C_n/(t_n − t_c) = ΔC/Δt

Where, Cs = Cost slope, ΔC = Increase in cost, Δt = Decrease in time

Calculation

Given data,

Normal activity duration = 28 days

Crash activity duration = 24 days

Cost slope = 1500/- , Indirect cost = 1000/- per day

Normal cost of activity = 55,000 /-

No. of days crashed = 28 - 24 = 4 days. 

The increase in direct cost for 4 days crashing = Cost Slope × 4 = 1500×4 = 6000 rupees

The decrease in indirect cost for 4 days = 1000×4 = 4000 rupees

Total project cost = Direct cost + Indirect cost

The total cost of the activities after crashing

= 55000 + 6000 – 4000 = 57000 rupees

∴ Total cost = 57,000 /-

Program Evaluation Review Technique (PERT) is a project management planning tool used to calculate the amount of time it will take to realistically finish a project. PERT charts are used to plan tasks within a project — making it easier to schedule and coordinate team members. PERT was developed by the U.S. Navy in the 1950s to help coordinate the thousands of contractors it had working on myriad projects. While PERT was originally a manual process, today there are computerized PERT systems that enable project charts to be created quickly.  PERT is still a popular project management tool today.

In his book Finding Lost Space (1986) Roger Trancik identifies three major approaches to theories of urban form: the figure-ground, linkage, and place theories.

Ekistics is the science of human settlements including regional, city, community planning and dwelling design. Its major incentive was the emergence of increasingly large and complex conurbations, tending even to a worldwide city.

Doxiadis defines 4 principles of Ekistics for solving the problems of human settlements:

(1) Unity of Purpose

(2) Hierarchy of functions (All functions should belong to a pyramid set in a hierarchical way)

(3) Four-dimensional synthesis

(4) Many masters creating many scales

Duration of an activity = Quantity ÷ (Manpower * Productivity).
Based on this the durations of all activities are calculated and tabulated below:

The project network for the given project will be: 

From the network, the critical path is B – C = 9 + 11 = 20 days
Project starts on January 29.
Project completion will be 20 days from January 29.
So, the project completion date will be February 17.   

Loss of water = 40%. So run-off coefficient = 0.6
Volume of stored water from runoff = C*i*A = 0.6 * 0.4 m * 500 sqm = 120 cu.m  
Daily water demand of the household = 3*200 = 600 liters = 0.6 cu.m
No. of days the stored water will be adequate for = 120/0.6 = 200 days 

The form of the folded solid is a cube with one corner cut-off as a triangular prism.
The solid is also a prismatic solid.
Volume for such solids is given by: Area of c/s * Ht  
Area of cross section for the given solid = (3*3) – (0.5*1*1) = 9 – 0.5 = 8.5 sqm
Volume of given solid = Area * Height = 8.5 * 3 = 25.5 cu.m

Let the number of HIG, MIG and LIG dwelling units be: 3a, 2a, and a.  
Total area of HIG = 3a * 100 = 300a
Total area of MIG = 2a * 60 = 120a
Total area of LIG = a * 30 = 30a
Ratio of the total built up area between (MIG + LIG) to HIG = (120a + 30a) ÷ 300a = 150a/300a = 1/2 

• Dolmens are a form of ancient architecture that consists of two or more upright stones topped by a capstone that creates a chamber. Stonehenge has several intact dolmens.
• A Menhir is a large human-made upright stone, typically dating from the European middle Bronze Age. They can be found individually as monoliths, or as part of a group of similar stones
• In Ancient Roman architecture, a basilica is a large public building with multiple functions, typically built alongside the town's forum. The basilica was in the Latin West equivalent to a stoa in the Greek East. The building gave its name to the architectural form of the basilica.
• An obelisk is a tall, four-sided, narrow tapering monument which ends in a pyramid-like shape or pyramidion at the top.

A tatami is a type of mat used as a flooring material in traditional Japanese-style rooms. Tatamis are made in standard sizes, twice as long as wide, about 0.9 m by 1.8 m depending on the region. 

Lothal is the only explored port-town of the Indus Valley civilisation, or Harappan Civilization, from the Bronze Age. But was not added to the list in the year 2022.

Alumina is the main constituent of clay. It acts as a cementing material in raw brick. Brick clay is plastic due to the presence of alumina. This plasticity ensures that bricks can be molded. An excess amount of alumina in clay may cause the bricks to shrink, warp or crack on drying and burning as any other cementing material.
brick composition: -
1) silica- silica is about 50 to 60% of dry weight of bricks
2) Alumina is about 20 to 30% of dry weight of bricks
3) lime is about 10 to 12% dry weight of bricks
4) ferric oxide is about less than 7% of dry weight of bricks
5) magnesia is about less than 1% of dry weight of bricks
6) Alkalies is about less than 10% of dry weight of bricks.

• Rusting of iron can be easily prevented by coating the surface with paint which protects iron from air and moisture.
• Galvanized metal is coated with a thin layer of zinc to protect it against corrosion. The zinc oxidizes when it is exposed to air creating a protective coating on the metal surface.
• Charring means to burn and blacken the outer layer of the lumber with flames. A preservative treatment like this helps prevent the lumber from rotting when exposed to severe moisture situations, or where it may come into direct contact with concrete or soil.
• Quench hardening is a mechanical process in which steel and cast-iron alloys are strengthened and hardened. These metals consist of ferrous metals and alloys. This is done by heating the material to a certain temperature, depending on the material.

Time taken for sound to reach first listener T₁ = Dist. / Velocity
T₁ = 680 / 340 = 2 sec
T₂ = 1360 / 340 = 4 sec
Time lag = T₁ – T₂ = 2 Sec 

To calculate the thermal transmittance (U) value, we need to determine the total thermal resistance (R_total) of the wall assembly. The resistances from the brick wall, cement plaster layers, and air films must be considered.

1. Convert the thermal resistivity of the brick wall to thermal conductivity, k_brick = 1 / 0.0143 = 69.93 kcal cm/(m² h deg C).
2. Calculate the resistance of the brick wall, R_brick = thickness / k_brick = 12 cm / 69.93 = 0.1717 m² h deg C/(kcal).
3. Calculate the resistance of one cement plaster layer, R_plaster = thickness / k_plaster = 1.5 cm / 81.8 = 0.0183 m² h deg C/(kcal).
4. Since there are two layers of cement plaster, the total resistance of the plaster layers is 2 × R_plaster = 2 × 0.0183 = 0.0366 m² h deg C/(kcal).
5. Add the air film resistances: R_air = 1/fi + 1/fo = 0.125 + 0.0515 = 0.1765 m² h deg C/(kcal).
6. Calculate the total resistance: R_total = R_brick + 2 × R_plaster + R_air = 0.1717 + 0.0366 + 0.1765 = 0.3848 m² h deg C/(kcal).
7. The thermal transmittance (U) value is the inverse of the total resistance: U = 1 / R_total = 1 / 0.3848 = 2.60 kcal/(m² h deg C).

Total self-weight (load) of brick wall = volume * density = (4 * 2 * 0.25) * (18 KN/m³) = 36 kN 
Total self-weight (load) of RCC beam = volume * density = (4 * 0.6 * 0.25) * (25 KN/m³) = 15 kN
Total UDL on the given beam = 36 + 15 = 51 kN 
Total UDL value on beam = Total load/span = 51/4 = 12.75
Maximum bending moment in the beam = wL²/8 = (12.75 * 4 * 4)/8 = 25.5 kNm