Architecture and Planning Miscellaneous-topic Difficult Questions and Answers | Page - 25

Architecture and Planning Miscellaneous-topic

A solar photo-voltaic system is proposed to be installed at the roof top of a hostel. The cost of installation and the annual maintenance are INR 2,40,000 and INR 6000 respectively. It is expected to generate 600 kWh of electricity per month. Assume unit price of electricity as INR 5. Ignoring the discount rate, the payback period of the investment in years is____________

1. 8 to 8
1. 6 to 6
1. 4 to 4
1. 2 to 2

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8 to 8
Let the payback time be ‘PB' year.
So, 240000 + 6000PB = 600 × 5 × 12 × PB
⇒ 240000 + 6000.PB = 36000.PB
⇒ 240000 = 30000.PB
⇒ PB = 8 Years

Correct Option: A

8 to 8
Let the payback time be ‘PB' year.
So, 240000 + 6000PB = 600 × 5 × 12 × PB
⇒ 240000 + 6000.PB = 36000.PB
⇒ 240000 = 30000.PB
⇒ PB = 8 Years

A 250 mm × 250 mm RCC column is reinforced with one percent steel. The permissible compressive stress of concrete and steel are 8 N/ mm² and 150 N/mm² respectively. The axial load carrying capacity of the column in kN is _____________

1. 585 to 595
1. 785 to 895
1. 58.9 to 5.75
1. 52.87 to 69.58

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585 to 595
P_u = 0.4 f_ck A_c + 0.67 f_y A_sc
P_u = 0.4 × 8 × 0.0625 (1-1/100) + 0.67 × 150 × 0.0625 × 1/100 = 0.198 + 0.063 = 0.261 = 261 kN
Where,
P_u = factored axial load on the member,
f_ck = characteristic compressive strength of the concrete,
A_c = area of concrete,
f_y = characteristic strength of the compression reinforcement, and
A_sc = area of longit udinal reinfor cement for columns.

Correct Option: A

585 to 595
P_u = 0.4 f_ck A_c + 0.67 f_y A_sc
P_u = 0.4 × 8 × 0.0625 (1-1/100) + 0.67 × 150 × 0.0625 × 1/100 = 0.198 + 0.063 = 0.261 = 261 kN
Where,
P_u = factored axial load on the member,
f_ck = characteristic compressive strength of the concrete,
A_c = area of concrete,
f_y = characteristic strength of the compression reinforcement, and
A_sc = area of longit udinal reinfor cement for columns.

In a 20 story building with 3m floor to floor height, a passenger lift is hoisted by a steel rope. Weight of the lift car is 750 kg and ultimate load the steel rope can carry is 39,000 kg. Assuming a factor of safety of 20 for the steel rope and an average passenger weight of 75 kg, the passenger capacity of the lift is ____________

1. 12 to 12
1. 14 to 14
1. 16 to 16
1. 18 to 18

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16 to 16
Usable strength of steel rope = 39000/20 = 1950kg
Strength of steel rope considering lift-car weight = 1950 – 750 = 1200kg
Capacity of the lift = 1200/75 = 16 persons.

Correct Option: C

16 to 16
Usable strength of steel rope = 39000/20 = 1950kg
Strength of steel rope considering lift-car weight = 1950 – 750 = 1200kg
Capacity of the lift = 1200/75 = 16 persons.

A room is mechanically ventilated through four air-conditioning ducts. The opening area of each duct is 0.35 sqm. The air velocity in the duct is 0.5 m/s. The temperature difference between the ambient air and supply air is 10°C. Volumetric specific heat of air is 1250 J/m³ °C. Assuming one Ton of refrigerat ion (TR) equals 3.5 kW, the cooling load of the room in TR will be _____________

1. 3.75 : 4.55
1. 5.65 : 6.75
1. 9.65 : 10.95
1. 2.45 : 2.55

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2.45 to 2.55
Total area of the duct = 4 × 0.35 = 1.4 m²
Total air volume in the room collected in 1 hr (3600 sec) = 1.4 × 0.5 × 3600 = 2520 m³
Total energy required = mcΔt = 2520 × 1250 × 10
= 31500000
Joules Load in TR = (31500000/3.4) = 2.5 ton
3600

Correct Option: D

2.45 to 2.55
Total area of the duct = 4 × 0.35 = 1.4 m²
Total air volume in the room collected in 1 hr (3600 sec) = 1.4 × 0.5 × 3600 = 2520 m³
Total energy required = mcΔt = 2520 × 1250 × 10
= 31500000
Joules Load in TR = (31500000/3.4) = 2.5 ton
3600

A steel I-beam section is subjected to a bending moment of 96 kN-m. The moment of inertia of the beam section is 24,000 cm⁴. The bending stress at 100 mm above the neutral axis of the beam in MPa will be ____________

1. 40 : 40
1. 30 : 30
1. 20 : 20
1. 10 : 10

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40 to 40
= 96 × 1000 × 1 40 × 10⁶Pa
(24000 × 10^-8

= 40 × 106 Pa = 40 MPa
Bending stress = (My)/I
M = the internal bending moment about the section's neutral axis
y = the perpendicular distance from the neutral axis to a point on the section
I = the moment of inertia of the section area about the neutral axis

Correct Option: A

40 to 40
= 96 × 1000 × 1 40 × 10⁶Pa
(24000 × 10^-8

= 40 × 106 Pa = 40 MPa
Bending stress = (My)/I
M = the internal bending moment about the section's neutral axis
y = the perpendicular distance from the neutral axis to a point on the section
I = the moment of inertia of the section area about the neutral axis