Test Level 2: Mensuration - 1 - CAT MCQ

Total surface area of the cone = πr(l + r) = 22/7 × 7(12 + 7) m² = 22 × 19 m² = 418 m²
Total cost incurred = Rs. 418 × 14/1000
= Rs. 58.52

Volume of the earth removed = Area of remaining field × Rise in Height
⇒10 × 4.5 × 3 = (180 – 45) × h
135/135 = h
⇒ h = 1

1 hectare = 10,000 m²
= 10,000 × 10,000 cm²
6 hectare = 6 × 108 cm²
1 m³ = 6 × 108 × thickness (cm³)
10,00,000 cm³ = 6 × 108 × thickness (cm³)
Thickness =  = 0.001666 = 0.0017 cm

Volume of the cylindrical column of increased height = Volume of the rectangular block
⇒ 22/7 × 42 cm × 42 cm × h = 24 cm × 21 cm × 11 cm
h = 1 cm

Let the width of the reservoir be 'w' m.
Volume of water removed = 20 × w ×
0.15 m³
= 3w m³
18 kl = 18 m³
3w = 18
w = 6
Width of the reservoir = 6 m

Diagonal of the cube =

Diameter of the sphere 

Radius of the sphere 

Volume of the hollow sphere = Volume of the hollow sphere of outer radius 10 - Volume of the inner sphere 76π/3 = 4π/3 [R³ - r³]
19 = 10³ - r³
r³ = 1000 - 19
r³ = 981
r = 9.93 ∽ 9.9 cm

Thickness of wood = 2.5 cm
Now, external dimensions of the wooden box are:
Length = 115 + (2 × 2.5) = 120 cm,
Width = 75 + (2 × 2.5) = 80 cm, Height = 35 + (2 × 2.5) = 40 cm
Volume of wood = 120 cm × 80 cm × 40 cm - 115 cm × 75 cm × 35 cm = 82,125 cm³

Bore = Internal diameter of the metallic pipe = 3 cm
Thickness = 1 cm
External diameter = 3 + (1 × 2) = 5 cm
Volume of the metallic pipe

Now, weight of 1 cm³ = 21 gm
Weight of 400π cm³ = 21 × 400 × 22/7 = 26.4 kg

The greatest volume of sphere