Test Level 1: Mensuration - 2 - CAT MCQ

Height of water in the cylinder before the cone is taken out = 24 cm
Volume of water in the cylinder before the cone is taken out = Volume of cone + Volume of water
Let H be the height of the water level after the cone is taken out.




H = 16 cm

Let the two spheres be A and B.

Volume of a sphere

Surface area = 4πr² 

Ratio of the volume to the surface area = r : 3

Given height and diameter:

Length of arc AXB = Circumference of the base of the cone

= 20.95 cm
Let the radius of the base be r.
So, 2πr = 20.95

44r = 146.65 
r = 146.65/44
r = 3.33 cm
Now, to find out height we can use Pythagorean theorem.
Let the height of the cone be h.
So, h² = R² - r²
h² = 100 - 11.08

h = 9.42 cm
Now, volume of the cone 


= 109.3 cm³

Radius of the sphere, r = 7 meters
Radius of the cone, r₂ = 7 centimeters = 0.07 meters
Height of the cone, h = 7 centimeters = 0.07 meters


Solving,
Number of cones = 4 × 10⁶

Volume of the conical vessel = (1/3) × π × (20)² × 24 = 3200π cm³
Volume flown in 1 minute = (π × (2.5/10)² × 1000) = 62.5π cm³
Therefore, time taken = 3200π/62.5π = 51 min 12 sec

Number of bricks required

= 6400

Outer diameter = do = 4.5 cm 
Inner diameter = di = 4 cm 


Weight of the pipe  

= 187 gm

Radius of semi-circle = 3.5 m
Volume of the building = Volume of cubical portion + Volume of vertical cross-section

Radius of the hemisphere = Base radius of the cone = 3.5 m

Let the height of the cone be h metres.

⇒  h = 4.67 m

Volume of the sphere = Volume of the wire (cylinder)

⇒ h = 24,300 cm = 243 metres

Let the radius of the original sphere be R and that of each smaller one be r.
⇒ r = R/8
Then, volumes of the larger and smaller spheres are 


Number of smaller spheres formed

Surface area of original sphere = 4πR² 
Surface area of each smaller sphere 

Surface area of all smaller spheres = Number of small spheres × Surface area of each smaller sphere 8³ × 0.8 × 4πR²/64 = 6.4 × (4πR²)
Therefore, ratio of the total surface area of all smaller spheres to the surface area of original sphere 

Volume of paint required = 6.4 × Volume of paint used on original sphere = 6.4 × 20 = 128 litres

Let hypotenuse = x cm
Then, by Pythagoras theorem:
x² = (48020)² + (36015)²
x ⇒ 60025 cm

Let inner radius = r; then 2π r = 440
∴ r = 70
Radius of outer circle = 70 + 14 = 84 cm ∴ Outer diameter = 2 × Radius = 2 × 84 = 168

Let outer radius = R; then inner radius = r = R – 7
2πR = 220 ⇒ R = 35 m;
r = 35 – 7 = 28 m
Area of track = π R² – πr² ⇒ p(R² – r²)
= 1386 m²

Volume of soil removed = l × b × h = 7.5 × 6 × 1.5 = 67.5 m³

Let the common ratio be = x
Then; length = 3x, breadth = 2x and height = x
Then; as per question 3x · 2x · x = 1296 ⇒ 6x³ = 1296
⇒ x = 6 m
Breadth = 2x = 12 m

Let the common ratio be = x
Then, length = 4x, breadth = 3x and height = 2x
As per question;
2(4x · 3x + 3x · 2x + 2x · 4x) = 8788
2(12x² + 6x² + 8x²) = 8788 ⇒ 52x² = 8788
⇒ x = 13 Length = 4x = 52 cm

Slant length 
Then curved surface area = πrl = π × 6 × 10⇒ 60p
And total surface area = πrl + πr² ⇒ π ((6 × 10) + 6²)
= 96π cm²