Test Level 2: Mensuration - 2 - CAT MCQ

Here, the height through which the bucket is to be raised = Number of revolutions of the drum × circumference

The figure formed on rotating a right-angled triangle is a right circular cone. First of all, we have to calculate its dimensions.
As shown in the adjoining figure, let the given dimensions of the right-angled triangle be x, x + 2 and 8.
(x + 2)² = x² + (8)²
⇒ x² + 4 + 4x = x² + (8)²
⇒ 4x = 60
⇒ x = 15

 Height of the cylinders = 42 cm
Radius of the outer cylinder = r₁ = 16/2 = 8 cm
Radius of the inner cylinder = r₂ = 14/2 = 7 cm
Therefore, volume of the cork dust required = Volume of the outer cylinder – Volume of the inner cylinder

= 1980 cm³

Average rate of tap = 2/3 drops/sec
Total number of drops in the completely filled vessel = 2080
So, time taken by the tap to drip 2080 drops 
i.e. Time required to fill the vessel = 3120 sec
∴ Number of times that the vessel be filled in this manner in 13 hours 

Lateral surface area (L. S. A.) = 2πrh
Total surface area = (T. S. A.) = 2πr(r + h)

5h = 3h + 3r
2h = 3r

r² = 196
r = 14 m

= 12,936 cubic m

Triangle ABC has area 
Triangles ABW, WBX, XBY, YBZ and ZBC have equal bases and altitudes, so the area of each of these small triangles is 1/5
Similarly, triangles ADW, WDX, XDY, YDZ and ZDC each have area 9/2.
Thus, the shaded region has area 

Diameter of the hemispherical cup = 14 m
Therefore, radius of the hemispherical cup = 7 m
We know that,
Volume of the hemisphere 
Volume of sphere 
Let r be the radius of a stone.
According to the question,
350 × volume of stone = volume of hemispherical cup

700r³ = (7)³
r³ = 0.49

Therefore, radius is 0.79 m.

Radius of cylindrical pipe (r) = 5/2 meters
Flow rate = 20 m/min
Diameter of conical vessel = 80 meters
Depth of the vessel (h) = 48 meters
According to the question,
Volume of the chemical flow through the pipe in one minute,

[here, h = 20 m]
= 125π m³
Volume of conical vessel,

= 25,600 m³
Therefore, time taken
 

Volume of water in the reservoir = Area of cross-section of pipe × Empty rate × Empty time
Let the empty time be t.

The resulting cuboid has dimensions:
ℓ = (6 + 6) cm = 12 cm
b = 6 cm
h = 6 cm
Therefore, surface area of the cuboid
= 2(ℓb + bh + hℓ)
= 2[12 × 6 + 6 × 6 + 6 × 12] cm² = 360 cm² 

Volume of given figure = Volume of cone + Volume of cylinder
Radius of cone = Radius of cylinder, r = 7 cm
Height of cone = h₁ = 9 cm
Height of cylinder, h = 30 - 9 = 21 cm

Volume of figure = 462 + 3234 = 3696 cm³

Total height = h
Number of turns = n
∴ Vertical spacing = h/h

Radius of the hemisphere = Radius of the cylinder = Height of the hemisphere = 3.5 cm
Height of the cylinder = Total height - 2 × (Radius of the hemisphere)
Height of the cylinder = 19 cm - 7 cm = 12 cm
Radius of the hemisphere = 3.5 cm
Surface area of the solid = Surface area of the cylinder + 2 × (Surface area of the hemisphere)
= (2πrh) + 2 × (2πr²)
= (2 × (22/7) × 3.5(12 + 7)) cm² = 22 × 19 sq. cm = 418 cm²

Radius of cone = 12 m
Height of cone = 5 m
Hence, slant height =

ℓ = 13 m
Curved surface area of cone = π rl

(i) Curved surface area of the frustum of the cone

(ii) Total surface area of the frustum of the cone

Let h be the height.
Given:
ℓ = slant height
'a' and 'b' are the two base radii of the frustum of the cone.

Let X be the vertex of the cone and let CX = P, r₃ = 8 cm, r₁ = 18 cm.
Let A, B and C be the centres of largest, middle and smallest spheres.
Let their radii be r₁ (largest), r₂ (middle) and r₃ (smallest), such that
XC = P ... (1)
XB = P + r₃ + r₂..(2)
AX = P + r₃ + 2r₂ + r₁ ... (3)
Applying rule of similarity triangle, we have

From equation (2) we get,

From equation (3) we get,
5K = 13 + r₂........ (5)
Solving equation (4) and equation (5), we get r₂ = 12 cm

Let the initial volume of the cone be

After increasing the radius by 6 units, new volume 


After increasing the height by 9 units, new volume


By putting the value of x in equation (i), we get:

9r² = h[36 + 12r]

By putting the value of h in above equation, we get: 
9r² = 3[36 + 12r]
3r² - 12r - 36 = 0
r² - 4r - 12 = 0
r² - 6r + 2r - 12 = 0
r = 6 units

We'll start by denoting the height of the large cylinder by h₁ and the height of the small cylinder by h₂. For simplicity, we'll suppose x = h₁ + h₂.
If the bottom cylinder is completely filled and the top cylinder is only partially filled, the top cylinder will have a cylindrical space that is not filled. This cylindrical space will have a height equal to x - 20 and a volume equal to π(1)²(x - 20).
Similarly, if we turn the bottle upside down, there will be cylindrical space unfilled that will have a height equal to x - 28 and a volume equal to π(3)²(x - 28).
Since these two unoccupied spaces must be equal, we can write:
π(1)²(x - 20) = π(3)²(x - 28)
x - 20 = 9x - 252
8x = 232
x = 29
Therefore, the total height is 29 cm.

Opening the pyramid, you will get

According to cosine formula,

We cut the can along a line passing through P, Q and R, and we stretch out the tank so that its side is in the shape of a rectangle. Then, P, Q and R appear as points on two opposite edges of the rectangle. The length of the rectangle is 2π × 2 = 4π and its height is 6. We draw two such rectangles beside each other and indicate where P, Q and R lie along the edges, as shown below.

The shortest path from P to Q to R is indicated above (the first time around the tank is indicated by the line segment from P to Q, and the second time around the tank is indicated by the line segment from Q to R). Given that the vertical distance from P to Q is 4 feet and the vertical distance from Q to R is 2 feet, we get from two applications of the Pythagorean theorem that the length of the shortest path is