PE 2 - Mensuration | Geometry - Mensuration
Join our Telegram Channel for CAT/MBA Preparation.
A 20-cm long cylindrical iron tank is filled with silver. The outer and the inner diameters of the iron tankare 12 cm and 9 cm respectively. Find the weight of the tank (in kg), if 1 cubic centimeter of iron weighs 28 gm and that of bronze weighs 21 gm. [π = 22/7]
Answer: 217.8
Join our Telegram Channel for CAT/MBA Preparation.
Explanation :
Volume of the iron tank = π × (R2 - r2) × h = π × (122 - 92) × 20 = 1260π cm3
Volume of the bronze filled = π × r2 × h = π × 92 × 20 = 1620π cm3
Now,
Weight of the iron tank = 1260 × 22/7 × 28 = 110,880 gms.
Weight of the bronze filled = 1620 × 22/7 × 21 = 106,920 gms.
Total weight of the tank = 110880 + 106920 = 217800 gms = 217.8 kgs.
Hence, 217.8.
Workspace:
A glass, in the shape of a frustum of a cone is full of water. The radii of the bases of the frustum are 6 cm cm and 4 cm and its height is 14 cm. The glass was emptied into another glass, which is conical with a base radius 4 cm. If the second tank is just full, find its height (in m).
- (a)
62.5
- (b)
60
- (c)
54
- (d)
66.5
Answer: Option D
Join our Telegram Channel for CAT/MBA Preparation.
Explanation :
Volume of the frustum of a cone = 1/3 × π × [R2 + Rr + r2] × h
= 1/3 × π × [62 + 6 × 4 + 42] × 14
= 1064/3 cm3.
Let the height of the conical glass be h.
∴ 1/3 × π × r2 × h = 1064/3
⇒ 1/3 × π × 42 × h = 1064/3
⇒ h = 66.5 cm.
Hence, option (d).
Workspace:
A water bottle, which contains water, has a cylinder of radius 2 cm on top of another cylinder of radius 3 cm. When the bottle is upright, the height of the water is 3 cm, as shown in the cross-section of the bottle in Figure 1. When the bottle is upside down, the height of the liquid is 8 cm, as shown in Figure 2. What is the total height of the bottle in cm?
- (a)
10
- (b)
11
- (c)
12
- (d)
13
- (e)
15
Answer: Option C
Join our Telegram Channel for CAT/MBA Preparation.
Explanation :
Let the total height of the bottle he H.
The empty volume in figure 1 = The empty volume in figure 2.
⇒ π × r2 × (H - 3) = π × R2 × (H - 8)
⇒ 22 × (H - 3) = 32 × (H - 8)
⇒ 4H - 12 = 9H - 72
⇒ 5H = 60
⇒ H = 12
Hence, option (c).
Workspace:
Feedback
Help us build a Free and Comprehensive Preparation portal for various competitive exams by providing us your valuable feedback about Apti4All and how it can be improved.