3D geometry is a crucial topic in many competitive exams, covering volume and surface area calculation, coordinate geometry and vector applications. Competitors should understand formulas and their applications to solve problems effectively. Practicing various problems is necessary to gain confidence in the subject and excel in exams.
Assignment:1
Que 1: What is the volume of water, in liters, pumped out in one hour through a circular pipe with an internal diameter of 7cm, if the flow rate of water is 12 cm per second?
(a) 1663.2
(b) 1576.2
(c) 1747.6
(d) 2024.6
Solution: 1 hour= 3600 seconds
h=3600×12 cm
V= 𝜋r2h=22/7×(7/2)2×3600×12
V= 22×7×3×3600 cm3
Since, 1 ltr= 1000 cm3
V=22×7×3×3.6 ltr
Unit digit=2
Only option (a) is divisible by 4
Option (a) is correct
Que 2: How long, in hours, will it take for the water level in a tank, 200m long and 150m wide, to reach 8m if water flows into the tank through a pipe of cross-section 0.3m × 0.2m at a speed of 20 km/hour?
(a) 50
(b) 120
(c) 150
(d) 200
Solution: 200×150×8=0.3×0.2×20000×k
k=200 hr
Option (d) is correct
Que 3: What is the ratio of volumes of solids V1, V2, V3, and V4 when a right circular cone is divided by three planes parallel to its base, which also divide the altitude into four equal parts?
(a) 1:2:3:4
(b) 1:6:15:29
(c) 1:7:19:37
(d) 1:8:27:64
Solution: V1: V1+V2: V1+V2+V3: V1+V2+V3+V4= 13:23:33:43= 1:8:27:64
V1: V2:V3:V4= 1:7:19:37
Option (c) is correct
Que 4: What is the volume of a pyramid with a regular hexagonal base of side length 2a cm, if each slant edge of the pyramid has a length of 5a/2 cm?
(a) 3a3 cm3
(b) 3√6 a3 cm3
(c) 3√3 a3 cm3
(d) 6a3 cm3
Solution: Let O be the center of the base and B be one of its vertices. A is the top of the pyramid.
In right angle triangle △AOB,
AO2=AB2-OB2 (Pythagoras theorem)
OB= Side length of hexagon=2a
AO= 5a/2
The height of the Pyramid (AO)=3a/2
Volume= 1/3×Base Area×Height=3√3 a3
Option (c) is correct
Que 5: What is the height of a right prism whose base is a triangle with a perimeter of 28 cm and an inscribed circle radius of 4 cm, given that the volume of the prism is 366 cc?
(a) 5.82 cm
(b) 4.37 cm
(c) 8.06 cm
(d) 6.53 cm
Solution: Area of the triangle (△)=rs= 4×14=56 cm2
The volume of the Prism= △×h=366
h=6.53 cm
Option (d) is correct
Que 6: If the total surface area of a solid right circular cylinder is twice that of a solid sphere, and they have the same radii, what is the ratio of the volume of the cylinder to that of the sphere?
(a) 9:4
(b) 16:9
(c) 3:1
(d) 9:16
Solution: 2𝜋rh+2𝜋r2=2×4𝜋r2
h=3r
𝜋r2h: 4/3×𝜋r3=9:4
Option (a) is correct
Que 7: What is the depth of water in a cylindrical can with an internal radius of 3.5 cm, before placing a solid sphere inside the can, if the sphere fits in the can exactly and the water just covers the sphere when it is placed inside the can?
(a) 35/3 cm
(b) 28/3 cm
(c) 14/3 cm
(d) 7/3 cm
Solution: The depth of water= d
The radius of the cylinder and sphere=r
Height of Cylinder=2r
The volume of Cylinder- Volume of Sphere= Volume of Water
𝜋r2×2r -4/3×𝜋r3=𝜋r2×d
d=2r/3
r=3.5 cm
d=7/3 cm
Option (d) is correct
Que 8: If the radius of a right circular cylinder is decreased by 50% and its volume is decreased by 60%, what is the percentage change in the height of the cylinder?
(a) +40%
(b) +60%
(c) -50%
(d) -66.67%
Solution: Radius⇒ 2:1
Volume⇒ 5: 2
Height= 5/22 : 2/12=5: 8
Option (b) is correct
Que 9: What is the volume of rainwater, in m³, that can be collected from 1.5 hectares of ground during a rainfall of 5 cm?
(a) 75
(b) 750
(c) 7500
(d) 75000
Solution: 1 hectare= 10000 m2
The volume of rain water= Base Area×Height= 15000×5/100=750 cubic meter
Option (b) is correct
Que 10: If the ratio of weights of two spheres made of different materials is 8:17, and the ratio of weights per 1 cc of material of each sphere is 289:64, what is the ratio of radii of the two spheres?
(a) 8:17
(b) 4:17
(c) 17:4
(d) 17:8
Solution: \frac{289\times x^3}{64\times y^3}=\frac{8}{17}.
x:y=8:17
Option (a) is correct