1. If the ratio of volume of two spheres is 1:8, then find the ratio of their surface area.
Solution:
Sol: Given, ratio of volumes = (1:8)
For spheres,
Volume ∝ r³
r₁³ / r₂³ = 1 / 8 ⇒ r₁ / r₂ = 1 / 2
Surface area of a sphere ∝ r²
Ratio of surface areas = 1² : 2² = 1:4
Ans: 1:4
2. Two cubes each of volume 64 cm3 are joined end to end. Find the surface area of the resulting cuboid.
Solution:
Sol: Volume of each cube = 64 cm³
Let side of cube = a
a³ = 64 ⇒ a = 4 cm
Two cubes joined end to end → dimensions of cuboid:
l = 8 cm, b = 4 cm, h = 4 cm
Surface area of cuboid:
= 2(lb + bh + hl)
= 2(8×4 + 4×4 + 8×4)
= 2(32 + 16 + 32) = 2×80 = 160
Ans: 160 cm²
3. Find the length of the longest rod that can be kept in a room of dimensions 12 cm × 9 cm × 8 cm. Hint: [Diagonal of Cuboid]
Solution:
Sol: Dimensions of room: 12 cm, 9 cm, 8 cm
Space diagonal:
d = √(l² + b² + h²)
= √(12² + 9² + 8²)
= √(144 + 81 + 64)
= √289
= 17
Ans: 17 cm
4. The decorative block shown in fig. is made of two solids-a cube and a hemisphere. The base of the block is a cube with edge 5 cm and the hemisphere fixed on the-top has a diameter of 4.2 cm. Find the total surface area of the block. [take π = 22/7]
Solution:
Sol: Cube edge = 5 cm
Surface area of cube:
= 6a² = 6×5² = 150 cm²
Hemisphere diameter = 4.2 cm
r = 2.1 cm
Area of circular base (covered part):
= πr²
Curved surface area of hemisphere:
= 2πr²
Total surface area of block:
= 150 - πr² + 2πr²
= 150 + πr²
Using π = 22/7, r = 2.1
πr² = (22/7) × (2.1)²
= (22/7) × 4.41
= 13.86
Total surface area = 150 + 13.86 = 163.86
Ans: 163.86 cm²
5. The radii of the circular ends of a bucket of height 15 cm are 14 cm and r cm (r < 14 cm). If the volume of bucket is 5390 cm3. Find the value of r. [Use π = 22/7]
Solution:
Sol: Height (h) = 15 cm
R = 14 cm, r = r (r < 14 cm)
Volume = 5390 cm³
Volume of frustum of cone:
V = (1/3) πh (R² + r² + Rr)
5390 = (1/3) × (22/7) × 15 × (14² + r² + 14r)
5390 = (1/3) × (22/7) × 15 × (196 + r² + 14r)
5390 = 110 × (196 + r² + 14r)
196 + r² + 14r = 5390 / 110 = 49
r² + 14r + 196 = 49
r² + 14r + 147 = 0
(r + 7)(r + 21) = 0
r = -7 or -21
Since radius cannot be negative,
Ans: r = 7 cm
6. From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid.
Solution:
Sol: Height (h) = 2.4 cm
Diameter = 1.4 cm ⇒ r = 0.7 cm
Slant height of cone:
l = √(h² + r²)
= √(2.4² + 0.7²)
= √(5.76 + 0.49)
= √6.25 = 2.5 cm
Total surface area = Curved surface area of cylinder + Area of base of cylinder + Curved surface area of cone
7. If the surface area of a sphere is 144 p cm2, then what is its radius?
Solution:
Sol: Surface area of sphere = 4πr²
4πr² = 144π
Divide both sides by 4π:
r² = 144π / 4π = 36
r = √36 = 6 cm
Ans: Radius = 6 cm
8. The radii of bases of cylinder and a cone are in the ratio 3 : 4 and their heights are in the ratio 2:3 then ratio between the volume of cylinder to volume of cone is?
Solution:
Sol: Radii ratio (cylinder : cone) = 3 : 4
Heights ratio (cylinder : cone) = 2 : 3
Volume of cylinder = πr²h
Volume of cone = (1/3)πr²h
Required ratio:
= π(3)²(2) : (1/3)π(4)²(3)
= π × 9 × 2 : (1/3)π × 16 × 3
= 18π : 16π
= 18 : 16 = 9 : 8
Ans: 9 : 8
9. If two identical cubes of side 'a' are joined end to end. Then the total surface area of resulting cuboid is ______.
Solution:
Sol: Two identical cubes of side a joined end to end
Dimensions of cuboid:
l = 2a, b = a, h = a
Total surface area of cuboid:
= 2(lb + bh + hl)
= 2(2a×a + a×a + 2a×a)
= 2(2a² + a² + 2a²)
= 2(5a²) = 10a²
Ans: 10a²
10. A well of diameter 4 m is dug 21 m deep. The earth taken out of it has been spread evenly all around it in the shape of a circular ring of width 3 m to form an embankment. Find the height of the embankment.
Solution:
Sol: Diameter of well = 4 m ⇒ radius r₁ = 2 m
Depth of well = 21 m
Volume of earth dug out (cylinder):
= πr₁²h
= π × 2² × 21
= 84π m³
Earth is spread as a circular ring
Width of ring = 3 m
Outer radius r₂ = 2 + 3 = 5 m
Let height of embankment = h
Volume of ring:
= π(r₂² - r₁²)h
= π(5² - 2²)h
= π(25 - 4)h
= 21πh
Since volumes are equal:
21πh = 84π
h = 84π / 21π = 4 m
Ans: Height = 4 m
11. The sum of the radius of base and height of a solid right circular cylinder is 37 cm. If the total surface area of the solid cylinder is 1628 sq. cm, find the volume of the cylinder, (use π=22/7)
Solution:
Sol: Given: r + h = 37 cm
Total surface area = 1628 cm²
Total surface area of cylinder:
= 2πr(r + h)
1628 = 2 × (22/7) × r × 37
1628 = (44/7) × 37 × r
1628 = (1628/7) r
Multiply both sides by 7:
11396 = 1628r
r = 7 cm
h = 37 - 7 = 30 cm
Volume of cylinder:
= πr²h
= (22/7) × 7² × 30
= (22/7) × 49 × 30
= 22 × 7 × 30
= 4620 cm³
Ans: Volume = 4620 cm³
12. A right circular cone of radius 3 cm, has a curved surface area of 47.1 cm2. Find the volume of the cone, (use π = 3.14)
Solution:
Sol: Radius r = 3 cm
Curved surface area = 47.1 cm²
Curved surface area of cone = πrl
47.1 = 3.14 × 3 × l
47.1 = 9.42 l
l = 47.1 / 9.42 = 5 cm
Height of cone:
l² = r² + h²
5² = 3² + h²
25 = 9 + h²
h² = 16 ⇒ h = 4 cm
Volume of cone:
= (1/3) π r² h
= (1/3) × 3.14 × 3² × 4
= (1/3) × 3.14 × 9 × 4
= (1/3) × 113.04
= 37.68 cm³
Ans: 37.68 cm³
13. If the total surface area of a solid hemisphere is 462 cm², find its volume.
Solution:
Sol: Total surface area of hemisphere = 462 cm²
Total surface area = 3πr²
3πr² = 462
Using π = 22/7:
3 × (22/7) × r² = 462
(66/7) r² = 462
r² = (462 × 7) / 66 = 49
r = 7 cm
Volume of hemisphere:
= (2/3) π r³
= (2/3) × (22/7) × 7³
= (2/3) × (22/7) × 343
= (2/3) × 22 × 49
= (44/3) × 49
= 2156 / 3
= 718.67 cm³ (approx)
Ans: 718.67 cm³
14. A 5 m wide cloth is used to make a conical tent of base diameter 14 m and height 24 m. Find the cost of cloth used at the rate 25 rupees/ meter.
Solution:
Sol: Diameter = 14 m ⇒ r = 7 m
Height h = 24 m
Width of cloth = 5 m
Slant height:
l = √(r² + h²)
= √(7² + 24²)
= √(49 + 576)
= √625 = 25 m
Curved surface area of cone (cloth needed):
= πrl
= (22/7) × 7 × 25
= 22 × 25 = 550 m²
Cloth is 5 m wide, so length required:
= Area / width = 550 / 5 = 110 m
Cost at ₹25 per meter:
= 110 × 25 = 2750 rupees
Ans: ₹2750
15. A toy is in the form of a cone mounted on a hemisphere of same radius 7 cm. If the total height of the toy is 31 cm, find its total surface area.
Solution:
Sol: Radius r = 7 cm
Total height of toy = 31 cm
Height of hemisphere = r = 7 cm
Height of cone:
h = 31 - 7 = 24 cm
Slant height of cone:
l = √(r² + h²)
= √(7² + 24²)
= √(49 + 576)
= √625 = 25 cm
Total surface area = Curved surface area of cone + Curved surface area of hemisphere
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