OP Malhotra Class 10 Maths Solutions Chapter 15 Three Dimensional Solids Ex 15(d)

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S Chand Class 10 ICSE Maths Solutions Chapter 15 Three Dimensional Solids Ex 15(d)

Question 1.
Find the volume and curved surface of the sphere if \(\left(\pi=\frac{22}{7}\right)\)
(i) radius = 1 cm
(ii) radius = 14 cm
(iii) radius = 28 cm
(iv) radius = 5\(\frac { 1 }{ 4 }\) cm
(v) diameter = 14 cm
(vi) diameter = 35 cm
Solution:
(i) Radius of sphere (r) = 1 cm
∴ volume = \(\frac { 4 }{ 3 }\)πr³ = \(\frac { 4 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × 1 × 1 × 1 cm³ = \(\frac { 88 }{ 21 }\)cm³
and curved surface area = 4πr²
= 4 × \(\frac { 22 }{ 7 }\) × 1 × 1 = \(\frac { 88 }{ 7 }\) cm²

(ii) Radius (r) = 14 cm
∴ volume = \(\frac { 4 }{ 3 }\)πr³
= \(\frac { 4 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × 14 × 14 × 14 cm³
= \(\frac { 34496 }{ 3 }\) = 11498\(\frac { 2 }{ 3 }\) cm³
curved surface area = 4πr²
= 4 × \(\frac { 22 }{ 7 }\) × 14 × 14 cm² = 2464 cm²

(iii) Radius (r) = 28 cm
volume = \(\frac { 4 }{ 3 }\)πr³ = \(\frac { 4 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × 28 × 28 × 28 cm³
= \(\frac { 275968 }{ 3 }\) = 91989\(\frac { 1 }{ 3 }\) cm³
curved surface area = 4πr²
= 4 × \(\frac { 22 }{ 7 }\) × 28 × 28 cm³ = 9856 cm²

(iv) Radius (r) = 5\(\frac { 1 }{ 4 }\) cm = \(\frac { 21 }{ 4 }\) cm
∴ volume = \(\frac { 4 }{ 3 }\)πr³
= \(\frac { 4 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × \(\frac { 21 }{ 4 }\) × \(\frac { 21 }{ 4 }\) × \(\frac { 21 }{ 4 }\) cm³ = \(\frac { 4851 }{ 8 }\) = 606\(\frac { 3 }{ 8 }\) cm³
curved surface area = 4πr²
= 4 × \(\frac { 22 }{ 7 }\) × \(\frac { 21 }{ 4 }\) × \(\frac { 21 }{ 4 }\) cm²
= \(\frac { 693 }{ 2 }\) = 346\(\frac { 1 }{ 2 }\) cm²

(v) Diameter = 14 cm
∴ Radius (r) = \(\frac { 14 }{ 2 }\) = 7 cm
∴ volume = \(\frac { 4 }{ 3 }\)πr³ = \(\frac { 4 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × 7 × 7 × 7 cm³
= \(\frac { 4312 }{ 3 }\) = 1437\(\frac { 1 }{ 3 }\) cm³ = 1437.3 cm³
curved surface area = 4πr²
= 4 × \(\frac { 22 }{ 7 }\) × 7 × 7 cm² = 616 cm²

(vi) Diameter = 35 cm
∴ Radius (r) = \(\frac { 35 }{ 2 }\) cm
∴ volume = \(\frac { 4 }{ 3 }\) πr³
= \(\frac { 4 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × \(\frac { 35 }{ 2 }\) × \(\frac { 35 }{ 2 }\) × \(\frac { 35 }{ 2 }\) cm³
= \(\frac { 67375 }{ 3 }\) = 22458\(\frac { 1 }{ 3 }\) cm³
and curved surface are = 4πr²
=4 × \(\frac { 22 }{ 7 }\) × × \(\frac { 35 }{ 2 }\) × × \(\frac { 35 }{ 2 }\) = 3850 cm²

Question 2.
Find the radius of the sphere whose surface area is equal to its volume.
Solution:
Let r be the radius of a sphere, then
Surface area = 4πr²
and volume = \(\frac { 4 }{ 3 }\)πr³
∵ Surface area = volune
∴ 4πr² = \(\frac { 4 }{ 3 }\)πr³
1 = \(\frac { 1 }{ 3 }\)r ⇒ r = 3
∴ Radius = 3 units

Question 3.
Find the diameter of the sphere whose surface area is equal to the area of a circle of diameter 2.8 cm.
Solution:
Let r be the radius of sphere = r
∴ Surface area = 4πr²
Diameter of a circle = 2.8 cm
Radius = 2.8 = 1.4 cm
∴ Area of the circle = πr² = π × 1.4 × 1.4 = 1.96 π
∵ The surface area of sphere = area of the circle
∴ 4πr² = 1.96π
r² = \(\frac{1.96 \pi}{4 \pi}\) = 0.49 = (0.7)²
∴ r = 0.7 cm
∴ Diameter of sphere = 2r = 2 × 0.7 = 1.4 cm

Question 4.
Find the radius of the sphere whose volume is \(\frac { 32 }{ 3 }\)π cm³.
Solution:
Volume of sphere = \(\frac { 32 }{ 3 }\)π cm³
Let r be the radius of the sphere, then
\(\frac { 4 }{ 3 }\)πr³ = \(\frac { 32 }{ 3 }\)π
r³ = \(\frac{32 \pi \times 3}{3 \times 4 \times \pi}\) = 8 = (2)³
∴ r = 2 cm
∴ Radius of the sphere = 2 cm

Question 5.
The volume of a sphere is 4.851 cm³. Find its surface area.
Solution:
Let r be the radius of the sphere
∴ Volume = \(\frac { 4 }{ 3 }\)πr³ = 4.851
∴ \(\frac { 4 }{ 3 }\)πr³ = 4.851
\(\frac { 4 }{ 3 }\) × \(\frac { 22 }{ 7 }\)r³ = 4.851 ⇒ r³ = \(\frac{4.851 \times 3 \times 7}{4 \times 22}\)
⇒ r³ = 1.157625 = (1.05)³
∴ r = 1.05 cm
Now surface area = 4πr²
= \(\frac{4 \times 22}{7}\) × (1.05)² cm² = 13.86 cm²

Question 6.
Find the volume of a hollow sphere whose outer diameter is 10 cm and the thickness of the material of which it is made is 1 cm.
Solution:
Outer diameter of hollow sphere = 10 cm
∴ Outer radius (R) = \(\frac { 10 }{ 2 }\) = 5 cm
Thickness of material = 1 cm
∴ Inner radius (r) = 5 – 1 = 4 cm
∴ Volume of hollow sphere = \(\frac { 4 }{ 3 }\)π [R³ – r³]
= \(\frac { 4 }{ 3 }\) × \(\frac { 22 }{ 7 }\) [(5)³ – (4)³]
= \(\frac { 88 }{ 21 }\) [125 – 64] = \(\frac { 88 }{ 21 }\) × 61 cm³
= 255.619 = 255.6 cm³

Question 7.
The surface areas of two spheres are in the ratio 4 : 9. Find the ratio of their volumes.
Solution:
Let r₁ and r₂ be the radii of two spheres
Then surface area of the first sphere = 4πr₁²
and surface area of the second = 4πr₂²
∴ 4πr₁² : 4πr₂² = 4 : 9
⇒ r₁² : r₂² = 4 : 9
⇒ \(\frac{r_1^2}{r_2^2}\) = \(\frac { 4 }{ 9 }\) = \(\left(\frac{r_1}{r_2}\right)^2\) = \(\left(\frac{2}{3}\right)^2\)
∴ \(\frac{r_1}{r_2}\) = \(\frac { 2 }{ 3 }\)
Now volume of the first square = \(\frac { 4 }{ 3 }\)πr₁³
and volume of the second sphere = \(\frac { 4 }{ 3 }\)πr₂³

∴ Ratio in the volumes of two spheres will be 8 : 27

Question 8.
The volumes of two spheres are in the ratio 64 : 27. Find their radii if the sum 4 of their radii is 21 cm.
Solution:
Let r₁ and r₂ be the radii of two spheres
Then volume of first sphere = \(\frac { 4 }{ 3 }\)πr₁³
and volume of second sphere = \(\frac { 4 }{ 3 }\)πr₂³

∴ Their radii are 12 cm and 9 cm

Question 9.
Find the cost of painting a hemispherical dome of diameter 10 m at the rate of Rs. 1.40 per m².
Solution:
Diameter of hemisphere dome = 10 m
∴ Radius (r) = \(\frac { 10 }{ 2 }\) = 5 m
Now curved surface area = 2πr
= 2 × \(\frac { 22 }{ 7 }\) × 5 × 5 = \(\frac { 1100 }{ 7 }\) m²
Rate of painting = Rs. 1.40 per m²
∴ Total cost of painting = \(\frac { 1100 }{ 7 }\) × 1.40
= \(\frac{1100 \times 140}{7 \times 100}\) = Rs. 220

Question 10.
A certain spherical ball of diameter 4 cm weighs 8 kg. Find the weight of a spherical ball of the same material whose inner and outer diameters are 8 cm and 10 cm respectively.
Solution:
Diameter of a ball = 4 cm
∴ Radius (r) = \(\frac { 4 }{ 2 }\) = 2 cm
and volume = \(\frac { 4 }{ 3 }\)πr³
= \(\frac { 4 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × 2 × 2 × 2 cm³ = \(\frac { 704 }{ 21 }\)cm³
Weight of ball = 8 kg
∴ weight of 1 cm³ = \(\frac{8 \times 21}{704}\) kg
= \(\frac{8 \times 21 \times 1000}{704}\) gm
= \(\frac{21 \times 1000}{88}\) = \(\frac{21 \times 125}{11}\) gm = \(\frac{2625}{11}\) gm
Now diameters of spherical ball are 8 cm and 10 cm
∴ Their radii will be 4 cm and 5 cm
i.e., R = 5 cm and r = 4 cm
∴ Volume of spherical ball
= \(\frac { 4 }{ 3 }\)π[R³ – r³] = \(\frac { 4 }{ 3 }\) × \(\frac{22}{7}\) [(5)³ – (4)³]
= \(\frac { 88 }{ 21 }\) [125 – 64]cm³ = \(\frac { 88 }{ 21 }\) × 61 cm³
∴ weight of the ball = \(\frac{88 \times 61}{21}\) × \(\frac{2625}{11}\)
= 61000 gm = \(\frac{61000}{1000}\) kg = 61 kg

Question 11.
(i) Prove that the surface area of a sphere of diameter d is πd² and the volume is \(\frac { 1 }{ 6 }\)πd³.
(ii) The volumes and diameters of a cone and sphere are equal. Prove that the height of the cone is twice the diameter of the sphere.
Solution:
(i) Diameter of a sphere = d
∴ Radius (r) = \(\frac { d }{ 2 }\)
∴ Surface area = 4πr² = 4π × \(\left(\frac{d}{2}\right)^2\)
= 4π × \(\frac{d^2}{4}\) = πd²
and volume = \(\frac { 4 }{ 2 }\)πr³ = \(\frac { 4 }{ 3 }\)π \(\left(\frac{d}{2}\right)^3\)
= \(\frac { 4 }{ 3 }\)π × \(\frac{d^3}{8}\) = \(\frac { 1 }{ 6 }\)πd³

(ii) Let diameter of cone and sphere be 2r
∴ Radius of each one = r
Now volume of cone = \(\frac { 1 }{ 3 }\)πr²h
and volume of sphere = \(\frac { 4 }{ 3 }\)πr³
∵ Their volume are equal
∴ \(\frac { 1 }{ 3 }\)πr²h = \(\frac { 4 }{ 3 }\)πr³
⇒ h = 4 r = 2(2r) = 2 × diameter
Hence height of cone = twice of the diameter of sphere
Hence proved.

Question 12.
The external diameter of a hollow metal sphere is 12 cm, and its thickness is 2 cm. Find the radius of a solid sphere made of the same material and having the same weight as the hollow sphere.
Solution:
External diameter of a hollow sphere = 12 cm
∴ Radius (r) = \(\frac { 12 }{ 2 }\) = 6 cm
Thickness of the metal used = 2 cm
∴ Internal radius = 6 – 2 = 4 cm
∴ Volume of the metal = \(\frac { 4 }{ 3 }\)π [R³ – r³]
= \(\frac { 4 }{ 3 }\)π [(6)³ – (4)³] cm³
= \(\frac { 4 }{ 3 }\)π (216 – 64) cm³
= \(\frac { 4 }{ 3 }\)π (512) cm³
= \(\frac { 608 }{ 3 }\) π cm³

Now the volume of solid sphere = \(\frac { 608 }{ 3 }\) π cm³
Let radius of the sphere = x
Then \(\frac { 4 }{ 3 }\) π x³ = \(\frac { 608 }{ 3 }\) π
x³ = \(\frac{608 \pi \times 3}{3 \times 4 \pi}\) = 152
∴ x = \(\sqrt[3]{152}\) = \((152)^{\frac{1}{3}}\) = 5.3368 = 5.34
∴ Radius of solid sphere = 5.34 cm

Question 13.
A hollow copper ball has an external diameter of 12 cm, and a thickness of 0.1 cm. Find
(i) the outer surface area of the ball;
(ii) the weight of the ball if 1 cm³ of copper weighs 8.88 g. (Take π to be 3.14).
Solution:
External diameter of a hollow spherical ball = 12 cm

∴ Outer radius (R) = \(\frac { 12 }{ 6 }\) = 6 cm
Thickness of the metal used = 0.1 cm
∴ Internal radius (r) = 6 – 0.1 = 5.9 cm

(i) Outer surface area = 4πR²
= 4 × 3.14 × (6)² cm²
= 4 × 3.14 × 36 = 452.16 cm²

(ii) Volume of metal used = \(\frac { 4 }{ 3 }\)π[R³ – r³]
= \(\frac { 4 }{ 3 }\) × (3.14) × [(6)³ – (5.9)³] cm³
= \(\frac { 4 }{ 3 }\) × (3.14) × [216 – 205.379] cm³
= \(\frac { 4 }{ 3 }\) × (3.14) × (10.621) cm²
= 44.467 cm³
weight of 1 cm³ = 8.88 cm
∴ Total weight of the ball = 44.467 × 8.88 gm = 394.87 gm

Question 14.
Marbles of diameter 1.4 cm, are dropped into a beaker containing some water and are fully submerged. The diameter of the beaker is 7 cm. Find how many marbles have been dropped in it if the water rises by 5.6 cm. \(\left(\pi=\frac{22}{7}\right)\)
Solution:
Diameter of each marble = 1.4 cm
∴ Radius of each marble (r) = \(\frac { 1.4 }{ 2 }\) = 0.7 cm
Volume of one marble = \(\frac { 4 }{ 3 }\)πr³
= \(\frac { 4 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × (0.7)³ = \(\frac { 88 }{ 21 }\) × 0.343 cm³
= 1.437333 cm³
Diameter of beaker = 7 cm
∴ Radius of the base of beaker (r₁) = \(\frac { 7 }{ 2 }\) cm
Height of water level = 5.6 cm
∴ Volume of water = πr²h
= \(\frac { 22 }{ 7 }\) × \(\frac { 7 }{ 2 }\) × \(\frac { 7 }{ 2 }\) × 5.6 cm³ = 215.6 cm³

Question 15.
A cylindrical tub of radius 5 cm and length 9.8 cm is full of water. A solid in the form of a right circular cone mounted on a hemisphere is immersed into the tub. If the radius of the hemisphere is 3.5 cm and height of the cone outside the hemisphere is 5 cm, find the volume of water left in the tub. (Take π = 22/7)
Solution:
Radius of tub (r) = 5 cm and height (length) (h) = 9.8 cm

∴ Volume of water filled in it = πr²h
= \(\frac { 22 }{ 7 }\) × 5 × 5 × 9.8 cm³ = 770 cm³
In the solid,
Radius of the hemisphere (r₁) = 3.5 cm and height of cone (h₁) = 5 cm
∴ Volume of solid = \(\frac { 2 }{ 3 }\)πr₁³ + \(\frac { 1 }{ 3 }\)πr₁²h₁
= \(\frac { 2 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × (3.5)³ + \(\frac { 1 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × (3.5)² × 5
= \(\frac { 2 }{ 3 }\) × \(\frac { 22 }{ 7 }\) (3.5)² [2 (3.5) + 5]
= \(\frac { 22 }{ 21 }\) × 12.25 × [7 + 5] cm³
= \(\frac { 22 }{ 21 }\) × 12.35 × 12 cm³ = 154 cm³
∴ Water left in the tub = (770 – 154) cm3 = 616 cm3

Question 16.
A toy is in the form of a cone mounted on a hemisphere of radius 3.5 cm. The total height of the toy is 15.5 cm. Find the total surface area.
Solution:
Total height of the toy = 15.5 cm
Radius of hemispherical part = 3.5 cm
∴ Height of conical part (h)=15.5 – 3.5 = 12 cm

∴ Slant height (l) = \(\sqrt{r^2+h^2}\) = \(\sqrt{(3.5)^2+(12)^2}\) = \(\sqrt{12.25+144}\) = \(\sqrt{156.25}\) = 12.5 cm
Now total surface area of the toy
= πrl + 2πr² = πr(l + 2r)
= \(\frac { 22 }{ 7 }\) × 3.5 (12.5 + 2 × 3.5) cm²
= 11 (12.5 + 7) = 11 × 19.5 cm²
= 214.5 cm²

Question 17.
The largest sphere is carved out of a cube of side 7 cm. Find the volume of the sphere.
Solution:
Side of cube = 7 cm
∴ Diameter of the largest sphere which is carved out of it = 7 cm

∴ Radius of the sphere = \(\frac { 7 }{ 2 }\) cm
Now volume of sphere = \(\frac { 4 }{ 3 }\)πr³
= \(\frac { 4 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × \(\left(\frac{7}{2}\right)^3\) cm³
= \(\frac { 88 }{ 21 }\) × \(\frac { 343 }{ 8 }\) = \(\frac { 549 }{ 3 }\) = 179.666
= 179.7 cm³

Question 18.
A cylindrical bucket, whose base has a radius of 15 cm, is filled with water up to a height of 20 cm. A heavy iron spherical ball of a radius 10 cm is dropped to submerge completely in water in the bucket. Find the increase in the level of water.
Solution:
Radius of the cylindrical bucket = 15 cm and height of water level = 20 cm

∴ Volume of water in the bucket
= πr²h = \(\frac { 22 }{ 7 }\) × 15 × 15 × 20 cm³
= \(\frac { 99000 }{ 7 }\) cm³
Radius of spherical ball = 10 cm
∴ Volume of ball = \(\frac { 4 }{ 3 }\)πr³
= \(\frac { 4 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × 10 × 10 × 10 cm³ = \(\frac { 88000 }{ 21 }\) cm³
Let the level of water arise upto h cm in the bucket
∴ Volume = πr²h
\(\frac { 88000 }{ 21 }\) = \(\frac { 22 }{ 7 }\) × 15 × 15h
⇒ \(\frac { 88000 }{ 21 }\) = \(\frac { 22 }{ 7 }\) × 225h
⇒ h = \(\frac { 88000 }{ 21 }\) × \(\frac{7}{22 \times 225}\)
= \(\frac { 160 }{ 27 }\) = 5\(\frac { 25 }{ 27 }\) cm
∴ Increase in water level = 5\(\frac { 25 }{ 27 }\) cm

Question 19.
The radius of the base of a cone and the radius of a sphere are the same, each being 8 cm. Given that the volumes of these two solids are also the same, calculate the slant height of the cone, correct to one place of decimal.
Solution:
Radius of the base of a cone (r) = 8 cm
and radius of the sphere (r) = 8 cm
Volume of cone = volume of sphere Let h be in height of the cone,
∴ \(\frac { 4 }{ 3 }\)πr³ = \(\frac { 1 }{ 3 }\)πr²h
h = \(\frac{4 \pi r^3 \times 3}{3 \times \pi r^2}\) = 4r = 4 × 8 = 32 cm
∴ Slant height (l) of cone = \(\sqrt{r^2+h^2}\) = \(\sqrt{(8)^2+(32)^2}\) = \(\sqrt{64+1024}\) cm = \(\sqrt{1088}\) cm = 32.98 cm = 33 cm

Question 20.
A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of its top which is open is 5 cm. It is filled with water up to the rim. When lead shots, each of which is a sphere of radius 0.5 cm, are dropped into the vessel, one- fourth of the water flows out. Find the number of lead shots dropped in the vessel. \(\left(\text { Take } \pi \text { to be } \frac{22}{7}\right)\)
Solution:
Radius of the conical vessal (r) = 5 cm
and height (h) = 8 cm
∴ Volume of water filled in it = \(\frac { 1 }{ 3 }\)πr²h
= \(\frac { 1 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × 5 × 5 × 8 cm³
= \(\frac { 4400 }{ 21 }\) cm³

∴ volume of lead shots = \(\frac { 1 }{ 4 }\) of \(\frac { 4400 }{ 21 }\) = \(\frac { 1100 }{ 21 }\) cm³
Radius of each lead shot (r₁) = 0.5 cm
∴Volume of one lead shot = \(\frac { 4 }{ 3 }\)πr₁³
= \(\frac { 4 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × 0.5 × 0.5 × 0.5
= \(\frac { 4 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × \(\frac { 1 }{ 2 }\) × \(\frac { 1 }{ 2 }\) × \(\frac { 1 }{ 2 }\) cm³ = \(\frac { 11 }{ 21 }\)cm³
∴ Number of lead shots = \(\frac { 1100 }{ 21 }\) ÷ \(\frac { 11 }{ 21 }\) = \(\frac { 1100 }{ 21 }\) × \(\frac { 21 }{ 11 }\) = 100

Question 21.
A buoy is made in the form of a hemisphere surmounted by a right cone whose circular base conicides with the plane surface of the hemisphere. The radius of the base of the cone is 3.5 metres and its volume is \(\frac { 2 }{ 3 }\) of the hemisphere. Calculate the height of the cone and the surface of the buoy correct to 2 places of decimals. \(\left(\text { Take } \pi=3 \frac{1}{7}\right)\)
Solution:
Radius of cone = 3.5 m
Volume of cone = \(\frac { 2 }{ 3 }\) of the volume of hemisphere

Let h be the height of the cone = h
volume of hemisphere = \(\frac { 2 }{ 3 }\)πr³
= \(\frac { 2 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × (3.5)³ m³
= \(\frac { 2 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × \(\frac { 7 }{ 2 }\) × \(\frac { 7 }{ 2 }\) × \(\frac { 7 }{ 2 }\) m³ = \(\frac { 539 }{ 6 }\) m³
∴ volume of cone = \(\frac { 2 }{ 3 }\) of volume of hemisphere
= \(\frac { 2 }{ 3 }\) × \(\frac { 539 }{ 6 }\) = \(\frac { 539 }{ 9 }\) m³
But volume of cone = \(\frac { 1 }{ 3 }\) πr²h
∴ \(\frac { 1 }{ 3 }\) πr²h = \(\frac { 539 }{ 9 }\)
⇒ \(\frac { 1 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × \(\frac { 7 }{ 2 }\) × \(\frac { 7 }{ 2 }\)h = \(\frac { 539 }{ 9 }\)
⇒ \(\frac { 77 }{ 6 }\)h = \(\frac { 539 }{ 9 }\)
⇒ h = \(\frac { 539 }{ 9 }\) × \(\frac { 6 }{ 77 }\)
⇒ h = \(\frac{7 \times 2}{3}\) = \(\frac { 14 }{ 3 }\) m = 4\(\frac { 2 }{ 3 }\) m
∴ Height of cone = 4\(\frac { 2 }{ 3 }\) m

(ii) slant height (l) = \(\sqrt{r^2+h^2}\)
= \(\sqrt{\left(\frac{7}{2}\right)^2+\left(\frac{14}{3}\right)^2}\) = \(\sqrt{\frac{49}{4}+\frac{196}{9}}\)
= \(\sqrt{\frac{441+784}{36}}\) = \(\sqrt{\frac{1225}{36}}\) = \(\frac{35}{6}\) m
∴ Surface area of the buoy = πrl + 2πr²
= \(\frac { 22 }{ 7 }\) × \(\frac { 7 }{ 2 }\) × \(\frac { 35 }{ 6 }\) + 2 × \(\frac { 22 }{ 7 }\) × \(\frac { 7 }{ 2 }\) × \(\frac { 7 }{ 2 }\) = \(\frac { 385 }{ 6 }\) + 77 = \(\frac{385+462}{6}\)
= \(\frac { 847 }{ 6 }\) = 141\(\frac { 1 }{ 6 }\) m²

Question 22.
A cylinder, whose height is equal to its diameter, has the same volume as a sphere of radius 4 cm. Calculate the radius of the base of the cylinder, correct to 1 decimal place.
Solution:
Height of a cylinder = its diameter
⇒ h = 2r
and volume of cylinder = volume of sphere Radius of sphere = 4 cm
∴ Volume of sphere = \(\frac { 4 }{ 3 }\)πr³
= \(\frac { 4 }{ 3 }\)π × 4 × 4 × 4 cm³ = \(\frac { 256 }{ 3 }\)π cm³
∴ Volume of cylinder = \(\frac{256 \pi}{3}\) cm³
But volume of cylinder = πr²h
= πr² × 2r = 2πr³
∴ 2πr³ = \(\frac { 256 }{ 3 }\)π
r³ = \(\frac{256 \pi}{3 \times 2 \pi}\) = \(\frac { 128 }{ 3 }\)
∴ r = \(\sqrt[3]{\frac{128}{3}}\) = \(\sqrt[3]{42.667}\) = 3.49 cm
∴ Radius = 3.49 cm = 3.5 cm