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

The availability of Class 10 ICSE Maths Solutions S Chand Chapter 15 Three Dimensional Solids Ex 15(e) encourages students to tackle difficult exercises.

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

Question 1.
The diameter of a metallic sphere is 6 cm. The sphere is melted and drawn into a wire of uniform circular cross section. If the length of the wire is 36 cm, find its radius.
Solution:
Diameter of metallic sphere = 6 cm
∴ Radius (R) = \(\frac { 6 }{ 2 }\) = 3 cm
and volume = \(\frac { 4 }{ 3 }\) πr³
= \(\frac { 4 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × 3 × 3 × 3 cm³ = \(\frac { 792 }{ 7 }\) cm³
Length of wire (h) = 36 cm
Let r be the radius of circular wire, then πr²h = \(\frac { 792 }{ 7 }\)
⇒ \(\frac { 22 }{ 7 }\) r² × 36 = \(\frac { 792 }{ 7 }\) ⇒ r² = \(\frac{792 \times 7}{7 \times 22 \times 36}\)
⇒ r² = 1 = (1)² ⇒ r = 1 cm
∴ Radius = 1 cm

Question 2.
How many lead balls, each of radius 1 cm, can be made from a sphere whose radius is 8 cm ?
Solution:
Radius of sphere (R) = 8 cm
∴ Volume = \(\frac { 4 }{ 3 }\) πR³ = \(\frac { 4 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × 8 × 8 × 8 cm³
= \(\frac { 45056 }{ 21 }\) cm³
Radius of one small lead ball = 1 cm
∴ Volume of one lead ball = \(\frac { 4 }{ 3 }\) πr³
= \(\frac { 4 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × 1 × 1 × 1 = \(\frac { 88 }{ 21 }\) cm³
∴ Number of ball = \(\frac { 45056 }{ 21 }\) ÷ \(\frac { 88 }{ 21 }\)
= \(\frac { 45056 }{ 21 }\) × \(\frac { 88 }{ 21 }\) = 512

Question 3.
A solid metallic sphere of diameter 21 cm is melted and recast into a number of smaller cones, each of diameter 7 cm and height 3 cm. Find the number of cones so formed.
Solution:
Diameter of sphere = 21 cm
∴ Radius (R) = \(\frac { 21 }{ 2 }\) cm
and volume = \(\frac { 4 }{ 3 }\)πR³
= \(\frac { 4 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × \(\frac { 21 }{ 2 }\) × \(\frac { 21 }{ 2 }\) × \(\frac { 21 }{ 2 }\) cm ³ = 4851 cm³
Diameter of a smaller cone = 7 cm
∴ Radius (r) = \(\frac { 7 }{ 2 }\) cm
and height (h) = 3 cm
∴ Volume of one cone = \(\frac { 1 }{ 3 }\)πr²h
= \(\frac { 1 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × \(\frac { 7 }{ 2 }\) × \(\frac { 7 }{ 2 }\) × 3 cm³ = \(\frac { 77 }{ 2 }\) cm³
∴ Number of cones formed = 4851 ÷ \(\frac { 77 }{ 2 }\)
= \(\frac{4851 \times 2}{77}\) = 63 × 2 = 126 cones

Question 4.
A metallic disc, in the shape of a right circular cylinder, is of height 2.5 mm, and base radius 12 cm. Metallic disc is melted and made into a sphere. Calculate the radius of the sphere.
Solution:
Radius of disc (R) = 12 cm
and height (H) = 2.5 mm = 0.25 cm
∴ Volume = πr²h = \(\frac { 22 }{ 7 }\) × 12 × 12 × 0.25 cm³
= \(\frac { 22 }{ 7 }\) × 144 × \(\frac { 1 }{ 4 }\) = \(\frac { 792 }{ 7 }\) cm³
Let radius of sphere r then
\(\frac { 4 }{ 3 }\)πr³ = \(\frac { 792 }{ 7 }\) ⇒ \(\frac { 4 }{ 3 }\) × \(\frac { 22 }{ 7 }\)r³ = \(\frac { 792 }{ 7 }\)
⇒ r³ = \(\frac { 792 }{ 7 }\) ⇒ \(\frac{3 \times 7}{4 \times 22}\) = 9 × 3 = 27 = (3)³
∴ r = 3
∴ Radius of the sphere = 3 cm

Question 5.
A hollow sphere of internal and external diameters 6 cm and 10 cm respectively is melted and recast into a cone of base diameter 14 cm. Find the height of the cone.
Solution:
External radius of hollow sphere (R)
= \(\frac { 10 }{ 2 }\) cm = 5 cm
and internal radius (r) = \(\frac { 6 }{ 2 }\) = 3 cm
∴ Volume of the metal used = \(\frac { 4 }{ 3 }\) (R³ – r³)
= \(\frac { 4 }{ 3 }\) × \(\frac { 22 }{ 7 }\) (5³ – 3³) cm³
= \(\frac { 88 }{ 21 }\)(125 – 27) = \(\frac { 88 }{ 21 }\) × 98 cm³
= \(\frac{88 \times 14}{3}\) = \(\frac{1232}{3}\) cm³
∴ Volume of cone = \(\frac{1232}{3}\) cm³
and diameter = 14 cm
∴ Radius (r) = \(\frac { 14 }{ 2 }\) = 7 cm
Let h be the height of the cone
∴ \(\frac { 1 }{ 3 }\)πr₁²h = \(\frac { 1232 }{ 3 }\)
⇒ \(\frac { 1 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × 7 × 7 × h = \(\frac { 1232 }{ 3 }\)
⇒ \(\frac { 154 }{ 3 }\) h = \(\frac { 1232 }{ 3 }\) ⇒ h = \(\frac{1232 \times 3}{3 \times 154}\)
∴ h = 8
∴ Height of the cone = 8 cm

Question 6.
A hollow metallic cylindrical tube has an internal radius of 3 cm and height 21 cm. The thickness of the metal tube is \(\frac { 1 }{ 2 }\) cm. The tube is melted and cast into a right circular cone of height 7 cm. Find the radius of cone correct to 1 decimal place. \(\left(\text { Take } \pi=\frac{22}{7}\right)\)
Solution:
Internal radius of hollow metallic cylindrical tube (r) = 3 cm
Height (h) = 21 cm
Thickness of metal = \(\frac { 1 }{ 2 }\) cm
∴ Outer radius (R) = 3 + \(\frac { 1 }{ 2 }\) = \(\frac { 7 }{ 2 }\) cm
and outer height (H) = 21
∴ Volume = πR²H – πr²h
= π [R²H – r²h] = \(\frac { 22 }{ 7 }\)

Question 7.
Three solid glass balls of radius 1, 6 and 8 cm, respectively are melted into a solid sphere. Find its radius.
Solution:
Radius of first ball (r₁) = 1 cm
Radius of second ball (r₂) = 6 cm
and radius of third ball (r₃) = 8 cm
∴ Volume of three balls
= \(\frac { 4 }{ 3 }\)πr₁³ + \(\frac { 4 }{ 3 }\)πr₂³ + \(\frac { 4 }{ 3 }\)πr₃³ = \(\frac { 4 }{ 3 }\)π (r₁³ + r₂³ + ₃³)
= \(\frac { 4 }{ 3 }\)π [(1)³ + (6)³ + (8)³] cm³
= \(\frac { 4 }{ 3 }\)π [1 + 216 + 512] = \(\frac { 4 }{ 3 }\)π × 729 cm³
Let the radius of the sphere = R
∴ \(\frac { 4 }{ 3 }\)πR³ = \(\frac { 4 }{ 3 }\)π × 729 R³
= \(\frac { 4 }{ 3 }\)π × 729 × \(\frac{3}{4 \times \pi}\)
⇒ R³ = 729 = (9)³ ⇒ R = 9
∴ Radius of the sphere = 9 cm

Question 8.
A hemispherical bowl of internal radius 9 cm contains a liquid. This liquid is to be filled into cylindrical shaped small bottles of diameter 3 cm and height 4 cm. How many bottles are necessary to empty the bowl ?
Solution:
Internal radius of hemispherical bowl (R) = 9 cm
Volume of liquid filled in it = \(\frac { 2 }{ 3 }\)πR³
= \(\frac { 2 }{ 3 }\)π × (9)³ cm³ = \(\frac { 2 }{ 3 }\)π 729 = 2 × 243π cm³
= 486π cm³
Radius of cylindrical bottle (l) = \(\frac { 3 }{ 2 }\) cm
and height (h) = 4 cm
∴ Volume of one bottle = πr²h
= π × \(\frac { 3 }{ 2 }\) × \(\frac { 3 }{ 2 }\) × 4 = 9π cm³
∴ Number of bottles filled with liquid
= \(\frac{486 \pi}{9 \pi}\) = 54 bottles

Question 9.
How many spherical lead shots, each 4.2 cm in diameter, can be obtained from a rectangular solid of lead of dimensions 66 cm × 42 cm × 21 cm. (Use π = 22/7)
Solution:
Dimensions of a rectangular solid of lead = 66 cm × 42 cm × 21 cm
∴ Volume = 66 × 42 × 21 cm³
Diameter of one spherical lead shot (d) = 4.2 cm
∴ Radius (r) = \(\frac { 4.2 }{ 2 }\) =2.1 cm
∴ Volume of one shot = \(\frac { 4 }{ 3 }\)πr³
= \(\frac { 4 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × (2.1)³ cm³
= \(\frac{4 \times 22}{21}\) × 2.1 × 2.1 × 2.1 cm³
= \(\frac { 88 }{ 21 }\) × \(\frac { 21 }{ 10 }\) × \(\frac { 21 }{ 10 }\) × \(\frac { 21 }{ 10 }\) cm³ = \(\frac { 38808 }{ 1000 }\) cm³
∴ Number of shots = \(\frac{\text { Volume of rect soild }}{\text { Volume of one shot }}\)
= \(\frac{66 \times 42 \times 21 \times 1000}{38808}\)
= \(\frac{3 \times 1000}{2}\) = 3 × 500 = 1500

Question 10.
The radius of a solid iron sphere is 8 cm. Eight rings of iron plate of external radius 6\(\frac { 2 }{ 3 }\) cm, thickness 3 cm are made by melting this sphere. Find the internal diameter of each ring.
Solution:

Radius of solid sphere (R) = 8 cm
∴ Volume = \(\frac { 4 }{ 3 }\)πR³ = \(\frac { 4 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × 8 × 8 × 8 cm³
= \(\frac { 45056 }{ 21 }\) cm³
outer radius of ring(r₁) = 6\(\frac { 2 }{ 3 }\) = \(\frac { 20 }{ 3 }\) cm

Let inner radius of the ring = r₂
Thickness of the ring (h) = 3 cm
∴ Volume of ring = π[R² – r²] × h

∴ Internal radius = 4 cm
and internal diameter = 2 × 4 = 8 cm

Self Evaluation And Revision (LATEST ICSE QUESTIONS)

Question 1.
The figure alongside shows the cross-section of an ice-cream cone consisting of a cone surmounted by a hemisphere. The radius of the hemisphere is 3.5 cm and the height of the cone is 10.5 cm. The outer shell ABCDEF is shaded and not filled with ice-cream AE = DC = 0.5 cm, AB || EF and BC || FD. Calculate:
(i) The volume of ice-cream in the cone (the unshaded portion including the hemisphere) in cm³.
(ii) The volume of the outer shell (the shaded portion ) in cm³. Give your answer correct to the nearest cm³.

Solution:
Diameter of hemisphere = 7 cm
Radius of hemisphere AC (R) = \(\frac { 7 }{ 2 }\) = 3.5 cm
Height of conical part NB = 10.5 cm
AE = DC = 0.5 cm, AB || EF and CB || DF
Let height NF = h
Shaded portion is not filled with ice-cream Radius of inner cone = 3.5 – 0.5 = 3.0 cm
In right △ADF and △ACB
∠A = ∠A (each 90°)
∠ADF = ∠ACB (corresponding angles)
∴ △ADF ~ △ACB (AA axiom)
∴ \(\frac { AD }{ AC }\) = \(\frac { EF }{ CB }\) ⇒ \(\frac { 3 }{ 3.5 }\) = \(\frac { h }{ 10.5 }\)
⇒ h = \(\frac{3 \times 10.5}{3.5}\) = 9 cm

(i) Volume of ice-cream = volume of inner cone + volume of hemisphere
= \(\frac { 1 }{ 3 }\)πr²h + \(\frac { 2 }{ 3 }\)πR³
=\(\frac { 1 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × 3 × 3 × 9 + \(\frac { 2 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × \(\frac { 7 }{ 2 }\) × \(\frac { 7 }{ 2 }\) × \(\frac { 7 }{ 2 }\) cm³ (nearest to cm³)

(ii) volume of outer shell (shaded portion) = volume of outer cone – volume of inner cone
= \(\frac { 1 }{ 3 }\)πR²H – \(\frac { 1 }{ 3 }\)πr²h
= \(\frac { 1 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × (3.5) (3.5) (3.5) – \(\frac { 1 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × 3 × 3 × 9 = 134.75 – \(\frac { 594 }{ 7 }\) = 134.75 – 84.857
= 49.893 cm³ = 50 cm³ (nearest to cm³)

Question 2.
The surface area of a solid metallic sphere is 1256 cm³. It is melted and recast into solid right circular cones of radius 2.5 cm and height 8 cm. Calculate :
(i) the radius of the solid sphere,
(ii) the number of cones recast. (π = 3.14)
Solution:
Surface area of a metallic sphere = 1256 cm²
(i) Let R be the radius
∴ 4πR² = 1256 ⇒ 4 × 3.14 R² = 1256
⇒ R² = \(\frac{1256}{4 \times 3.14}\) = \(\frac{1256 \times 100}{4 \times 314}\) = 100 = (10)²
∴ R = 10 cm

(ii) Volume of the sphere = \(\frac { 4 }{ 3 }\) πr³
= \(\frac { 4 }{ 3 }\) (3.14) × (10)³ cm³
= \(\frac{4 \times 314 \times 1000}{3 \times 100}\) cm³ = \(\frac { 12560 }{ 3 }\) cm³
Radius of cone (r) = 2.5 cm
and height (h) = 8 cm
∴ volume of cone = \(\frac { 1 }{ 3 }\)πr²h
= \(\frac { 1 }{ 3 }\) (3.14) × (2.5)² × 8 cm³
= \(\frac{3.14 \times 6.25 \times 8}{3}\) = \(\frac{157}{3}\) cm³
Number of cones formed = \(\frac { 12560 }{ 3 }\) ÷ \(\frac { 157 }{ 3 }\)
= \(\frac { 12560 }{ 3 }\) × \(\frac { 3 }{ 157 }\) = 80

Question 3.
An exhibition tent is in the form of a cylinder surmounted by a cone. The height of the tent above the ground is 85 m and the height of the cylindrical part is 50 m. If the diameter of the base is 168 m, find the quantity of canvas required to make the tent. Allow 20% extra for folds and stitching. Give your answer to the nearest m².
Solution:
Diameter of the base = 168 m
∴ Radius (r) = = 84 m
Total height of the tent = 85 m
Height of cylinderical portion (h₁) = 50 m

∴ Height of conical portion = 85 – 50 = 35 m
Total curved surface = curved surface area of cylinder + curved surface area of cone = 2πrh₁ + πrl
= 2πrh₁ + πr\(\left(\sqrt{r^2+h^2}\right)\) \((∴l \sqrt{r^2+h^2})\)
= 2 × \(\frac { 22 }{ 7 }\) × 84 × 50 + \(\frac { 22 }{ 7 }\) × 84 × \(\sqrt{\left(84^2+35^2\right)}\)
= 26400 + 264 × \((\sqrt{7056+1225})\)
= 26400 + 264 × \(\sqrt{8281}\)
= 26400 + 264 × 91 = 26400 + 24024 = 50424 m²
Extra area of 20% for stitching etc.
= 50424 m²
Extra area of 20% for stitching etc.
= 50424 + 10084.8 m² = 60508.8 m²
= 60509 m² (nearest to m²)

Question 4.
A hollow sphere of internal and external diameters 4 cm and 8 cm respectively is melted into a cone of base diameter 8 cm. Find the height of the cone.
Solution:
Internal diameter of hollow sphere = 4 cm
and external diameter = 8 cm

∴ Internal radius (r) = \(\frac { 4 }{ 2 }\) = 2 cm
and external radius (R) = \(\frac { 8 }{ 2 }\) = 4 cm
∴ Volume of the hollow sphere
= \(\frac { 4 }{ 3 }\) π (R³ – r³) = \(\frac { 4 }{ 3 }\) × \(\frac { 22 }{ 7 }\) [4³ – 2³]
= \(\frac { 88 }{ 21 }\) × (64 – 8) = \(\frac { 88 }{ 21 }\) × 56 cm³ = \(\frac { 704 }{ 3 }\) cm³
Base diameter of cone = 8 cm
∴ Radius (r₁) = \(\frac { 8 }{ 2 }\) = 4 cm
Let h be the height of the cone
Then volume of cone = \(\frac { 1 }{ 3 }\) πr²h

∴ \(\frac { 1 }{ 3 }\) πr²h = \(\frac { 704 }{ 3 }\)
\(\frac { 1 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × 4 × 4 × h = \(\frac { 704 }{ 3 }\)
\(\frac { 352 }{ 21 }\) h = \(\frac { 704 }{ 3 }\) ⇒ h = \(\frac{704 \times 21}{3 \times 352}\) = 14
∴ Height of the cone = 14 cm

Question 5.
A vessel is in the form of an inverted cone. Its height is 11 cm and the radius of its top which is open is 2.5 cm. It is filled with water up to the rim. When lead shots, each of which is a sphere of radius 0.25 cm, are dropped into the vessel, \(\frac { 2 }{ 5 }\) of the water flows out. Find the number of lead shots dropped in the vessel.
Solution:
Radius of the cone = 2.5 cm
Height (h) = 11 cm

∴ Volume of the cone = \(\frac { 1 }{ 3 }\)πr²h
= \(\frac { 1 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × 2.5 × 2.5 × 11 cm³
= \(\frac { 1512.5 }{ 21 }\) cm³
Radius of spherical lead = 0.25 cm
∴ Volume of one lead = \(\frac { 4 }{ 3 }\) πr³
= \(\frac { 4 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × (0.25)³ cm³
= \(\frac { 88 }{ 21 }\) × 0.015625 cm³ = \(\frac { 1.375 }{ 21 }\) cm³
volume of water flows out of the vessel after
putting the shots = \(\frac { 2 }{ 5 }\) 0f \(\frac { 1512.5 }{ 21 }\) = \(\frac { 605.0 }{ 21 }\) cm³
∴ Number of shots = \(\frac { 605.0 }{ 21 }\) ÷ \(\frac { 1.375 }{ 21 }\)
= \(\frac { 605.0 }{ 21 }\) × \(\frac { 1.375 }{ 21 }\) = 440

Question 6.
A girl fills a cylindrical bucket 32 cm in height and 18 cm in radius with sand. She empties the bucket on the ground and makes a conical heap of sand. If the height of the conical heap is 24 cm, find:
(i) its radius and
(ii) its slant height (leave your answer in square root form)
Solution:
Radius of the cylindrical bucket (r) = 18 cm
Height (h) = 32 cm
∴ Volume of sand filled in it = πr²h
= π (18)² × 32 = π × 18 × 18 × 32 cm³
= 10368 π cm³

(i) Volume of sand in a conical shape heap = 10368π cm³
Height (h₁)=24 cm
∴ Let r₁, be the radius, then
\(\frac { 1 }{ 3 }\) πr₁²h = 10368 π
⇒ \(\frac { 1 }{ 3 }\) πr₁² × 24 = 10368π ⇒ 8πr₁² = 10638π
⇒ r₁² = \(\frac { 10368π }{ 8π }\) =1296 = (36)²
∴ r₁ = 36 cm

(ii) and slant height = \(\sqrt{r_1^2+h_1^2}\)
= \(\sqrt{(36)^2+(24)^2}\) cm
= \(\sqrt{1296+576}\) cm = \(\sqrt{1872}\) cm

Question 7.
A metallic sphere of radius 10.5 cm is melted and then recast into small cones, each of radius 3.5 cm and height 3 cm. Find the number of cones thus obtained.
Solution:
Radius of a sphere (R) = 10.5 cm
∴ Volume = \(\frac { 4 }{ 3 }\)πR³ = \(\frac { 4 }{ 3 }\)π (10.5)³ cm³
= \(\frac { 4 }{ 3 }\)π × 1157.625 cm³
Radius of each cone (r) = 3.5 cm
and height (h) = 3 cm
∴ Volume of one cone = \(\frac { 1 }{ 3 }\)πr³h
= \(\frac { 1 }{ 3 }\)π (3.5)² × 3 cm³
= π × 12.25 = 12.25 = 12.25π cm³
∴ Number of cone will be \(\frac{4 \pi \times 1157.625}{3 \times 12.25 \pi}\)
= 126 cones

Question 8.
A vessel in the form of an inverted cone is filled with water to the brim. Its height is 20 cm and diameter is 16.8 cm. Two equal solid cones are dropped in it so that they are fully submerged. As a result, one-third of the water in the original cone overflows. What is the volume of each of the solid copes submerged ?
Solution:
Diameter of a conical vessel = 16.8 cm
∴ Radius (R) = \(\frac { 16.8 }{ 2 }\) = 8.4 cm
and height (H) = 20 cm
∴ volume of water filled

in it = \(\frac { 1 }{ 3 }\)πR²H
= \(\frac { 1 }{ 3 }\)π × (8.4)² × 20 cm³
= \(\frac { 1 }{ 3 }\)π × 8.4 × 8.4 × 20 cm³ = 470.4π cm³
By dropping two soild cones in it, volume of water overflows = \(\frac { 1 }{ 3 }\) of 470.4π = 156.8π cm³
∴ volume of two small solid cones = 156.8π cm³
and volume of one soild cone = \(\frac{156.8 \pi}{2}\) cm³
= 78.4π cm³ = 78.4 × \(\frac { 22 }{ 7 }\) = 246.4 cm³

Question 9.
The surface area of a solid metallic sphere is 616 cm². It is melted and recast into smaller spheres of diameter 3.5 cm. How many such spheres can be obtained ?
Solution:
Surface area of a metallic solid sphere = 616 cm²
Let radius of the sphere = R
Then 4πR² = 616
⇒ 4 × \(\frac { 22 }{ 7 }\)R² = 616
⇒ R² = \(\frac{616 \times 7}{4 \times 22}\) = 7 × 7 = 49 cm
∴R² = 49 = (7)²
⇒ R = 7 cm
∴ volume = \(\frac { 4 }{ 3 }\)πR³
= \(\frac { 4 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × 7 × 7 × 7 cm³ = \(\frac { 4312 }{ 3 }\)cm³
Diameter of each smaller spheres = 3.5 cm
Radius (r) = \(\frac { 3.5 }{ 2 }\) = 1.75 cm
volume of one smaller sphere = \(\frac { 4 }{ 3 }\)πr³
= \(\frac { 4 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × (1.75)³ cm³
= \(\frac { 88 }{ 21 }\) × 5.359375 cm³
∴ Number of small spheres will be
= \(\frac { 4312 }{ 3 }\) ÷ \(\frac{88 \times 5.359375}{21}\)
= \(\frac { 4312 }{ 3 }\) × \(\frac{21}{88 \times 5.359375}\)
= \(\frac{49 \times 7 \times 1000000}{5359375}\) = \(\frac{1000000}{15625}\) = 64

Question 10.
The volume of a conical tent is 1232 m³ and the area of the base floor is 154 m². Calculate the :
(i) radius of the floor;
(ii) height of the tent;
(iii) length of canvas required to cover this conical tent, if its width is 2 m.
Solution:
Volume of conical tent = 1232 m³
and area of its floor = 154 m²
Let r be the radius and h be the height of the tent
(i) Then πr² = 154 ⇒ \(\frac { 22 }{ 7 }\)r² = 154
⇒ r² = \(\frac{154 \times 7}{22}\) = 7 × 7 = (7)²
∴ r = 7 m

(ii) Volume = \(\frac { 1 }{ 3 }\)πr²h
⇒ \(\frac { 1 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × 7 × 7 × h = 1232
⇒ \(\frac { 154 }{ 3 }\)h = 1232 ⇒ h = \(\frac{1232 \times 3}{154}\) = 24
(i) ∴ Radius = 7 m and (ii) height = 24 cm

(iii) Slant height (l) = \(\sqrt{r^2+h^2}\) = \(\sqrt{(7)^2+(24)^2}\) = \(\sqrt{49+576}\) = \(\sqrt{625}\) = 25
∴ l = 25 m
∴ Curved surface area = πrl
= \(\frac { 22 }{ 7 }\) × 7 × 25 = 550 m²
width of canvas = 2 m
∴ Length of canvas = \(\frac { Area }{ Width }\) = \(\frac { 550 }{ 2 }\) = 275 m

Question 11.
The given figure represents a hemisphere surmounted by a conical block of wood. The diameter of their bases is 6 cm each and the slant height of the cone is 5 cm. Calculate :
(i) the height of the cone ;
(ii) the volume of the solid.
Solution:
(i) Diameter of cone = 6 cm

Radius (r) = \(\frac { 6 }{ 2 }\) = 3 cm
and slant height (l) = 5 cm
Let h be the height of the cone height = \(\sqrt{l^2-r^2}\) = \(\sqrt{(5)^2-(3)^3}\)
= \(\sqrt{25-9}\) = \(\sqrt{16}\) = 4 cm

(ii) Volume = \(\frac { 1 }{ 3 }\)πr²h + \(\frac { 2 }{ 3 }\)πr³ = \(\frac { 1 }{ 3 }\)πr²(h + 2r)
= \(\frac { 1 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × (3)² (4 + 6)
=\(\frac { 1 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × 9 × 10
= \(\frac { 660 }{ 7 }\) = 94.29 cm³

Question 12.
A hemispherical bowl of diameter 7.2 cm is filled completely with chocolate sauce. This sauce is poured into an inverted cone of radius 4.8 cm. Find the height of the cone.
Solution:
Radius of hemispherical bowl = \(\frac { 7.2 }{ 2 }\) = 3.6 cm
volume of hemispherical bowl = \(\frac { 2 }{ 3 }\)πr³
= \(\frac { 2 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × 3.6 × 3.6 × 3.6 = 97.755 cm³
∴ Volume of chocolate sauce = 97.755 cm³
It is to poured in cone of radius = 4.8 cm
Volume of inverted cone = \(\frac { 1 }{ 3 }\)πr²h
97.755 cm³ = \(\frac { 1 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × 4.8 × 4.8 × h
⇒ h = \(\frac{97.755 \times 7 \times 3}{4.8 \times 4.8 \times 22}\) = \(\frac{2052.855}{506.88}\)
⇒ h = 4.05 cm

Question 13.
A solid cone of radius 5 cm and height 8 cm is melted and made into small spheres of radius 0.5 cm. Find the number of spheres formed.
Solution:
Volume of a cone = \(\frac { 1 }{ 3 }\)πr²h
= \(\frac { 1 }{ 3 }\)π × (5)² × 8
= \(\frac { 1 }{ 3 }\)π × 25 × 8 = \(\frac{200 \pi}{3}\) cm³
Volume of a sphere = \(\frac { 1 }{ 3 }\)πr³ = \(\frac { 4 }{ 3 }\) × π × (0.5)³
= \(\frac { 4 }{ 3 }\) × π × 0.125 = \(\frac { 0.5 }{ 3 }\) π = \(\frac { π }{ 6 }\) cm³
Number of spheres formed = \(\frac{200 \pi}{3}\) × \(\frac{6}{\pi}\) = 400

Question 14.
A hollow sphere of internal and external radii 6 cm and 8 cm respectively is melted and recast into small cones of base radius 2 cm and height 8 cm. Find the number of cones.
Solution:
Volume of metal in the sphere

Question 15.
A solid sphere of radius 15 cm is melted and recast into solid right circular cones of radius 2.5 cm and height 8 cm. Calculate the number of cones recast.
Solution:
volume of sphere = \(\frac { 4 }{ 3 }\)πr³
= \(\frac { 4 }{ 3 }\)π × (15)³
volume of cone = \(\frac { 1 }{ 3 }\)πr²h
= \(\frac { 1 }{ 3 }\)π × (2.5)² × 8
∴ No. of cones = \(\frac{\text { Vol. of spheres }}{\text { Vol.of cone }}\)

Question 16.
The surface area of a solid metallic sphere is 2464 cm² . It is melted and recast into solid right circular cones of radius 3.5 cm and height 7 cm. Calculate:
(i) the radius of the sphere.
(ii) the number of cones recast. (Take π = 22/7)
Solution:
(i) Surface area = 47πr²= 2464 cm² (given)
r² = \(\frac{2464}{4 \pi}\)
r² = \(\frac{2464 \times 7}{4 \times 22}\) = 196 ⇒ r = \(\sqrt{196}\)
⇒ r = \(\sqrt{14 \times 14}\)
r = 14 cm

(ii) Volume of sphere = \(\frac { 4 }{ 3 }\)πr³ = \(\frac { 4 }{ 3 }\)π (14)³
Volume of cone = \(\frac { 1 }{ 3 }\)πr²h
= \(\frac { 1 }{ 3 }\)π (3.5)² × 7

Question 17.
Two solid spheres of radii 2 cm and 4 cm are melted and recast into a cone of height 8 cm. Find the radius of the cone so formed.
Solution:
Radius of small sphere r = 2 cm
Radius of big sphere R = 4 cm
Volume of small sphere = \(\frac { 4 }{ 3 }\)πr³
= \(\frac { 4π }{ 3 }\) × (2)³ = \(\frac { 32π }{ 3 }\) cm³
Volume of big sphere = \(\frac { 4 }{ 3 }\)πr³
= \(\frac{256 \pi}{3}\) cm³
Volume of both sphere
= \(\frac{32 \pi}{3}\) + \(\frac{256 \pi}{3}\) = \(\frac{288 \pi}{3}\)cm³
Volume of the cone = \(\frac { 1 }{ 3 }\)πR₁²h
We need to find R₁
h = 8 cm (Given)
Volume of the cone = \(\frac { 1 }{ 3 }\)πR₁² × (8)
Volume of the cone = Volume of both the sphere
\(\frac { 1 }{ 3 }\)πR₁² × (8) = \(\frac{288 \pi}{3}\)
⇒ R₁² × (8) = 288
⇒ R₁² = \(\frac { 288 }{ 8 }\)
⇒ R₁² = 36 ⇒ R₁² = (6)²
⇒ R₁ = 6 cm

Question 18.
A certain number of metallic cones, each of radius 2 cm and height 3 cm are melted and recast into a solid sphere of radius 6 cm. Find the number of cones.
Solution:
Let the number of cones be n,
Let radius of the sphere be r_s, radius of a cone be r_c and h be the height of the cone. Volume of sphere = n(Volume of a metallic cone)

Hence, the number of cones is 72.