Students can track their progress and improvement through regular use of Class 10 ICSE Maths Solutions S Chand Chapter 15 Three Dimensional Solids Ex 15(c)
S Chand Class 10 ICSE Maths Solutions Chapter 15 Three Dimensional Solids Ex 15(c)
Question 1.
Complete the following table. Measurements of the cone are in centimetres. Do not substitute the value of π.
Solution:
(i) Base radius (r) = 3 cm
Height (h) = 4 cm
∴ Slant height (l) = \(\sqrt{r^2+h^2}\) = \(\sqrt{(3)^2+(4)^2}\) = \(\sqrt{9+16}\) = \(\sqrt{25}\) = 5 cm
Curved surface area = πrl
= π × 3 × 5 = 15π cm2
Area of the base = πr2
= π × 3 × 3 = 9π cm2
Total surface area = 15π + 9π = 24π cm2
volume = \(\frac { 1 }{ 3 }\)πr2h = \(\frac { 1 }{ 3 }\)π (3)2 × 4
= \(\frac { 1 }{ 3 }\) × π × 9 × 4 cm3 = 12π cm3
(ii) Base radius (r) = 20 cm
slant height (l) = 25
∴ Height (h) = \(\sqrt{l^2-r^2}\) = \(\sqrt{(25)^2-(20)^2}\) = \(\sqrt{625-400}\) = \(\sqrt{225}\) = 15 cm
Curved surface area = πrl = π × 20 × 25 cm2 = 500π cm2
Area of the base = πr2 = π × (20)2cm2 = 400π cm2
Total surface area = πrl + πr2 = 500π + 400π = 900π cm2
Volume = \(\frac { 1 }{ 3 }\)πr2h = \(\frac { 1 }{ 3 }\)π (20)2 × 15 cm3
= \(\frac { 1 }{ 3 }\)π × 400 × 15 = 2000π cm3
(iii) Base radius = ?
Height (h) = 18
Slant height (l) = 30
∴ Radius (r) = \(\sqrt{l^2-h^2}\) = \(\sqrt{(30)^2-(18)^2}\) = \(\sqrt{900-324}\) = \(\sqrt{576}\) = 24 cm
Curved surface area = πrl = π × 24 × 30 cm2 = 720π cm2
Area of the base = πr2 = π (24)2 = 576π cm2
Total surface area = 720π + 576π = 1296π cm2
Volume = \(\frac { 1 }{ 3 }\)πr2h = \(\frac { 1 }{ 3 }\)π (24)2 × 18 cm3
= \(\frac { 1 }{ 3 }\)π × 576 × 18 = 3456π cm3
(iv) Base radius (r) = 27
Height (h) = 36
∴ Slant height (l) = \(\sqrt{r^2+h^2}\)
= \(\sqrt{(27)^2+(36)^2}\) = \(\sqrt{729+1296}\)
= \(\sqrt{2025}\) = 45 cm
Curved surface area = πrl = π × 27 × 45 = 1215π cm2
Area of base = πr2 = π(27)2 = π × 729 = 729 π cm2
Total surface area = 1215π + 729π = 1944π cm2
volume = \(\frac { 1 }{ 3 }\)πr2h = \(\frac { 1 }{ 3 }\)π × (27)2 × 36 cm3
= \(\frac { 1 }{ 3 }\)π × 729 × 36 = 8748π cm3
(v) Base radius (r) = 5 cm
Curved surface area πrl = 65π
Slant height (l) = \(\frac{65 \pi}{5 \pi}\) = 13 cm
and height (h) = \(\sqrt{l^2-r^2}\) = \(\sqrt{(13)^2-(5)^2}\) = \(\sqrt{169-25}\) = \(\sqrt{144}\) = 12 cm
Area of base = πr2 = π(5)2 = 25π cm2
Total surface area = 65π + 25π = 90π cm2
Volume = \(\frac { 1 }{ 3 }\)πr2h = \(\frac { 1 }{ 3 }\)π (5)2 × 12 cm3
= \(\frac { 1 }{ 3 }\)π × 25 × 12 = 100π cm3
(vi) Base radius (r) = 35 cm
Total surface area =πrl + πr2 = 4410π
⇒ π × 35 l + π × (35)2 = 4410π
⇒ π(35 l + 1225) = 4410π
⇒ 35 l + 1225 = 4410
⇒ 35 l = 4410 – 1225 = 3185
l = \(\frac { 3185 }{ 35 }\) = 91
∴ Slant height = 91 cm
Height (h) = \(\sqrt{l^2-r^2}\) = \(\sqrt{91^2-35^2}\) = \(\sqrt{8281-1225}\) = \(\sqrt{7056}\) = 84 cm
Curved surface = πrl = π × 35 × 91 = 3185π
Area of base = πr2 = π(35)2 = 1225π
Total surface area = 4410π
volume = \(\frac { 1 }{ 3 }\)πr2h = \(\frac { 1 }{ 3 }\)π (35)2 × 84 cm3
= \(\frac { 1 }{ 3 }\)π × 1225 × 84 = 34300π cm3
Question 2.
Find the volume of the cone, given :
(i) Height 8 m; area of base 156 m2.
(ii) Slant height 17 cm, radius 8 cm.
(iii) Height 8 cm, slant length 10 cm.
(iv) Height 5 cm, perimeter of base 8 cm.
Solution:
(i) Area of base (πr2) = 156 m2
Height (h) = 8 m
∴ Volume = \(\frac { 1 }{ 3 }\)πr2h = Area of base × \(\frac { 1 }{ 3 }\)h
= \(\frac{156 \times 8}{3}\) = \(\frac{1248}{3}\) m3 = 416 m3
(ii) Slant height (l) = 17 cm
Radius (r) = 8 cm
Height (h) = \(\sqrt{l^2-r^2}\) = \(\sqrt{(17)^2-(8)^2}\) = \(\sqrt{289-64}\) = \(\sqrt{225}\) = 15 cm
∴ Volume = \(\frac { 1 }{ 3 }\)πr2h = \(\frac { 1 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × (8)2 × 15 cm3
= \(\frac { 22 }{ 21 }\) × 64 × 15 = 1005.71 cm3
(iii) Height (h) = 8 cm
and slant height (l) = 10 cm
∴ Radius (r) = \(\sqrt{l^2-h^2}\) = \(\sqrt{(10)^2-(8)^2}\) = \(\sqrt{100-64}\) = \(\sqrt{36}\) = 6 cm
∴ volume = \(\frac { 1 }{ 3 }\)πr2h = \(\frac { 1 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × (6)2 × 8 cm3
= \(\frac { 1 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × 36 × 8 = \(\frac { 2112 }{ 7 }\) = 301.71 cm3
(iv) Height (h) = 5 cm
Perimeter of base (2πr) = 8 cm
Radius (r) = \(\frac{8}{2 \pi}\) = \(\frac{4}{\pi}\) = 4 × \(\frac { 7 }{ 22 }\) = \(\frac { 14 }{ 11 }\) cm
∴ volume = \(\frac { 1 }{ 3 }\)πr2h = \(\frac { 1 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × \(\left(\frac{14}{11}\right)^2\) × 5 cm3
= \(\frac { 1 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × \(\frac { 196 }{ 121 }\) × 5 = \(\frac { 280 }{ 33 }\) cm3
= 8.4848 = 8.49 cm3
Question 3.
Find the area of curved surface of the circular cone, given :
(i) Height 8 m, slant height 10 m.
(ii) Perimeter of the base 88 cm, slant height 2 dm.
(iii) Area of the base 154 cm2, height 24 cm.
Solution:
(i) Height (h) = 8 m
Slant height (l) = 10 m
∴ Radius (r) = \(\sqrt{l^2-h^2}\) = \(\sqrt{(10)^2-(8)^2}\) = \(\sqrt{100-64}\) = \(\sqrt{36}\) = 6 m
∴ Curved surface area = πrl = \(\frac { 22 }{ 7 }\) × 6 × 10 m2
= \(\frac { 1320 }{ 7 }\) = 188.57 m2 = 188.6 m2
(ii) Perimeter of the base (2πr) = 88 cm
Radius (r) = \(\frac{88}{2 \pi}\) = \(\frac{88 \times 7}{2 \times 22}\) = 14 cm
Slant height (l) = 2 dm = 20 cm
∴ Height (h) = \(\sqrt{l^2-r^2}\) = \(\sqrt{(20)^2-(14)^2}\) = \(\sqrt{400-196}\) = \(\sqrt{204}\) cm
∴ Curved surface area = πrl
= \(\frac { 22 }{ 7 }\) × 14 × 20 cm2 = 880 cm2
(iii) Area of the base \(\left(\pi r^2\right)\) = 154 cm
Height (h) = 24 cm
Now πr2 = 154 ⇒ r2 = \(\frac{154}{\pi}\)
⇒ r2 = \(\sqrt{r^2+h^2}\) = \(\sqrt{(7)^2+(24)^2}\) = \(\sqrt{49+576}\) = \(\sqrt{625}\) = 25 cm
∴ Curved surface area = πrl
= \(\frac { 22 }{ 7 }\) × 7 × 25 cm2 = 550 cm2
Question 4.
Find the height of the cone whose base- radius is 5 cm and volume 50π cm3.
Solution:
Radius (r) = 5 cm
volume \(\left(\frac{1}{3} \pi r^2 h\right)\) = 50π cm3
⇒ \(\frac { 1 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × (5)2h = 50π ⇒ \(\frac { 22 }{ 21 }\) × 25h = 50π
h = \(\frac{50 . \pi \times 21}{22 \times 25}\) = \(\frac{50 \times 22 \times 21}{22 \times 7 \times 25}\) = 6 cm
∴ Height = 6 cm
Question 5.
The curved surface (area) of a right circular cone of radius 11.3 cm is 710 cm2. what is the slant height of the cone? \(\left(\text { Take } \pi=\frac{355}{113}\right)\)
Solution:
Radius of a cone (r) = 11.3 cm
Curved surface area = 710 cm2
∴ πrl = 710
\(\frac { 355 }{ 113 }\) × 11.3 × l = 710
⇒ l = \(\frac{710 \times 113 \times 10}{355 \times 113}\) = 20 cm
∴ Slant height = 20 cm
Question 6.
If the radius of the base of a circular cone is halved, keeping the height same, what is the ratio of the volume of the reduced cone to that of the original cone ?
Solution:
Let in first cone, radius = r and height = h
∴ volume = \(\frac { 1 }{ 3 }\)πr2h
In second case,
The radius = \(\frac { r }{ 2 }\)
and height = h
∴ volume = \(\frac { 1 }{ 3 }\)π \(\left(\frac{r}{2}\right)^2\) h = \(\frac{\frac{1}{3} \pi r^2 h}{4}\) = \(\frac{1}{12}\) πr2h
∴ Ratio \(\frac { 1 }{ 12 }\)πr2h : \(\frac { 1 }{ 3 }\)πr2h
⇒ \(\frac { 1 }{ 12 }\) : \(\frac { 1 }{ 3 }\) : \(\frac { 1 }{ 4 }\) : 1 ⇒ 1 : 4
Question 7.
A conical tent requires 264 m2 of canvas. If the slant height is 12 m, find the vertical height.
Solution:
Area of canvas = 264 m2
and slant height (l) = 12 m
∴πrl = 264
\(\frac { 22 }{ 7 }\) × r × 12 = 264
r = \(\frac{264 \times 7}{22 \times 12}\) = 7 cm
∴ vertical heiht (h) = \(\sqrt{l^2-r^2}\) = \(\sqrt{(12)^2-(7)^2}\) = \(\sqrt{144-49}\) = \(\sqrt{95}\) cm
= 9.746 = 9.75 cm
Question 8.
The vertical height of a right circular cone is twice its diameter and its volume is 36π cm3. Find the height.
Solution:
Let height of the cone = h
∴ Diameter = \(\frac { h }{ 2 }\)
∴ Radius (r) = \(\frac { h }{ 4 }\)
Volume = 36π cm3
⇒ \(\frac { 1 }{ 3 }\) πr2h = 36π ⇒ \(\frac { 1 }{ 3 }\) r2h = 36
r2h = 36 × 3
\(\left(\frac{h}{4}\right)^2\) × h = 108 ⇒ \(\frac{h^2}{16}\) × h = 108 (∵ h = 2d or 4r)
⇒ h3 = 108 × 16 = 1728 = (12)3
∴ h = 12 cm
∴ Height = 12 cm
Question 9.
The radius and height of a cone are in the ratio 3 : 4. If its volume is 301.44 cm3, what is its radius? What is its slant height ? (Take π = 3.14)
Solution:
Ratio in radius and height of a cone = 3 : 4
Let radius (r) =3x
Then height (h) = 4 x
Volume = \(\frac { 1 }{ 3 }\)πr2h
⇒ 301.44 =\(\frac { 1 }{ 3 }\) × 3.14 × (3x)2 × 4x
⇒ \(\frac{301.44 \times 3}{3.14}\) = 36x3
⇒ x3 = \(\frac{301.44 \times 3}{3.14 \times 36}\) = 8 = (2)3
∴ x = 2
∴ Radius (r) = 3x = 3 × 2 = 6 cm
and height (h) = 4x = 4 × 2 = 8 cm
∴ Slant height (l) = \(\sqrt{r^2+h^2}\) = \(\sqrt{(6)^2+(8)^2}\) = \(\sqrt{36+64}\) = \(\sqrt{100}\) = 10 cm
Question 10.
The radius and slant height of a cone are in the ratio of 4 : 7. If its curved surface area is 792 cm2, find its radius. (Use π = 22/7)
Solution:
Ratio in radius and slant height = 4 : 7
Let radius (r) =4 x
The slant height (l) = 7 x
Curved surface area = πrl
⇒ 792 = \(\frac { 22 }{ 7 }\) × 4x × 7x
⇒ 792 = 88x2 ⇒ x2 = \(\frac { 792 }{ 88 }\) = 9 = (3)2
∴ x = 3
∴ Radius = 4x = 4 × 3 = 12 cm
Question 11.
The base radii of two right circular cones of the same height are in the ratio 3 : 5. Find the ratio of their volumes.
Solution:
Ratio between the radii of two cones r1 : r2 = 3 : 5
Let r1 = 3 x and r2 = 5 x and height in each case =h
∴ Volume of first cone =\(\frac { 1 }{ 3 }\)πr12h = \(\frac { 1 }{ 3 }\)π (3x)2h
= \(\frac { 1 }{ 3 }\)π × 9x2h = 3πx2h
and volume of second cone
= \(\frac { 1 }{ 3 }\)πr2h = \(\frac { 1 }{ 3 }\)π(5x)2h = \(\frac { 1 }{ 3 }\)π × 25x2h = \(\frac { 25 }{ 3 }\)πr2h
∴ Ratio between their volumes
= 3πr2h : \(\frac { 25 }{ 3 }\)πr2h = 9πr2h : 25πr2h = 9 : 25
Question 12.
The circumference of the base of a 10 m high conical tent is 44 metres. Calculate the length of canvas used in making the tent if width of canvas is 2 cm. (Use π = 22/7)
Solution:
Height of tent (h) = 10 m
Circumference of the base = 44 m
i.e. 2nr = 44 ⇒ \(\frac{2 \times 22}{7}\)r = 44
⇒ r = \(\frac{44 \times 7}{2 \times 22}\) = 7 m
Slant height (l) = \(\sqrt{r^2+h^2}\) = \(\sqrt{(7)^2+(10)^2}\) = \(\sqrt{49+100}\) = \(\sqrt{149}\) m
∴ Curved surface area = πrl
= \(\frac { 22 }{ 7 }\) × 7 × 12.2 = 268.4 m2
Width of canvas = 2 m = 12.2 m
∴ Length = 268.4 ÷ 2 = 134.2 m
∴ Length of canvas = 134.2 m
Question 13.
How many metres of cloth 5 m wide will be required to make to conical tent, the radius of whose base is 7 m and whose height is 24 m ? (Take π = 22/7)
Solution:
Radius of the base of the tent (r) = 7 m and height (h) = 24 m
∴ Slant height (l) = \(\sqrt{r^2+h^2}\) = \(\sqrt{(7)^2+(24)^2}\) = \(\sqrt{49+576}\) = \(\sqrt{625}\) = 25
∴ Curved surface area = πrl
= \(\frac { 22 }{ 7 }\) × 7 × 25 m2 = 550 m2
width of the cunvass = 5 m
∴ Length of convas required = \(\frac { 550 }{ 5 }\) = 110 m
Question 14.
A conical tent of capacity 1232 m2 stands on a circular base of area 154 m2. Find in m2 the area of the canvas.
Solution:
Capacity of the tent = 1232 m2 and area of base of the tent = 154 m2
Let r be the radius h be the height of the tent, then
πr2 = 154 ⇒ \(\frac { 22 }{ 7 }\) r2 = 154
⇒ r2 = \(\frac{154 \times 7}{22}\) = 49 = (7)2
∴ r = 7 m
∴ volume = 1232
∴ \(\frac { 1 }{ 3 }\) πr2h = 1232
⇒ \(\frac { 1 }{ 3 }\) × \(\frac { 22 }{ 7 }\) (7)2h = 1232
⇒ \(\frac { 1 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × 49h = 1232
⇒ h = \(\frac{1232 \times 3 \times 7}{22 \times 49}\) = 24 m
Now slant height (l) = \(\sqrt{r^2+h^2}\) = \(\) = \(\) = \(\) = 25 m
∴ Curved surface area = πrl
= \(\frac { 22 }{ 7 }\) × 7 × 25 = 550 m2
∴ Area of convas = 550 m2
Question 15.
A conical tent is to accommodate 11 persons, each person must have 4 m2 of space on the ground and 20 m3 of air to breathe. Find the height of the cone.
Solution:
In the tent, accommodation is available for = 11 persons
Space required for each person on the ground = 4 m2
∴ Area of the base of the tent =11 × 4 = 44 m2
Air is required for each person = 20 m3
∴ Volume of the air of the tent = 20 × 11 = 220 m3
Let r be the radius and h be the height of the tent
∴ πtr2 = 44
⇒ \(\frac { 22 }{ 7 }\) r2 = 44 ⇒ r2 = \(\frac{44 \times 7}{22}\) = 14
and \(\frac { 1 }{ 3 }\) πr2h = 220
⇒ \(\frac { 1 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × 14h = 220 (∵ r2 = 14)
⇒ h = \(\frac{220 \times 3 \times 7}{22 \times 14}\) = 15
∴ Height of the conical tent = 15 m
Question 16.
Find out whether the following statement is true or false :
The volume of a cone is one-half of the volume of the cylinder of the same radius and height.
Solution:
It is false as the volume of a cone is one third (\(\frac { 1 }{ 3 }\)) of the cylinder of the same radius and height.
Question 17.
The volume of a cone is the same as that of a cylinder whose height is 9 cm and diameter 40 cm. Find the radius of the base of cone if its height is 108 cm. \(\left(\text { Take } \pi \text { to be } \frac{22}{7}\right)\)
Solution:
Diameter of cylinder = 40 cm
∴ Radius (r1) = \(\frac { 40 }{ 2 }\) = 20 cm
Height (h1) = 9 cm
∴ Volume = πr12h1 = π(20)2 × 9 cm3 = 3600π cm3
Height (h2) = 108 cm
Let r2 be the radius, then
\(\frac { 1 }{ 3 }\) πr22h = 3600π
\(\frac { 1 }{ 3 }\) πr22 × 108 = 3600π
r22 = \(\frac{3600 \pi \times 3}{108 \pi}\) = 100 = (10)2
∴ r2 = 10 cm
∴ Radius of the base of the cone = 10 cm
Question 18.
A canvas tent is in the shape of a cylinder surmounted by a conical roof. The common diameter of the cone and cylinder is 14 m. The height of the cylinderical part is 8 m and the height of the conical roof is 4 m. Find the area of the canvas used to make the tent. Give your answer in m2 correct to one decimal place. \(\left(\text { Take } \pi \text { to be } \frac{22}{7}\right)\)
Solution:
Diameter of the tent = 14 m
∴ Radius (r) = \(\frac { 14 }{ 2 }\) = 7 m
Height cf cylinderical part (h1) = 8 m
and height of conical part (h2) = 4 m
∴ Slant height (l) = \(\sqrt{r^2+h_2^2}\) = \(\sqrt{(7)^2+(4)^2}\) = \(\sqrt{49+16}\) m = \(\sqrt{65}\) m = 8.06 m
Now area of convas = curved surface area of cylindrical part + curved surface area of conical part = 2πrh + πrl
= 2 × \(\frac { 22 }{ 7 }\) × 7 × 8 + \(\frac { 22 }{ 7 }\) × 7 × 8.06 m2
= 352 + 177.3 = 529.3 m2
Question 19.
A circus tent is cylindrical to a height 3 m and conical above it. If its diameter is 105 m and slant height of the cone is 53 m, calculate the total area of the canvas
required. \(\left(\text { Use } \pi=\frac{22}{7}\right)\)
Solution:
Diameter of the tent = 105 m
∴ Radius (r) = \(\frac { 105 }{ 2 }\) m
Height of the cylinderical part (h) = 3 m
and slant height of the conical part (l) = 53 m
Total area of the convas used to make it = 2 πrh + πrl
= 2 × \(\frac { 22 }{ 7 }\) × \(\frac { 105 }{ 2 }\) × 3 + \(\frac { 22 }{ 7 }\) × \(\frac { 105 }{ 2 }\) × 53
= 990 + 8745 = 9735 m2
Question 20.
Find the volume of the largest circular cone that can be cut out of a cube whose edge is 9 cm.
Solution:
Edge of cube = 9 cm
∴ Diameter of the largest cone = 9 cm (equal to edge of cube)
and radius (r) = \(\frac { 9 }{ 2 }\) cm
Height (h) = 9 cm (equal to edge of cube)
∴ Volume of the largest cone = \(\frac { 1 }{ 3 }\) πr2h
= \(\frac { 1 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × \(\frac { 9 }{ 2 }\) × \(\frac { 9 }{ 2 }\) × 9 cm3
= 190.9285 = 190.93 cm3
Question 21.
A tent is of the shape of a right circular cylinder upto a height of 3 metres and then becomes a right circular cone with a maximum height of 13.5 metres above the ground. Calculate the cost of painting the inner side of the tent at the rate of Rs. 2 per square metre, if the radius of the base is 14 metres.
Solution:
Total height of the tent = 13.5 m
Radius of the base of the tent (r) = 14 m
Height of the cylindrical part (h1) = 3 m and height of the conical part (h2)
= 13.5 – 3.0 = 10.5 m
Now slant height (l) = \(\sqrt{r^2+h_2^2}\) = \(\sqrt{(14)^2+(10.5)^2}\) = \(\sqrt{196+110.25}\) = \(\sqrt{306.25}\) = 17.5 m
Now curved surface area of the tent = πrl + 2πrh1 = πr (l + 2h1)
= \(\frac { 22 }{ 7 }\) × 14 (17.5 + 2 × 3) m2
= 44(23.5) = 1034 m2
Rate of painting the inner side = Rs. 2 per m2
∴ Tota cost = 1034 × 2 = Rs. 2068
Question 22.
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 the sand. If the height of the conical heap is 24 cm, find (i) the radius and (ii) the slant height of the heap. Give your answer correct to one place of decimal.
Solution:
Radius of the bucket (r) = 18 cm
Height (h) = 32 cm
∴ Volume of sand in it = πr2h
= π × 18 × 18 × 32 cm3 = 10368π cm3
∴ Volume of conical heap of the sand = 10368π cm3
Height of heap = 24 cm
∴ \(\frac { 1 }{ 3 }\)πr2h = 10368π
\(\frac { 1 }{ 3 }\)πr2 × 24 = 10368π
r2 = \(\frac{10368 \pi \times 3}{\pi \times 24}\) = 1296 = (36)2
∴ Radius of cone = 36 cm and slant heigth (l) = \(\sqrt{r^2+h^2}\)
= \(\sqrt{(36)^2+(24)^2}\)
= \(\sqrt{1296+576}\) cm
= \(\sqrt{1872}\) cm = 43.266 cm
= 43.3 cm
Question 23.
From a solid cylinder whose height is 8 cm and radius is 6 cm, a conical cavity of height 8 cm and of base radius 6 cm is hollowed out. Find the volume of the remaining solid correct to 4 places of decimals. (π = 3.1416)
Solution:
Radius of a cylinder (r) = 6 cm and height (h) = 8 cm
∴ Volume of the cylinder = πr22h
= 3.1416 × (6)2 × 8 cm3
= 3.1416 × 36 × 8 cm3 = 904.7808 cm3
Radius of cone (r) = 6 cm and height (h) = 8 cm
∴ Volume of cone carved = \(\frac { 1 }{ 3 }\)πr2h
= \(\frac { 1 }{ 3 }\) × 3.1416 × 36 × 8 cm2
= 301.5936 cm3
∴ Volume of the remaining solid
= 904.7808 – 301.5936 = 603.1872 cm3
Question 24.
A metallic cylinder has radius 3 cm and height 5 cm. It is made of a metal A. To reduce its weight, a conical hole is drilled in the cylinder as shown in the figure and it is completely filled with a lighter metal B. The conical hole has a radius of \(\frac { 3 }{ 2 }\) cm and its depth is \(\frac { 8 }{ 9 }\) cm. Calculate the ratio of the volume of the metal A to the volume of the metal B in the solid.
Solution:
Radius of cylinder (R) = 3 cm
and height (H) = 5 cm
Volume of whole cylindrical metal = πR2H
= π(3)2 × 5 cm3 = 45π cm3
Radius of conical portion (r) = \(\frac { 3 }{ 2 }\) cm and height (h) = \(\frac { 8 }{ 9 }\)
∴ volume = \(\frac { 1 }{ 3 }\)πr22h = \(\frac { 1 }{ 3 }\)π \(\left(\frac{3}{2}\right)^2\) × \(\frac { 8 }{ 9 }\)cm3
= \(\frac { 1 }{ 3 }\)π × \(\frac { 9 }{ 4 }\) × \(\frac { 8 }{ 9 }\) = \(\frac { 2 }{ 3 }\)π cm3
Metal in the remaining portion
= 45π – \(\frac { 2 }{ 3 }\)π = \(\frac{(135-2) \pi}{3}\) = \(\frac { 133 }{ 3 }\)π cm
∴ Ratio = \(\frac { 133 }{ 3 }\)π : \(\frac { 2 }{ 3 }\)π = 133 : 2
Question 25.
An open cylindrical vessel of internal diameter 7 cm and height 8 cm stands on a horizontal table. Inside this is placed a solid metallic right circular cone, the diameter of whose base is \(\frac { 7 }{ 2 }\) cm and height 8 cm. Find the volume of water required to fill the vessel. If the cone is replaced by another cone, whose height is 1\(\frac { 3 }{ 4 }\) cm and the radius of whose base is 2 cm, find the drop in the water level.
Solution:
Diameter of the cylindrical vessel = 7 cm
∴ Radius (r1) = \(\frac { 7 }{ 2 }\) cm
Height (h1) = 8 cm
In First case,
Radius of cone (r2) = \(\frac { 7 }{ 4 }\) cm
and heigth (h2) = 8 cm
∴ Volume of cylinder = πr12h1
= \(\frac { 22 }{ 7 }\) × \(\left(\frac{7}{2}\right)^2\) × 8 cm
= \(\frac { 22 }{ 7 }\) × \(\frac { 7 }{ 2 }\) × \(\frac { 7 }{ 2 }\) × 8 = 308 cm3
and volume of cone = \(\frac { 1 }{ 3 }\)πr22h2
= \(\frac { 1 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × \(\frac { 7 }{ 4 }\) × \(\frac { 7 }{ 4 }\) × 8 = \(\frac { 77 }{ 3 }\) cm3
∴ volume of water to be filled in the vessel = \(\frac { 308 }{ 1 }\) – \(\frac { 77 }{ 3 }\) = \(\frac { 924 – 77 }{ 3 }\)
= \(\frac { 847 }{ 3 }\) = 282\(\frac { 1 }{ 3 }\) cm3
In second case, when the cone is replaced by another one of the height = 1\(\frac { 3 }{ 4 }\) cm and radius = 2 cm
∴ volume = \(\frac { 1 }{ 3 }\) πr3h3
= \(\frac { 1 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × (2)2 × \(\frac { 7 }{ 4 }\) cm3
= \(\frac { 1 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × 4 × \(\frac { 7 }{ 4 }\) = \(\frac { 22 }{ 3 }\) cm3
∴ Change in volume of the cones
= \(\frac { 77 }{ 3 }\) – \(\frac { 22 }{ 3 }\) = \(\frac { 55 }{ 3 }\) cm3
Let the drop in water level be x cm, then
\(\frac { 22 }{ 7 }\) × \(\left(\frac{7}{2}\right)^2\) × x = \(\frac { 55 }{ 3 }\)
⇒ \(\frac { 22 }{ 7 }\) × \(\frac { 49 }{ 4 }\) x = \(\frac { 55 }{ 3 }\)
⇒ \(\frac { 77 }{ 2 }\)x = \(\frac { 55 }{ 3 }\)
⇒ x = \(\frac { 55 }{ 3 }\) × \(\frac { 2 }{ 77 }\) = \(\frac { 10 }{ 21 }\) cm
∴ Fall in water level = \(\frac { 10 }{ 21 }\) cm