问题13.船只是一个空心圆柱体,在同一基座的半球形底部装有一个圆柱体。圆柱体的深度为14/3,半球的直径为3.5 m。计算固体的体积和内表面积。
解决方案:
According to the question
Diameter of the hemisphere(d) = 3.5 m,
So, the radius of the hemisphere(r) = 1.75 m,
Height of the cylinder(h) = 14/3 m,
Now we find the volume of the cylinder
V1 = πr2 h1
= π(1.75)2 x 14/3 m3
Now we find the volume of the hemisphere
V2= 2/3 × 22/7 × r3
V2 = 2/3 × 22/7 × 1.753 m3
So, the total volume of the vessel is
V = V1 + V2
V = π(1.75)2 x 14/3 + 2/3 × 22/7 × 1.753
V = 56 m3
Now we find the internal surface area of solid
S = 2πrh1 + 2πr2
= 2 π(1.75)(143) + 2 π(1.75)2 = 70.51 m3
Hence, the internal surface area of the solid is 70.51 m3 and the volume is 56 m3
问题14.考虑一个由具有半球形末端的圆柱体组成的固体。如果固体的完整长度为104厘米,每个半球形末端的半径为7厘米。找到以Rs速率抛光其表面的成本。每dm 2 10。
解决方案:
According to the question
Radius of the hemispherical end (r) = 7 cm,
Height of the solid = h + 2r = 104 cm, h = 90 cm
Now we find the curved surface area of the cylinder
A1 = 2 πr h
= 2 π(7) h
= 2 π(7)(90)
= 3948.40 cm2
Now we find the curved surface area of the two Hemisphere
A2 = 2 (2πr2)
= 22π(7)2 = 615.75 cm2
Hence, the total curved curface area of the solid is
A = A1 + A2
= 3948.40 + 615.75 = 4571.8 cm2 = 45.718 dm2
So, he cost of polishing the 1 dm2 surface of the solid is Rs. 15
Therefore, the cost of polishing the 45.718 dm2 surface of the solid = 10 45.718 = Rs. 457.18
Hence, the cost of polishing is Rs. 457.18.
问题15:直径为14厘米,高度为42厘米的圆柱形容器对称地固定在直径为16厘米,高度为42厘米的类似容器内。两个容器之间的整个空间都充满了软木粉尘,以达到隔热的目的。发现需要多少立方厘米的软木粉尘?
解决方案:
According to the question
Depth of the vessel = Height of the vessel = h = 42 cm,
Inner diameter of the vessel = 14 cm,
So, inner radius of the vessel = r1 = 14/2 = 7 cm
Outer diameter of the vessel = 16 cm,
So, the outer radius of the vessel = r2 = 16/2 = 8 cm
Now we find the volume of the vessel
V = π(r22 – r12)h
= π(82 – 72) x 42 = 1980 cm3
Hence, the volume of the vessel is 1980 cm3 which is equal to the amount of cork dust required.
问题16:铁制的圆柱形压路机长1 m。它的内径为54厘米,用于制造滚筒的铁片的厚度为9厘米。如果1 cm 3的铁的质量为7.8 gm,则求出压路机的质量。
解决方案:
According to the question
Height of the road roller (h) = 1 m = 100 cm,
Internal Diameter of the road roller = 54 cm,
So, the internal radius of the road roller (r)= 27 cm
Thickness of the road roller (T) = 9 cm,
Let us considered R be the outer radii of the road roller. So,
T = R – r
9 = R – 27
R = 27 + 9 = 36 cm
Now we find the volume of the Iron Sheet
V = π × (R2 − r2) × h
= π × (362 − 272) × 100 = 1780.38 cm3
Hence the mass of 1 cm3 of the iron sheet = 7.8 gm [Given]
Therefore, the mass of 1780.38 cm3 of the iron sheet = 1388696.4 gm = 1388.7 kg
Hence, the mass of the road roller is 1388.7 kg
问题17:通过空心圆柱体安装的空心半球形式的容器。半球的直径为14厘米,血管的总高度为13厘米。找到容器的内表面积。
解决方案:
According to the question
Diameter of the hemisphere = 14 cm,
So, the radius of the hemisphere = 7 cm
And the total height of the vessel = = h + r = 13 cm
Now we find the inner surface area of the vessel
A = 2πr (h + r)
= 2 x 22/7 x (13) x (7)
= 572 cm2
Hence, the inner surface area of the vessel is 572 cm2
问题18:一个玩具是一个半径为3.5厘米的圆锥体形式,安装在相同半径的半球上。玩具的总高度为15.5厘米。找到玩具的总表面积。
解决方案:
According to the question
Radius of the conical part of the toy(r) = 3.5 cm
Total height of the toy (h) = 15.5 cm
Slant height of the cone (L) = 15.5 – 3.5 = 12 cm
Now we find the curved surface area of the cone
A1 = πrL
= π(3.5)(12) = 131.94 cm2
Now we find the curved surface area of the hemisphere
A2 = 2πr2
= 2π(3.5)2 = 76.96 cm2
Hence, the total surface area of the toy
A = A1 + A2
= 131.94 + 76.96 = 208.90 cm2
Hence, the total surface area of the toy is 209 cm2
问题19:14 cm长的圆柱形金属管的内外表面积之差为44 dm 2 。如果管道由99 cm 2的金属制成。查找管道的外半径和内半径。
解决方案:
According to the question
Length of the cylinder (h) = 14 cm
Difference between the outer and the inner surface area = 44 dm2
Volume of the metal used = 99 cm2,
Let’s assume that the inner radius of the pipe be r1 and r2 be the outer radius of the pipe
Now, the surface area of the cylinder is = 2π x 14 x (r2 – r1) = 44
(r2 − r1) = 1/2 —————-(i)
Volume of the cylinder is
πh(r22 – r12) = 99
πh(r2 – r1) (r2 + r1) = 99
= 22/7 x 14 x 1/2 x (r2 + r1) = 99
(r2 + r1) = 9/2 ————–(ii)
On Solving eq(i) and (ii), we get,
r2 = 5/2 cm
r1 = 2 cm
Hence, the inner radius is 2cm and outer radius of the pipe is 5/2 cm.
问题20.直径为12厘米,高度为15厘米的右圆柱是完整的冰淇淋。冰淇淋要装在高度为12厘米,直径为6厘米,顶部为半球形的圆锥形中。找到可以装满冰淇淋的这种锥体的数量。
解决方案:
According to the question
Radius of cylinder (r1) = 6 cm,
Radius of hemisphere (r2) = 3 cm,
Height of cylinder (h) = 15 cm,
slant height of the cones (l) = 12 cm,
Now we find the volume of cylinder
V = πr12h
= π × 62 ×15 ————–(i)
The volume of each ice cream cone = Volume of cone + Volume of hemisphere
= 1/3πr22h + 2/3πr23
= 1/3π x 62 x 12 + 2/3π x 33 —————-(ii)
Let’s assume that number of cones be ‘n’
n(Volume of each ice cream cone) = Volume of cylinder
n(1/3 π x 32 x 12 + 2/3 π x 33) = π(6)2 x15
n = 50/5 = 10
Hence, the number of cones being filled with ice-cream is 10
问题21.考虑一个实心铁杆,其圆柱形部分高110厘米,底直径12厘米,上方是9厘米的圆锥体。找到极点的质量。假设1 cm 3的铁杆质量为8 gm。
解决方案:
According to the question
Base diameter of the cylinder = 12 cm,
So, the radius of the cylinder (r) = 6 cm
Height of the cylinder (h) = 110 cm,
Slant height of the cone (L) = 9 cm,
Now we find the volume of the cylinder
V1 = π × r2 × h
= π × 62 × 110 cm3
Now we find the volume of the cone
V2 = 1/3 × πr2L
= 1/3 × π x 62 x 12 = 108π cm3
Hence, the volume of the pole (V)
V = V1 + V2
= 108π + π(6)2 110 = 12785.14 cm3
So, the mass of 1 cm3 of the iron pole = 8 gm [Given]
Then, the mass of 12785.14 cm3 of the iron pole = 8
12785.14 = 102281.12 gm = 102.2 kg
Hence, the mass of the iron pole is 102.2 kg
问题22:一个固体玩具是一个半球形,上面覆盖着一个右圆锥形。锥体的高度为2厘米,底座的直径为4厘米。如果右圆柱体围住了玩具,请问该玩具将覆盖多少空间?
解决方案:
According to the question
Radius of the cone, cylinder, and hemisphere (r) = 2 cm
Height of the cone (l) = 2 cm
Height of the cylinder (h) = 4 cm
Now we find the volume of the cylinder
V1 = π × r2 × h
= π × 22 × 4 cm3
Now we find the volume of the cone
V2 = 1/3 π r2l
= 1/3 × π × 22 × 2
= 1/3 × π x 4 × 2 cm3
Now we find the volume of the hemisphere
V3 = 2/3 π r3
= 2/3 × π × 23 cm3
= 2/3 π × 8 cm3
Therefore, the remaining volume of the cylinder when the toy is inserted to it is
V = V1 – (V2 + V3)
= 16π – 8π = 8π cm3
Hence, remaining volume of the cylinder when toy is inserted into it is 8π cm3
问题23。考虑一个由高度为120 cm,半径为60 cm的直圆锥构成的实体,直立在半径为60 cm的半球上,将其垂直放置在充满水的右圆柱中,使其接触底部。如果圆柱体的半径为60厘米,高度为180厘米,则求出圆柱体内剩余的水量。
解决方案:
According to the question
Radius of the circular cone (r) = 60 cm,
Height of the circular cone (L) = 120 cm,
Radius of the hemisphere (r) = 60 cm,
Radius of the cylinder (R) = 60 cm,
Height of the cylinder (H) = 180 cm,
Now we find the volume of the circular cone
V1 = 1/3 × πr2l
= 1/3 × π × 602 × 120 = 452571.429 cm3
Now we find the volume of the hemisphere
V2 = 2/3 × πr3
= 2/3 × π × 603 = 452571.429 cm3
Now we find the volume of the cylinder
V3 = π × R2 × H
= π × 602 × 180 = 2036571.43 cm3
Hence, the volume of water left in the cylinder
V = V3 – (V1 + V2)
= 2036571.43 – (452571.429 + 452571.429)
= 2036571.43 – 905142.858 = 1131428.57 cm3 = 1.1314 m3
Hence, the volume of the water left in the cylinder is 1.1314 m3
问题24.考虑一个内径10厘米,高10.5厘米的圆柱形容器装满水。底部直径为8厘米,高度为6厘米的实心圆锥完全浸入水中。在以下情况下求水的价值:
(i)从气缸中移出
(ii)留在气瓶中
解决方案:
According to the question
Internal diameter of the cylindrical vessel (D)= 10 cm,
Radius of the cylindrical vessel (r) = 5 cm
Height of the cylindrical vessel (h) = 10.5 cm,
Base diameter of the solid cone = 7 cm,
Radius of the solid cone (R) = 3.5 cm
Height of the cone (L) = 6 cm,
(i) We find the volume of water displaced out from the cylinder which is equal to the volume of the cone
So, V1 = 1/3 × πR2L
V1 = 1/3 × π3.52 × 6 = 77 cm3
Hence, the volume of the water displaced out of the cylinder is 77 cm3
(ii) First we find the volume of the cylindrical vessel is
V2 = π × r2 × h
= π × 52 × 10.5
= 824.6 cm3 = 825 cm3
Now we find the volume of the water left in the cylinder
V = V2 – V1
V = 825 – 77 = 748 cm3
Hence, the volume of the water left in cylinder is 748 cm3