问题17.右圆柱的高度为10.5 m。其两个圆形表面的面积之和的三倍是曲面面积的两倍。找到圆柱体的体积。
解决方案:
Given that
Height of the cylinder = 10.5 m
According to the question
3(2πr2) = 2(2πrh)
3r = 2h
r = 2/3 x h
r = 2/3 x 10.5 = 7cm
So, the radius of the well is 7cm
Now we find the volume of the cylinder
V = πr2 h
= 22/7 x 7 x 7 x 10.5
= 154 x 10.5 = 1617 cm3
Hence, the volume of the cylinder is 1617 cm3
问题18.必须挖出多少立方米的土壤才能沉入21m深,6m直径的井中?找出内表面的抹灰成本为每m 2 9.50卢比。
解决方案:
Given that
Height of the well = 21 m
Diameter of the well = 6 m
So, the radius of the well = 3 m
Now,
Volume of the cylinder = πr2 h
= 22/7 x 3 x 3 x 21
= 66 x 9 = 594 cm3
Cost of plastering = 9.5 per m3
Cost of plastering inner surface = (594 x 9.50) = Rs. 5643
问题19:一棵树的树干是圆柱形的,周长为176厘米。如果树的长度是3 m。找到可以从树干上获得的木材量。
解决方案:
Given that,
The length of the tree = 3m = 300cm
The circumference of the trunk = 176 cm
We have to find the volume of the timber that can be obtained from the trunk
So, According to the circumference formula
C= 2πr
176 = 2πr
r = 176/2π = 28 cm
Hence, the radius of the trunk = 28 cm
Now we find the volume of timber
V= πr2 h
= 22/7 x 28 x 28 x 300
= 44 x 8400 = 739200 cm3 or 0.7392 m3
Hence, the volume of timber is 739200 cm3 or 0.7392 m3
问题20.直径为14 m的一口井的深度为8 m。从中取出的泥土已均匀地分布在其周围,宽度为21 m,形成了路堤。找到路堤的高度。
解决方案:
Given that
The diameter of well = 14 m
so the radius = 7 m
Height of the well = 8 m
So,
Volume of well = πr2 h
= 22/7 x 7 x 7 x 8 = 22 x 56
= 1232 m3
Now, Let’s assume that r1 be the radius of embankment, and h1 be the height of embankment
So, Volume of well = Volume of embankment
1232 m3 = π x r1 x h1
1232 = 22/7 x (282 − 72) x h1
h1 = 1232 x 7 / 22(784 – 49)
h1 = 0.533 m
Hence, the height of the embankment is 0.533 m or 53.3 cm
问题21.圆柱管的内外表面之间的差为14厘米长,为88平方厘米。如果管的体积为176立方厘米,请找到管的内半径和外半径。
解决方案:
Let’s assume that R be the outer radius and r be the inner radius
Given that
The height of the cylindrical tube = 14 cm
The difference between inside and outside surfaces of a cylindrical tube = 88 cm2
2πRh – 2πrh = 88
2πh(R – r) = 88
2 x 22/7 x 14(R – r) = 88
(R – r) = 1cm ——————-(i)
Now,
Volume of tube = πR2 h – πr2 h
176 = πh(R2 – r2)
176 = 22/7 x 14(R2 – r2)
(R2 – r2) = 4
(R + r)(R – r) = 4
Put the value of (R – r) from eq(1)
(R + r) (1) = 4
(R + r) = 4 cm
R = 4 – r ————(ii)
Here, R – r = 1
R = 1 + r
Now substitute value of R in eq(ii)
1 + r = 4 – r
2r = 3
r = 3/2 = 1.5 cm
Substitute value of r in eq(i)
R – 1.5 = 1
R = 1 + 1.5 = 2.5 cm
Hence, the value of inner radii is 1.5 cm and radius of outer radii is 2.5 cm.
问题22。水通过内径为2厘米的圆形管道以每秒6米的速度流入圆柱形水箱。将水收集在底部直径为60厘米的圆柱形容器半径内。在30分钟内发现水位上升?
解决方案:
Given that,
The internal diameter of the pipe = 2 cm,
So, the radius of the pipe = 1 cm = 1/100 m
Water flow rate through the pipe in the cylindrical tank = 6 m/sec,
Radius of the cylindrical tank = 60 cm = 60/100 m
Time = 30 minutes.
Now we find the volume of water that flows in 1 sec
= 22/7 × 1/100 × 1/100 × 6
Now we have to find the volume of water flows in 30 minute.
= 22/7 × 1/100 × 1/100 × 6 × 30 × 60
So, the volume of water collected in the tank after 30 minutes = Volume of water that flows through the pipe in 30 minutes
22/7 × 60/100 × 60/100 × h = 22/7 × 1/100 × 1/100 × 6 × 30 × 60
h = 3 m
Hence, the height of the tank is 3 meter.
问题23.底部直径为56厘米的圆柱形容器装有足够的水,可以浸没尺寸为32厘米×22厘米×14厘米的矩形铁固体。当固体完全浸没时,求出水位的升高。
解决方案:
Given that,
The diameter of the cylindrical container = 56 cm
So, the radius of cylindrical container = 28 cm
The dimensions of rectangular block = 32 cm × 22 cm × 14 cm
Find: the raise in the level of water in the cylinder.
Let’s assume that the raise in the level of water be h, then
So, the volume of cylinder of height h and radius 28 cm = Volume of the rectangular block
22/7 × 28 × 28 × h = 32 × 22 × 14
h = 4 cm
Hence, the rise in the level of the water when the solid is completely submerged is 4 cm.
问题24.两端开口的圆柱管由金属制成。管的内径为10.4厘米,长度为25厘米。各处的金属厚度均为8毫米。计算金属的体积。
解决方案:
Given that,
Internal diameter of the tube = 10.4 cm,
So, the radius of the tube = 10.4/2 = 5.2 cm
Thickness of the metal of the tube = 8 mm = 0.8 cm
Length of the pipe = 25 cm,
Find: the volume of the metal used in the pipe.
Let us assume that the external radius be ‘R’.
So, now we find the external radius
R = 5.2 + 0.8 = 6 cm
So, the volume of metal in the pipe
= π(R2 − r2)h
= 22/7 × (62 − 5.22) × 25
= 704 cm3
Hence, the volume of metal present in the hollow pipe is 704 cm3.
问题25.水从内径0.75厘米的水龙头以每秒7 m的速度流动。在1小时内找到通过管道输送的水的升量。
解决方案:
Given that,
Radius of the tap = 0.75 cm
Water flow rate = 7 m/sec = 700 cm/sec
So height of the cylinder = 7 m
Time = 1 hour = 60 min = 3600 sec
Find: the volume of water that flows through the pipe for 1 hour.
Now we find the volume of water delivered in 1 second
= 22/7 × 0.75 × 0.75 × 700
Now, we have to find the volume of water delivered in 1 hour
= 22/7 × 0.75 × 0.75 × 700 × 3600 = 4455000 cm3 = 4455 liters
Hence, the volume of water delivered by the pipe is 4455 liters.
问题26.一个直径为1.4 m,高度为2.1 m的圆柱形水箱由直径为3.5 cm的管子供水,水以每秒2米的速度流过。多少时间可以将水箱装满?
解决方案:
Given that,
Diameter of the tank = 1.4 m,
So, the radius = 1.4/2 = 0.7 m,
Height of the tank = 2.1 m,
Diameter of the pipe = 3.5 cm,
So, the radius of pipe = 3.5/2 cm = 3.5/200 m
Water flow rate = 2 m/sec,
Find: the time required to fill the tank using the pipe.
So, the volume of the tank
V = πr2h
= 22/7 × 0.7 × 0.7 × 2.1
Now we find the volume of water that flows through the pipe in 1 second
= 22/7 × 3.5/200 × 3.5/200 × 2
Let’s assume that the time taken to fill the tank be x seconds,
So, the volume of water that flows through the pipe in x seconds
= 22/7 × 3.5/200 × 3.5/200 × 2 × x
As we know that the volume of water that flows through the pipe in x seconds = Volume of the tank
22/7 × 3.5/200 × 3.5/200 × 2 × x = 22/7 × 0.7 × 0.7 × 2.1
So, x = 1680 seconds = 1680/60 minutes = 28 minutes
Hence, the tank is filled in 28 minutes.
问题27.可以通过两种方式将一张30厘米×18厘米的矩形纸转换为直圆柱的曲面,即通过沿其长度滚动或沿其宽度滚动。求出由此形成的两个圆柱体的体积比。
解决方案:
Given that,
The dimensions of the rectangular sheet of paper = 30 cm × 18 cm
Let’s assume that V1 be the volume of the cylinder which is formed by rolling the sheet along its length.
So,
2πr1 = 30
r1 = 15/π
h1 = 18 cm
So, the volume is
V1 = π × 15/π × 15/π × 18
V1 = 225/π × 18 cm3
Let’s assume that V2 be the volume of the cylinder formed by rolling the sheet along its width.
So, 2πr2 = 18
r2 = 9/π
h2 = 30 cm
So, the volume is
V2 = π r22 h2
= π x (9/π)2 x 30
V2 = 81 x 30 / π
Now we find the ratio of volumes of two cylinders:
V1 / V2 = 225 x 18 / 81 x 30
V1 / V2 = 5/3
Hence, the ratio of the volumes of the two cylinders is 5 : 3.
问题28.如果管道中水的速度为30厘米/秒,在一分钟内有多少升水从横截面为5厘米2的管道中流出?
解决方案:
Given that,
Area of cross-section of the pipe = 5 cm2
Speed of water = 30 cm/sec
So, the volume of water that flows through the pipe in one second
= 5 × 30 = 150 cm3
Now we find the volume of water that flows through the pipe in one minute
= 150 × 60 = 9000 cm3 = 9 liter
Hence, the volume of water that flows through the given pipe in 1 minute is 9 liter.
问题29.底座的半径和实心圆柱体的高度之和为37 m。如果实心圆柱体的总表面积为1628 cm 2 。找到圆柱体的体积。
解决方案:
Given that,
The total surface area of the cylinder = 1628 cm2
The sum of the radius of the base and height of a solid cylinder = 37 m
h + r = 37 cm …(1)
Find: the volume of the cylinder.
Now we find the radius and height of the cylinder
So, total surface area of the cylinder = 2πr(h + r) = 1628
Therefore, 2πr × 37 = 1628
2 × 22/7 × r × 37 = 1628
r = 7 cm
Put the value of r in eq(1)
h = 30 cm
Now lets calculate volume of the cylinder
Volume = πr2h
Volume = 22/7 × 7 × 7 × 30 = 4620 cm3
Hence, the volume of the given cylinder is 4620 cm3.
问题30.寻找下沉直径为3 m的280 m深管井的成本,每立方米3.60卢比。还可以发现其内部曲面的粘合成本为每平方米2.50卢比。
解决方案:
Given that,
Height of the tube well = 280 m,
Diameter the tube well = 3 m,
So, the radius of tube well = 3/2 m
Rate of sinking of the tube well = Rs. 3.60/m3,
Rate of cementing = Rs. 2.50/m2k
Now we find the volume of the tube well
V = πr2h
= 22/7 × 3/2 × 3/2 × 280 = 1980 m2
Cost of Sinking the tube well = Volume of the tube well × Rate of sinking the tube well
= 1980 × 3.60 = Rs 7128
Now we find the curved surface area
CSA = 2πrh
= 2 × 22/7 × 3/2 × 280 = 2640 m2
Cost of cementing = Curved Surface area × Rate of cementing
= 2640 × 2.50 = Rs 6600
Hence, the total cost of sinking the tube well is Rs.7128 and
the total cost of cementing its inner surface is Rs. 6600.
问题31.当1立方厘米的铜重8.4克时,找到直径为4毫米的13.2千克铜线的长度。
解决方案:
Given that,
Weight of copper wire = 13.2 kg = 13.2 × 1000 gm = 13200 gm
Diameter of copper wire = 4 mm,
So, the radius of the wire = 2 mm = 210 cm
Density = 8.4 gm / cm3,
Find: the length of the copper wire.
So, Volume × Density = Weight
Therefore, πr2h × 8.4 = 13.2
22/7 × 2/10 × 2/10 × h × 8 .4
h = 12500 cm = 125 m
Hence, the length of the copper wire is 125 meter.
问题32.内径为10 m的一口井的深度为8.4 m。从中取出的土在其周围扩展至7.5 m的宽度,形成路堤。找到路堤的高度。
解决方案:
Given that,
Inner diameter of the well = 10 m,
So, the radius of well = 5 m
The height of the well = 8.4 m,
Width of embankment = 7.5 m,
Find: the height of the embankment.
So, the outer radius of the embankment,
R = Inner radius of the well + width of the embankment
= 5 + 7.5 = 12.5 m
Let’s assume that H be the height of the embankment,
Volume of embankment = Volume of earth dug out
π(R2 − r2)H = πr2h
22/7 × (12.52 − 52)H = 22/7 × 5 × 5 × 8.4
H = 1.6 m
Hence, height of the embankment is 1.6 m.