第 10 类 NCERT 解决方案 - 第 13 章表面积和体积 - 练习 13.3

问题 1. 将一个半径为 4.2 厘米的金属球熔化并重铸成半径为 6 厘米的圆柱体。求圆柱体的高度。

解决方案：

Radius of sphere (R)= 4.2 cm

Radius of cylinder (r)= 6 cm

In recasting process the volume will be same, so

Volume of cylinder = Volume of sphere

πr²h = $\frac{4}{3}$πR³

π(6)²h = $\frac{4}{3}$π(4.2)³                                                 (cancel π from both side)

36h = $\frac{4}{3}$(4.2 × 4.2 × 4.2)

h = $\frac{4 \times 4.2 \times 4.2 \times 4.2} {3 \times 36}$ (cm)

h = 1.4 × 1.4 × 1.4 (cm)

h = 2.74 cm

问题 2. 半径分别为 6 cm、8 cm 和 10 cm 的金属球被熔化形成一个单一的固体球。找到生成的球体的半径。

解决方案：

Radius of sphere 1 (r₁)= 6 cm

Radius of sphere 2 (r₂)= 8 cm

Radius of sphere 3 (r₃)= 10 cm

Let Radius of resulting sphere = R

In recasting process the volume will be same, so

Volume of Resulting sphere = Volume of sphere 1 + Volume of sphere 2 + Volume of sphere 3

$\frac{4}{3}$π(R)³ = $\frac{4}{3}$π(r₁)³ + $\frac{4}{3}$π(r₂)³ +$\frac{4}{3}$π(r₃)³                           (cancel $\frac{4}{3}$π from both side)

R³ = (r1)³ + (r2)³ + (r3)³

R³ = (6)³ + (8)³ + (10)³

R³ = 216 + 512 + 1000

R³ = 1728

R = (1728) ^1/3  

R = 12 cm

问题 3. 挖一口 20m 深、直径 7m 的井，将挖出的土均匀铺开，形成一个 22m×14m 的平台。求平台的高度。

解决方案：

So basically here, the digging of cylindrical shape is changed to cuboidal shape

Given values, 

Diameter of cylinder = 7 m

Radius of cylinder (r)= $\frac{7}{2}$m

Height of cylinder (H)= 20 m

Length of Cuboid (l) = 22 m

Breadth of Cuboid (b) =14 m

Let Height of Cuboid = h

In this process the volume will be same, so

Volume of Cuboid = Volume of Cylinder 

l × b × h = πr²H

22 × 14 × h = π × $(\frac{7}{2})^2$ × 20

h = $\frac{\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times 20}{22 \times 14}$ m                                  (taking π =$\frac{22}{7}$)

h = $\frac{5}{2}$ m

h = 2.5 m

问题 4. 直径 3 m 的井被挖 14 m 深。从中取出的泥土均匀地散布在它周围，形成一个宽4m的圆环，形成一个堤坝。求路堤的高度。

解决方案：

So basically here, the digging of cylindrical shape is changed to another Hollow cylindrical shape

Given values,

Diameter of cylinder = 3 m

Radius of cylinder (r) = $\frac{3}{2}$m

Height of cylinder (h) = 14 m

Width of embankment = 4 m

Outer Radius of embankment R₁= radius of cylinder + width = 3/2 + 4 = $\frac{11}{2}$m

Inner Radius of embankment R₂= radius of cylinder = $\frac{3}{2}$m

Height of embankment = H 

In this process the volume will be same, so

Volume of Embankment = Volume of Cylinder  

(πR₁²H) – (πR₂²H) = πr²h

(R₁² – R₂²)H = ($\frac{3}{2}$)²× 14                                                               (cancel π from both side)

H = $\frac{(\frac{3}{2})^2 × 14}{(\frac{11}{2})^2 - (\frac{3}{2})^2}$

H = $\frac{\frac{63}{2}}{(\frac{121}{4} - \frac{9}{4})}$

H = $\frac{\frac{63}{2}} {\frac{112}{4}}$

H = $\frac{\frac{63}{2}} {28}$

H = $\frac{63}{2×28}$

H = $\frac{9}{8}$ m

H = 1.125 m

问题 5. 一个直径 12 厘米，高 15 厘米的直圆柱容器装满冰淇淋。冰淇淋将被装入高 12 厘米、直径 6 厘米的圆锥体中，顶部呈半球形。找出可以装满冰淇淋的这种锥体的数量。

解决方案：

Given values,

Radius of cylinder (r) = 6 cm

Height of cylinder (h) = 15 cm

Radius of each cone (R) = 3 cm

Height of each cone (H) = 12 cm

Let n be the total number of ice creams

In this process the volume will be same, so

n × (Volume of each Cone + Volume of each hemisphere) = Volume of Cylinder  

n × ($\frac{1}{3}$πR²H + $\frac{2}{3}$πR³) = πr²h

n = $\frac{π(6^2×15)}{π(\frac{1}{3}3^2 ×12 + \frac{2}{3}3^3)}$                                      (cancel π from both side)

n = $\frac{36 \times 15}{\frac{1\times 9 \times12 + 2 \times 27}{3}}$

n = $\frac{36 \times15}{\frac{162}{3}}$

n = $\frac{540}{54}$

n = 10

Hence, 10 numbers of cones filled with ice-cream.

问题 6. 有多少银币，直径 1.75 厘米，厚 2 毫米，必须熔化成一个 5.5 厘米 × 10 厘米 × 3.5 厘米的长方体？

解决方案：

Given values,

Radius of cylindrical coin (r) = $\frac{1.75}{2} cm = \frac{175}{200} cm$

Height of cylindrical coin (H) = 2 mm = $\frac{2}{10}$ cm

Length of Cuboid (l) = 5.5 cm

Breadth of Cuboid (b) =10 cm

Height of Cuboid (h)= 3.5 cm 

Let n be the total number of coins

In this process the volume will be same, so

n × (Volume of each Coin) = Volume of Cylinder  

n × (πR²H) = l × b × h

n × π × ($\frac{175}{200}$)²× $\frac{2}{10}$= 5.5 × 10 × 3.5

n = $\frac{\frac{55}{10} × 10 × \frac{35}{10}}{\frac{22}{7} × (\frac{7}{8})^2× \frac{2}{10}}$                   (taking π = $\frac{22}{7}$)

n = $\frac{55 \times 10 \times 35 \times 8 \times 8 \times 10 \times 7}{22 \times 7 \times 7 \times 2 \times10 \times 10}$

n = 400

Hence, 400 silver coins must be melted to form this cuboid.

问题 7. 一个圆柱形桶，高 32 厘米，底部半径 18 厘米，装满沙子。这个桶倒在地上，就形成了一个圆锥形的沙堆。如果圆锥堆的高度是 24 厘米，求堆的半径和斜高。

解决方案：

So basically here, the cylindrical shape is changed to conical shape

Given values,

Radius of cylinder (r) = 18 cm

Height of cylinder (h) = 32 cm

Height of cone (H) = 24 cm

Let Radius of cone = R

In this process the volume will be same, so

Volume of Cone = Volume of Cylinder  

$\frac{1}{3}$πR²H = πr²h

$\frac{1}{3}$R² × (24) = 18² × 32                                           (cancel π from both side)

R² = $\frac{18 \times 18 \times 32 \times 3}{24}$

R = √(18 × 18 × 4)

R = 36 cm

Slant height (l) = √H² + R²

l = √(24² + 36²)

l = √(12×2)² + (12×3)²

l =12 √(4+9)

l = 12 √13 cm

问题 8. 一条宽 6 m、深 1.5 m 的运河中的水以 10 km/h 的速度流动。如果需要 8 厘米的积水，它会在 30 分钟内灌溉多少面积？

解决方案：

Given values,

Width of canal (w) = 6 m

Depth of canal (h) = 1.5 m

Speed of canal = 10 km/hr = (10,000 m/hr)

For 1hr (60 mins), we can take length (l) as = 10,000 m and 

Volume in 1hr = (l × w × h)

 = (10,000 × 6 × 1.5) m³

= 90,000 m³

So for 30 mins volume will be = $\frac{90000}{30}$ m³

= 45,000 m³

Area irrigated in 30 minutes will be as: 

Area × length of standing = Volume of canal in 30mins

Area = $\frac{45000}{\frac{8}{100}}$

Area = 562500 m²

问题 9. 一位农民将一根内径为 20 cm 的管道从一条运河连接到她的田地中的一个圆柱形水箱中，该水箱直径为 10 m，深 2 m。如果水以 3 公里/小时的速度流过管道，水箱需要多长时间才能充满？

解决方案：

Given values,

Radius of Pipe (R) = 10 cm = $\frac{10}{100}$ m

Radius of Cylindrical tank (r) = 5 m

Depth of Cylindrical tank (h) = 2 m

Speed of water flows in pipe = 3 km/hr = (3,000 m/hr)

Let Time to fill cylindrical tank = t

For 1hr (60 mins), we can take height of Pipe (H) as = 3,000 m and

Volume in 1hr = (πR²H)

= (π × ($\frac{1}{10}$)² × 3,000) m³

= 30π m³

In this process the volume will be same, so

t (in hr) × (Volume of Pipe in 1hr) = Volume of Cylindrical Tank

t × (30π) = πr²h

t × (30π) = π × 5² × 2

t = $\frac{5 \times 5 \times 2}{30}$

t = $\frac{5}{3}$hr

t = $\frac{5}{3}$× 60 mins

t = 100 minutes

Hence, 100 minutes will be taken to fill the tank.