Mensuration — Chapter Test

A cylindrical container is to be made of tin sheet. The height of the container is 1 m and its diameter is 70 cm. If the container is open at the top and the tin sheet costs ₹ 300 per m², find the cost of the tin for making the container.

Answer

Since, container is open at the top hence total surface area of container (S) = curved surface area + area of base.

Radius of container = $\dfrac{70}{2}$ = 35 cm = 0.35 m.

$\Rightarrow S = 2πrh + πr^2 \\[1em] = 2 \times \dfrac{22}{7} \times 0.35 \times 1 + \dfrac{22}{7} \times 0.35 \times 0.35 \\[1em] = 2 \times 22 \times 0.05 + 22 \times 0.05 \times 0.35 \\[1em] = 2.2 + 0.385 \\[1em] = 2.585 \text{ m}^2$

Cost of 1 m² sheet = ₹ 300.

Total cost = ₹ 300 × 2.585 = ₹ 775.50

Hence, the cost of tin for making the container is ₹ 775.50

A cylinder of maximum volume is cut out from a wooden cuboid of length 30 cm and cross section of square of side 14 cm. Find the volume of the cylinder and the volume of wood wasted.

Answer

Largest size of cylinder cut out of the wooden cuboid will be of diameter = 14 cm, radius = $\dfrac{14}{2}$ = 7 cm and height = 30 cm.

Volume of cylinder = πr²h = $\dfrac{22}{7} \times 7 \times 7 \times 30$ = 22 × 7 × 30 = 4620 cm³.

Volume of cuboid = lbh = 30 × 14 × 14 = 5880 cm³.

Volume of wooden wasted = Volume of cuboid - Volume of cylinder = 5880 - 4620 = 1260 cm³.

Hence, the volume of cylinder = 4620 cm³ and volume of wooden wasted = 1260 cm³.

Find the volume and the total surface area of a cone having slant height 17 cm and base diameter 30 cm. Take π = 3.14

Answer

Radius of cone = $\dfrac{30}{2}$ = 15 cm.

l = $\sqrt{r^2 + h^2}$

Putting values we get,

$\Rightarrow 17 = \sqrt{15^2 + h^2} \\[1em] \Rightarrow 17^2 = 15^2 + h^2 \\[1em] \Rightarrow 289 = 225 + h^2 \\[1em] \Rightarrow h^2 = 289 - 225 \\[1em] \Rightarrow h^2 = 64 \\[1em] \Rightarrow h = \sqrt{64} = 8 \text{ cm}.$

Volume of cone (V) = $\dfrac{1}{3}πr^2h$

Putting values we get,

$V = \dfrac{1}{3} \times 3.14 \times 15^2 \times 8 \\[1em] = \dfrac{3.14 \times 225 \times 8}{3} \\[1em] = 1884 \text{ cm}^3$

Total surface area of cone (S) = $πrl + πr^2 = πr(l + r)$

Putting values we get,

$S = 3.14 \times 15 \times (17 + 15) \\[1em] = 3.14 \times 15 \times 32 \\[1em] = 1507.2 \text{ cm}^2$

Hence, the volume of cone = 1884 cm³ and surface area of cone = 1507.2 cm².

Find the volume of a cone given that its height is 8 cm and the area of base 156 cm².

Answer

Height of cone = 8 cm.

Area of base = 156 cm².

Volume of cone (V) = $\dfrac{1}{3} \times$ Area of base × Height

Putting values we get,

$V = \dfrac{1}{3} \times 156 \times 8 \\[1em] = \dfrac{1248}{3} \\[1em] = 416 \text{ cm}^3.$

Hence, the volume of cone = 416 cm³.

The circumference of the edge of a hemispherical bowl is 132 cm. Find the capacity of the bowl.

Answer

Circumference of the edge of bowl = 132 cm.

$\therefore 2πr = 132 \\[1em] \Rightarrow 2 \times \dfrac{22}{7} \times r = 132 \\[1em] r = \dfrac{132 \times 7}{22 \times 2} \\[1em] r = 3 \times 7 \\[1em] r = 21 \text{ cm}.$

Volume of hemispherical bowl (V) = $\dfrac{2}{3}πr^3$

Putting values we get,

$V = \dfrac{2}{3} \times \dfrac{22}{7} \times (21)^3 \\[1em] = \dfrac{2 \times 22 \times 21^3}{21} \\[1em] = 2 \times 22 \times 441 \\[1em] = 19404 \text{ cm}^3.$

Hence, the capacity of bowl = 19404 cm³.

The volume of a hemisphere is $2425\dfrac{1}{2}$ cm³. Find its curved surface area.

Answer

Let radius of hemisphere be r cm.

Volume of hemisphere (V) = $\dfrac{2}{3}πr^3$

Given, V = $2425\dfrac{1}{2} = \dfrac{4851}{2}$.

$\therefore \dfrac{2}{3} \times \dfrac{22}{7} \times r^3 = \dfrac{4851}{2} \\[1em] \Rightarrow r^3 = \dfrac{4851 \times 3 \times 7}{2 \times 22 \times 2} \\[1em] \Rightarrow r^3 = \dfrac{441 \times 3 \times 7}{2 \times 2 \times 2} \\[1em] \Rightarrow r^3 = \dfrac{9261}{8} \\[1em] \Rightarrow r^3 = \Big(\dfrac{21}{2}\Big)^3 \\[1em] \Rightarrow r = \dfrac{21}{2} \text{ cm}.$

Curved surface area = 2πr²

$= 2 \times \dfrac{22}{7} \times \dfrac{21}{2} \times \dfrac{21}{2} \\[1em] = \dfrac{44 \times 441}{28} \\[1em] = 693 \text{ cm}^2.$

Hence, curved surface area of hemisphere = 693 cm².

A solid wooden toy is in the shape of a right circular cone mounted on a hemisphere. If the radius of the hemisphere is 4.2 cm and the total height of the toy is 10.2 cm, find the volume of the toy.

Answer

Given,

The solid wooden toy is in the shape of a right circular cone mounted on a hemisphere.

Radius of hemisphere (r) = 4.2 cm

Total height (h) = 10.2 cm.

Height of conical part (h₁) = 10.2 - 4.2 = 6 cm.

Volume of toy (V) = Volume of cone + Volume of hemisphere

= $\dfrac{1}{3}πr^2h + \dfrac{2}{3}πr^3$

Putting values we get,

$V = \dfrac{1}{3}πr^2\Big(h + 2r\Big) \\[1em] = \dfrac{1}{3} \times \dfrac{22}{7} \times (4.2)^2 \times (6 + 2(4.2)) \\[1em] = \dfrac{22 \times 17.64 \times 14.4}{21} \\[1em] = \dfrac{5588.352}{21} \\[1em] = 266.112 \text{ cm}^3.$

Hence, the volume of the toy = 266.112 cm³.

A solid is in the form of a cylinder with hemispherical ends. The total height of the solid is 19 cm and the diameter of the cylinder is 7 cm. Find the volume and the total surface area of the solid.

Answer

From figure,

r = Radius of cylinder = Radius of hemisphere = $\dfrac{7}{2}$.

Height of cylinder (h) = Total height - (2 x Radius of hemisphere)

= 19 - $\Big(2 \times \dfrac{7}{2}\Big)$

= 19 - 7 = 12 cm.

Total Volume of solid (V) = 2 × Volume of hemisphere + Volume of cylinder

$\therefore V = 2 \times \dfrac{2}{3}πr^3 + πr^2h \\[1em] = πr^2\Big(\dfrac{4r}{3} + h\Big) \\[1em] = \dfrac{22}{7} \times \dfrac{7}{2} \times \dfrac{7}{2} \Big(\dfrac{4}{3} \times \dfrac{7}{2} + 12\Big) \\[1em] = \dfrac{11 \times 7}{2} \times \Big(\dfrac{14}{3} + 12\Big) \\[1em] = \dfrac{77}{2} \times \Big(\dfrac{14}{3} + 12\Big) \\[1em] = \dfrac{77}{2} \times \dfrac{14}{3} + \dfrac{77}{2} \times 12 \\[1em] = \dfrac{539}{3} + 462 \\[1em] = \dfrac{539 + 1386}{3} \\[1em] = \dfrac{1925}{3} \\[1em] = 641\dfrac{2}{3} \text{ cm}^3.$

Surface area of solid (S) = 2 × Surface area of each hemisphere + Surface area of cylinder.

$\therefore S = 2 \times 2πr^2 + 2πrh \\[1em] = πr(4r + 2h) \\[1em] = \dfrac{22}{7} \times \dfrac{7}{2} \times \Big(4 \times \dfrac{7}{2} + 2 \times 12\Big) \\[1em] = 11 \times (14 + 24) \\[1em] = 11 \times 38 \\[1em] = 418 \text{ cm}^2$

Hence, the volume of the solid = $641\dfrac{2}{3}$ cm³ and surface area of solid = 418 cm².

The radius and height of a right circular cone are in the ratio 5 : 12. If its volume is 2512 cm³, find its slant height. (Take π = 3.14)

Answer

Let radius of cone (r) = 5a and height (h) = 12a.

Volume of cone (V) = $\dfrac{1}{3}πr^2h$

Given, V = 2512 cm³.

$\therefore \dfrac{1}{3}πr^2h = 2512 \\[1em] \Rightarrow \dfrac{1}{3} \times 3.14 \times (5a)^2 \times 12a = 2512 \\[1em] \Rightarrow \dfrac{3.14 \times 25a^2 \times 12a}{3} = 2512 \\[1em] \Rightarrow \dfrac{942a^3}{3} = 2512 \\[1em] \Rightarrow 314a^3 = 2512 \\[1em] \Rightarrow a^3 = \dfrac{2512}{314} \\[1em] \Rightarrow a^3 = 8 \\[1em] \Rightarrow a^3 = (2)^3 \\[1em] \Rightarrow a = 2 \text{ cm}.$

r = 5a = 5 × 2 = 10 cm.

h = 12a = 12 × 2 = 24 cm.

$\text{l} = \sqrt{r^2 + h^2} \\[1em] = \sqrt{10^2 + 24^2} \\[1em] = \sqrt{100 + 576} \\[1em] = \sqrt{676} \\[1em] = 26 \text{ cm}.$

Hence, the slant height of cone is 26 cm.

A cone and a cylinder are of the same height. If diameters of their bases are in the ratio 3 : 2, find the ratio of their volumes.

Answer

Let height of cone and cylinder = h.

Diameter of base of cone = 3a, (r₁) = $\dfrac{3a}{2}$.

Diameter of base of cylinder = 2a, (r₂) = $\dfrac{2a}{2}$.

Volume of cone (V₁) = $\dfrac{1}{3}πr_1^2h$.

Volume of cylinder (V₂) = $πr_2^2h$.

$\therefore \dfrac{V_1}{V_2} = \dfrac{\dfrac{1}{3}πr_1^2h}{πr_2^2h} \\[1em] = \dfrac{1}{3} \times \dfrac{r_1^2}{r_2^2} \\[1em] = \dfrac{1}{3} \times \dfrac{\Big(\dfrac{3a}{2}\Big)^2}{\Big(\dfrac{2a}{2}\Big)^2} \\[1em] = \dfrac{1}{3} \times \dfrac{9a^2 \times 2^2}{4a^2 \times 2^2} \\[1em] = \dfrac{1}{3} \times \dfrac{9}{4} \\[1em] = \dfrac{3}{4} \\[1em] = 3 : 4.$

A solid cone of base radius 9 cm and height 10 cm is lowered into a cylindrical jar of radius 10 cm, which contains water sufficient to submerge the cone completely. Find the rise in water level in the jar.

Answer

Radius of cone (r) = 9 cm.

Height of cone (h) = 10 cm.

Volume of water filled in cone = $\dfrac{1}{3}πr^2h$

Let h₁ be the rise in height of water in the jar.

Radius of jar (r₁) = 10.

Since, cone is submerged completely hence, volume of water level rise = volume of cone.

$\therefore πr_1^2h_1 = \dfrac{1}{3}πr^2h \\[1em] \Rightarrow π(10)^2 \times h_1 = \dfrac{1}{3}π(9)^2 \times 10 \\[1em] \Rightarrow h_1 = \dfrac{\dfrac{1}{3}π \times 81 \times 10}{π \times (10)^2} \\[1em] \Rightarrow h_1 = \dfrac{\cancel{π} \times \overset{27}{\cancel{81}} \times \cancel{10}}{\cancel{3} \times \cancel{π} \times \cancel{10} \times 10} \\[1em] \Rightarrow h_1 = \dfrac{27}{10} \\[1em] \Rightarrow h_1 = 2.7 \text{ cm}.$

Hence, the rise in height of water in jar is 2.7 cm.

An iron pillar has some part in the form of a right circular cylinder and the remaining in the form of a right circular cone. The radius of the base of each of cone and cylinder is 8 cm. The cylindrical part is 240 cm high and the conical part is 36 cm high. Find the weight of the pillar if one cu. cm of iron weighs 7.8 grams.

Answer

Radius of base of cone (r) = 8 cm,

Radius of cylinder (r) = 8 cm

Height of cylindrical part (h₁) = 240 cm

Height of conical part (h₂) = 36 cm.

Volume of pillar (V) = Volume of cylinder + Volume of cone.

$V = πr^2h_1 + \dfrac{1}{3}πr^2h_2 \\[1em] = πr^2(h_1 + \dfrac{h_2}{3}) \\[1em] = \dfrac{22}{7} \times (8)^2 \times \Big(240 + \dfrac{36}{3}\Big) \\[1em] = \dfrac{22}{7} \times 64 \times \Big(240 + 12\Big) \\[1em] = \dfrac{22}{7} \times 64 \times 252 \\[1em] = 22 \times 64 \times 36 \\[1em] = 50688 \text{ cm}^3$

Weight of 1 cm³ of iron = 7.8 g

∴ Weight of 50688 cm³ of iron = 50688 × 7.8 = 395366.4 g

Converting it into Kg,

395366.4 g = $\dfrac{395366.4}{1000}$ Kg

= 395.3664 Kg

Hence, the weight of the pillar is 395.3664 kg.

A circus tent is made of canvas and is in the form of right circular cylinder and a right circular cone above it. The diameter and height of cylindrical part of tent are 126 m and 5 m respectively. The total height of the tent is 21 m. Find the total cost of tent if canvas used costs ₹36 per square metre.

Answer

The below figure shows the circus tent:

Diameter of cylindrical part = 126 m.

Radius of cone = Radius of cylinder = r = $\dfrac{126}{2}$ = 63 m.

Height of cylindrical part (h₁) = 5 m,

Total height of tent = 21 m,

Height of conical part (h) = 21 - 5 = 16 m.

Slant height of cone (l) = $\sqrt{r^2 + h^2}$

$= \sqrt{63^2 + 16^2} \\[1em] = \sqrt{3969 + 256} \\[1em] = \sqrt{4225} \\[1em] = 65 \text{ m}.$

Surface area of tent (S) = Surface area of cylinder + Surface area of cone.

$S = 2πrh_1 + πrl \\[1em] = πr(2h_1 + l) \\[1em] = \dfrac{22}{7} \times 63 \times (2 \times 5 + 65) \\[1em] = 22 \times 9 \times (10 + 65) \\[1em] = 198 \times 75 \\[1em] = 14850 \text{ m}^2.$

Cost of one 1 m² cloth = ₹ 36.

Total cost = 14850 × 36 = ₹ 534600.

Hence, the total cost of tent is ₹ 534600.

The entire surface of a solid cone of base radius 3 cm and height 4 cm is equal to entire surface of a solid right circular cylinder of diameter 4 cm. Find the ratio of their

(i) curved surfaces

(ii) volumes.

Answer

Radius of base of cone (r₁) = 3 cm,

Height of cone (h₁) = 4 cm.

Slant height of cone (l) = $\sqrt{r_1^2 + h_1^2}$

$\sqrt{3^2 + 4^2} \\[1em] = \sqrt{9 + 16} \\[1em] = \sqrt{25} \\[1em] = 5 \text{ cm}.$

Let height of cylinder be h₂ cm and radius be r₂ cm.

r₂ = $\dfrac{4}{2}$ = 2 cm.

Given, total surface area of cylinder = total surface area of cone.

⇒ 2πr(r₂ + h₂) = πr₁(l + r₁)

⇒ 2π × 2 × (2 + h₂) = π × 3 × (5 + 3)

⇒ π(8 + 4h₂) = 24π

Dividing both sides by π,

⇒ 4h₂ = 24 - 8

⇒ 4h₂ = 16

⇒ h₂ = 4 cm.

(i) Ratio between curved surface area of cone and cylinder (Ratio) = $\dfrac{πr_1l}{2πr_2h}$

Putting values we get,

$\dfrac{\text{Curved Surface of Cone}}{\text{Curved Surface of Cylinder}} = \dfrac{π \times 3 \times 5}{2 \times π \times 2 \times 4} \\[1em] = \dfrac{15π}{16π} \\[1em] = 15 : 16.$

Hence, the ratio between curved surface area of cone and cylinder 15 : 16.

(ii) Ratio between their volumes = $\dfrac{\text{Vol. of Cone}}{\text{Vol. of Cylinder}}$

$= \dfrac{\dfrac{1}{3}πr_1^2h_1}{πr_2^2h_2} \\[1em] = \dfrac{r_1^2h_1}{3r_2^2h_2} \\[1em] = \dfrac{3^2 \times 4}{3 \times 2^2 \times 4} \\[1em] = \dfrac{3}{2^2} \\[1em] = \dfrac{3}{4} \\[1em] = 3 : 4.$

Hence, the ratio between volumes of cones and cylinder is 3 : 4.

A cone is 8.4 cm high and the radius of its base is 2.1 cm. It is melted and recast into a sphere. Find the radius of the sphere.

Answer

Radius of base of cone (r) = 2.1 cm and height (h) = 8.4 cm.

Let radius of sphere be R cm.

Since, cone is recasted into sphere hence, their volume will be equal.

Volume of cone = Volume of sphere.

$\therefore \dfrac{1}{3}πr^2h = \dfrac{4}{3}πR^3$

Multiplying both sides by $\dfrac{3}{π}$

$\Rightarrow r^2h = 4R^3 \\[1em] \Rightarrow R^3 = \dfrac{r^2h}{4} \\[1em] \Rightarrow R^3 = \dfrac{(2.1)^2 \times 8.4}{4} \\[1em] \Rightarrow R^3 = (2.1)^2 \times (2.1) \\[1em] \Rightarrow R^3 = (2.1)^3 \\[1em] \Rightarrow R = 2.1 \text{ cm}.$

Hence, the radius of the sphere = 2.1 cm.

Find the least number of coins of diameter 2.5 cm and height 3 mm which are to be melted to form a solid cylinder of radius 3 cm and height 5 cm.

Answer

Given,

Radius of coin (r) = $\dfrac{2.5}{2}$= 1.25 cm,

Height of coin (h) = 3 mm = 0.3 cm,

Radius of cylinder (R) = 3 cm.

Height of cylinder (H) = 5 cm

Let no. of coins required to be melted to form cylinder be n.

Volume of cylinder = n × Volume of each coin.

$\therefore πR^2H = n \times πr^2h \\[1em] n = \dfrac{πR^2H}{πr^2h} \\[1em] n = \dfrac{3^2 \times 5}{1.25^2 \times 0.3} \\[1em] n = \dfrac{45}{0.46875} \\[1em] n = 96.$

Hence, 96 coins are required to form a solid cylinder.

A hemisphere of lead of radius 8 cm is cast into a right circular cone of base radius 6 cm. Determine the height of the cone correct to 2 places of decimal.

Answer

Given,

Radius of hemisphere (r) = 8 cm.

Radius of cone (R) = 6 cm.

Let height of cone be h cm.

Since, hemisphere is casted into cone,

Volume of hemisphere = Volume of cone.

$\therefore \dfrac{2}{3}πr^3 = \dfrac{1}{3}πR^2h \\[1em] \Rightarrow 2r^3 = R^2h \\[1em] \Rightarrow h = \dfrac{2r^3}{R^2} \\[1em] \Rightarrow h = \dfrac{2 \times 8^3}{6^2} \\[1em] \Rightarrow h = \dfrac{1024}{36} \\[1em] \Rightarrow h = 28.44 \text{ cm}.$

Hence, the height of the cone is 28.44 cm.

A vessel in the form of a hemispherical bowl is full of water. The contents are emptied into a cylinder. The internal radii of the bowl and cylinder are respectively 6 cm and 4 cm. Find the height of water in the cylinder.

Answer

Radius of hemispherical bowl (r) = 6 cm.

Radius of cylinder (R) = 4 cm.

Let h be the height of water in cylinder.

Volume of hemispherical bowl = Volume of water in cylinder.

$\Rightarrow \dfrac{2}{3}πr^3 = πR^2h \\[1em] \Rightarrow \dfrac{2}{3}r^3 = R^2h \\[1em] \Rightarrow \dfrac{2}{3} \times 6^3 = 4^2 \times h \\[1em] \Rightarrow 2 \times 2 \times 36 = 16h \\[1em] \Rightarrow h = \dfrac{144}{16} \\[1em] \Rightarrow h = 9 \text{ cm}.$

Hence, the height of water in cylinder is 9 cm.

A sphere of diameter 6 cm is dropped into a right circular cylinder vessel partly filled with water. The diameter of the cylindrical vessel is 12 cm. If the sphere is completely submerged in water, by how much will the level of water rise in the cylindrical vessel ?

Answer

Radius of sphere (r) = $\dfrac{6}{2}$ = 3 cm.

Radius of cylinder (R) = $\dfrac{12}{2}$ = 6 cm.

Let height of water raised be h cm.

Volume of sphere = Volume of water rise in cylinder

$\dfrac{4}{3}πr^3 = πR^2h \\[1em] \dfrac{4}{3} \times 3^3 = 6^2 \times h \\[1em] 4 \times 3^2 = 6^2 \times h \\[1em] h = \dfrac{36}{36} = 1 \text{ cm}.$

Hence, the height of water raised is 1 cm.

A solid sphere of radius 6 cm is melted into a hollow cylinder of uniform thickness. If the external radius of the base of the cylinder is 5 cm and its height is 32 cm, find the uniform thickness of the cylinder.

Answer

Radius of solid sphere (r₁) = 6 cm

Volume of solid sphere (V) = $\dfrac{4}{3}πr_1^3$

$= \dfrac{4}{3}π \times 6^3 \\[1em] = 4π \times 2 \times 6^2 \\[1em] = 288π \text{ cm}^3$

External radius of cylinder (R) = 5 cm, height (h) = 32 cm.

Let r be inner radius of cylinder.

Volume of cylinder = Volume of sphere.

$\therefore V = π(R^2 - r^2)h \\[1em] \Rightarrow 288π = π(5^2 - r^2) \times 32 \\[1em] \Rightarrow 288 = 32(25 - r^2) \\[1em] \Rightarrow \dfrac{288}{32} = 25 - r^2 \\[1em] \Rightarrow 9 = 25 - r^2 \\[1em] \Rightarrow r^2 = 25 - 9 \\[1em] \Rightarrow r^2 = 16 \\[1em] \Rightarrow r = 4 \text{ cm}.$

Thickness of hollow cylinder = R - r = 5 - 4 = 1 cm.

Hence, the thickness of the cylinder = 1 cm.

In the adjoining diagram, a tilted right circular cylindrical vessel with base diameter 7 cm contains a liquid. When placed vertically, the height of the liquid in the vessel is the mean of two heights shown in the diagram. Find the area of wet surface, when the cylinder is placed vertically on a horizontal surface. (Use $\pi = \dfrac{22}{7}$ )

Answer

When vertically placed,

Height of liquid (h) = $\dfrac{1 + 6}{2} = \dfrac{7}{2}$ cm

Diameter of base = 7 cm

Radius (r) = $\dfrac{7}{2}$ cm

$\text{Area of wet surface} = πr^2 + 2πrh \\[1em] = πr(r + 2h) \\[1em] = \dfrac{22}{7} \times \dfrac{7}{2}\Big(\dfrac{7}{2} + 2\times \dfrac{7}{2}\Big) \\[1em] = 11 \times (3.5 + 7) \\[1em] = 11 \times 10.5 \\[1em] = 115.5 \text{ cm}^2.$

Hence, area of wet surface = 115.5 cm².

A manufacturing company prepares spherical ball bearings, each of radius 7 mm and mass 4 g. These ball bearings are packed into boxes. Each box can have maximum of 2156 cm³ of ball bearings. Find the :

(a) maximum number of ball bearings that each box can have.

(b) mass of each box of ball bearings in kg.

$\Big(\text{use } \pi = \dfrac{22}{7}\Big)$

Answer

(a) Given,

Radius of ball bearings = 7 mm

Volume of box = 2156 cm³ = 2156 × 10³ mm³

Number of ball bearings that each box can have (N)

= $\dfrac{\text{Volume of box}}{\text{Volume of each ball}}$

Substituting values we get :

$N = \dfrac{2156 \times 10^3}{\dfrac{4}{3} \times \dfrac{22}{7} \times 7^3} \\[1em] = \dfrac{2156 \times 10^3}{\dfrac{88}{3} \times 7^2} \\[1em] = \dfrac{2156 \times 3 \times 1000}{88 \times 49} \\[1em] = \dfrac{6468000}{4312} \\[1em] = 1500.$

Hence, maximum no. of ball bearings in a box = 1500.

(b) Mass of each box = No. of balls × Mass of each ball

= 1500 × 4 g

= 6000 g

= $\dfrac{6000}{1000}$ = 6 kg.

Hence, mass of each box = 6 kg.