Cylinder, Cone & Sphere (Surface Area & Volume) — Exercise 20(E)

In the given figure, the radius of the cone is same as its height equal to 2 cm each. The volume of the whole body is :

1. 8π cm³

2. 16 cm³

3. 32π cm³

4. 32 cm³

Answer

Given,

Radius of cone (r) = 2 cm

Height of cone (h) = 2 cm

From figure,

Radius of hemisphere = Radius of cone = 2 cm.

⇒ Volume of body = Volume of cone + Volume of hemisphere

⇒ Volume of body = $\dfrac{1}{3}πr^2h + \dfrac{2}{3}πr^3$

$= \dfrac{πr^2}{3}(h + 2r) \\[1em] = \dfrac{π \times 2^2}{3} \times (2 + 2 × 2) \\[1em] = \dfrac{4π}{3} \times (2 + 4) \\[1em] = \dfrac{4π}{3} \times 6 \\[1em] = \dfrac{24π}{3} \\[1em] = 8π \text{ cm}^3.$

Hence, Option 1 is the correct option.

In the given figure, height of the conical part is 3 cm. The radius and the height of the cylindrical part are 1 cm each. The total volume of the body is :

1. 9π cm³

2. 18π cm³

3. 2π cm³

4. 4π cm³

Answer

Given,

Radius of cylindrical part (r) = 1 cm

Height of cylindrical part (h) = 1 cm

Height of conical part (H) = 3 cm

From figure,

Radius of conical part = Radius of cylindrical part (r) = 1 cm

Volume of body = Volume of conical part + Volume of cylindrical part

$= \dfrac{1}{3}πr^2H + πr^2h \\[1em] = \dfrac{1}{3}π \times 1^2 \times 3 + π \times 1^2 \times 1 \\[1em] = π + π \\[1em] = 2π \text{ cm}^3.$

Hence, Option 3 is the correct option.

The given figure shows a solid cylinder of height h cm and radius r cm. If the slant height of the conical cavity is l cm, the total surface area of the remaining solid is :

1. πrl + 2πrh + 2πr²

2. πr² + πrl + 2πrh

3. 2πr² - πrl + 2πrh

4. (πrl + πr²) + 2πr² + 2πrh

Answer

From figure,

Total surface area of remaining solid = Curved surface area of cylinder + Area of upper surface of cylinder + Curved surface area of cone

= 2πrh + πr² + πrl.

Hence, Option 2 is the correct option.

In the given figure, a solid cone is kept inverted in a closed cylindrical container such that the height of cone = height of cylinder = h, radius of cone = radius of cylinder = r and slant height of cone = l. If the remaining of cylinder is completely filled with water, the wetted surface area of the whole body is :

1. 2πrh + πr² + πrl

2. 2πrh + πrl

3. 2πrh + 2πr² + πrl

4. 2πrh - πr² + πrl

Answer

From figure,

Total wetted surface area of body = Curved surface area of cylinder + Area of upper surface of cylinder + Curved surface area of cone

= 2πrh + πr² + πrl.

Hence, Option 1 is the correct option.

The given figure shows a solid sphere and a closed cylindrical container, both having the same height and same radii. The volume of air left in the cylinder is :

1. 6πr³

2. $\dfrac{3}{2}πr^3$

3. $\dfrac{2}{3}πr^3$

4. 4πr³

Answer

From figure,

Radius of sphere = Radius of cylinder = r

Height of cylinder (h) = 2r

Volume of air left = Volume of cylinder - Volume of sphere

$= πr^2h - \dfrac{4}{3}πr^3 \\[1em] = πr^2.2r - \dfrac{4}{3}πr^3 \\[1em] = 2πr^3 - \dfrac{4}{3}πr^3 \\[1em] = \dfrac{6πr^3 - 4πr^3}{3} \\[1em] = \dfrac{2πr^3}{3} \\[1em] = \dfrac{2}{3}πr^3.$

Hence, Option 3 is the correct option.

A cone of height 15 cm and diameter 7 cm is mounted on a hemisphere of same diameter. Determine the volume of the solid thus formed.

Answer

Given,

Height of the cone (h) = 15 cm

Diameter of the cone = 7 cm

Radius of cone (r) = $\dfrac{7}{2}$ = 3.5 cm

So, radius of the hemisphere (R) = 3.5 cm

From figure,

Volume of the solid (V) = Volume of the cone + Volume of the hemisphere

$V = \dfrac{1}{3}πr^2h + \dfrac{2}{3}πR^3 \\[1em] = \dfrac{1}{3} \times \dfrac{22}{7} \times (3.5)^2 \times 15 + \dfrac{2}{3} \times \dfrac{22}{7} \times (3.5)^3 \\[1em] = \dfrac{1}{3} \times \dfrac{22}{7} \times 12.25 \times 15 + \dfrac{2}{3} \times \dfrac{22}{7} \times 42.875 \\[1em] = 22 \times 1.75 \times 5 + \dfrac{1886.5}{21} \\[1em] = 192.5 + 89.833 \\[1em] = 282.33\text{ cm}^3.$

Hence, volume of solid formed = 282.33 cm³.

A buoy is made in the form of a hemisphere surmounted by a right cone whose circular base coincides with the plane surface of the hemisphere. The radius of the base of the cone is 3.5 metres and its volume is two-third of the hemisphere. Calculate the height of the cone and the surface area of the buoy, correct to two places of decimal.

Answer

From figure,

Radius of hemisphere (r) = Radius of cone (R) = 3.5 m.

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

= $\dfrac{2}{3} \times \dfrac{22}{7} \times (3.5)^3$

= $\dfrac{2}{3} \times 22 \times 0.5 \times 3.5 \times 3.5$

= $\dfrac{269.5}{3}$ m³.

Let height of the cone = h.

By formula,

Volume of conical part = $\dfrac{1}{3}πR^2h$ ........(1)

Given,

Volume of cone = $\dfrac{2}{3}$ Volume of hemisphere

= $\dfrac{2}{3} \times \dfrac{269.5}{3} = \dfrac{539}{9}$ m³ ...........(2)

From (1) and (2), we get :

$\Rightarrow \dfrac{1}{3}πR^2h = \dfrac{539}{9} \\[1em] \Rightarrow \dfrac{1}{3} \times \dfrac{22}{7} \times (3.5)^2 \times h = \dfrac{539}{9} \\[1em] \Rightarrow h = \dfrac{539 \times 7 \times 3}{9 \times (3.5)^2 \times 22} \\[1em] \Rightarrow h = \dfrac{11319}{2425.5} \\[1em] \Rightarrow h = 4.67 \text{ m}.$

By formula,

⇒ l² = R² + h²

⇒ l² = (3.5)² + (4.67)²

⇒ l² = 12.25 + 21.81

⇒ l² = 34.06

⇒ l = $\sqrt{34.06}$

⇒ l = 5.83 cm.

Surface area of buoy = Surface area of hemisphere + Surface area of cone = 2πr² + πRl

$= 2 \times \dfrac{22}{7} \times (3.5)^2 + \dfrac{22}{7} \times 3.5 \times 5.83 \\[1em] = 2 \times 22 \times 0.5 \times 3.5 + 22 \times 0.5 \times 5.83 \\[1em] = 77 + 64.13 \\[1em] = 141.13 \text{ m}^2.$

Hence, height = 4.67 and surface area of buoy = 141.13 m².

From a rectangular solid of metal 42 cm by 30 cm by 20 cm, a conical cavity of diameter 14 cm and depth 24 cm is drilled out. Find:

(i) the surface area of the remaining solid

(ii) the volume of remaining solid

(iii) the weight of the material drilled out if it weighs 7 gm per cm³.

Answer

Given,

Dimensions of rectangular solid are:

l = 42 cm, b = 30 cm and h = 20 cm.

Conical cavity’s diameter = 14 cm

So, its radius (r) = 7 cm

Depth (h) = 24 cm

(i) Total surface area of cuboid = 2(lb + bh + lh)

= 2 (42 x 30 + 30 x 20 + 20 x 42)

= 2 (1260 + 600 + 840)

= 2 (2700)

= 5400 cm²

Area of circular base of conical cavity = πr² = $\dfrac{22}{7}$ x 7 x 7 = 154 cm².

By formula,

⇒ l² = r² + h²

⇒ l² = (7)² + (24)²

⇒ l² = 49 + 576

⇒ l² = 625

⇒ l = $\sqrt{625}$

⇒ l = 25 cm.

Area of curved surface area of cone = πrl

= $\dfrac{22}{7} \times 7 \times 25$ = 22 x 25 = 550 cm²

Surface area of remaining part = Surface area of rectangular solid + Surface area of cone - Area of base of conical cavity

= 5400 + 550 - 154

= 5796 cm².

Hence, surface area of remaining part = 5796 cm².

(ii) Volume of the rectangular solid = lbh

= (42 x 30 x 20) cm³

= 25200 cm³

Radius of conical cavity (r) = 7 cm

Depth (h) = 24 cm

Volume of cone = $\dfrac{1}{3}πr^2h$

= $\dfrac{1}{3} \times \dfrac{22}{7} \times 7 \times 7 \times 24$

= 22 × 7 × 8

= 1232 cm³.

Volume of remaining solid = Volume of rectangular solid - Volume of cone

= 25200 - 1232

= 23968 cm³.

Hence, volume of remaining solid = 23968 cm³.

(iii) Volume of material drilled out = Volume of cone = 1232 cm³.

Weight of material drilled out = 1232 × 7 = 8624 g = 8.624 kg.

Hence, weight of material drilled out = 8.624 kg.

The cubical block of side 7 cm is surmounted by a hemisphere of the largest size. Find the surface area of the resulting solid.

Answer

It’s known that, the diameter of the largest hemisphere that can be placed on the face of a cube of side 7 cm will be 7 cm.

So, it’s radius (r) = $\dfrac{7}{2}$ = 3.5 cm

Curved surface area = 2πr²

= $2 \times \dfrac{22}{7} \times (3.5)^2$ = 77 cm².

Area of base of hemisphere = πr²

= $\dfrac{22}{7} \times (3.5)^2$

= 38.5 cm².

From figure,

Surface area of the resulting solid = Surface area of the cube – Area of the base of the hemisphere + Curved surface area of hemisphere

= 6(side)² - 38.5 + 77

= 6(7)² + 38.5

= 294 + 38.5

= 332.5 cm².

Hence, surface area of resulting solid = 332.5 cm³.

A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of the top which is open is 5 cm. It is filled with water up to the rim. When lead shots, each of which is a sphere of radius 0.5 cm, are dropped into the vessel, one-fourth of the water flows out. Find the number of lead shots dropped in the vessel.

Answer

Given,

Height of cone (H) = 8 cm

Radius of cone (R) = 5 cm

Radius of sphere (r) = 0.5 cm.

Let no. of lead shots dropped in vessel be n.

According to question,

One-fourth of the water flows out after dropping lead shots.

∴ $\dfrac{1}{4}$ x Volume of cone = n × Volume of each sphere

$\Rightarrow \dfrac{1}{4} \times \dfrac{1}{3}πR^2H = n \times \dfrac{4}{3}πr^3 \\[1em] \Rightarrow \dfrac{R^2H}{4} = n \times 4r^3 \\[1em] \Rightarrow n = \dfrac{R^2H}{16r^3} \\[1em] \Rightarrow n = \dfrac{5^2 \times 8}{16 \times (0.5)^3} \\[1em] \Rightarrow n = \dfrac{25 \times 8}{16 \times 0.125} \\[1em] \Rightarrow n = \dfrac{200}{2} \\[1em] \Rightarrow n = 100.$

Hence, no. of shots dropped = 100.

A hemi-spherical bowl has negligible thickness and the length of its circumference is 198 cm. Find the capacity of the bowl.

Answer

Given,

Circumference = 198 cm

∴ 2πr = 198

⇒ r = $\dfrac{198}{2π} = \dfrac{198}{2 \times \dfrac{22}{7}} = \dfrac{198 \times 7}{44} = \dfrac{1386}{44}$ = 31.5 cm.

By formula,

$\text{Capacity of bowl} = \dfrac{2}{3}πr^3 \\[1em] = \dfrac{2}{3} \times \dfrac{22}{7} \times (31.5)^3 \\[1em] = \dfrac{2 \times 22 \times 31255.875}{21} \\[1em] = \dfrac{1375258.5}{21} \\[1em] = 65488.5 \text{ cm}^3.$

Hence, capacity of bowl = 65488.5 cm³.

Find the maximum volume of a cone that can be carved out of a solid hemisphere of radius r cm.

Answer

The max height and radius of cone, inside a hemisphere of radius r cm can be r cm.

Volume of cone = $\dfrac{1}{3}πr^2h$

= $\dfrac{1}{3} \times π \times r^2 \times r$

= $\dfrac{1}{3}πr^3 \text{ cm}^3.$

Hence, maximum volume of cone that can be carved out of solid hemisphere of radius r cm = $\dfrac{1}{3}πr^3$ cm³.

The radii of the bases of two right circular cones of same height are r₁ and r₂ respectively. The cones are melted and recast into a solid sphere of radius R. Find the height of each cone in terms of r₁, r₂ and R.

Answer

Let height of each cone be h cm.

Given,

Cones are melted and recasted into a sphere.

∴ Volume of cone with radii r₁ + Volume of cone with radii r₂ = Volume of sphere

$\Rightarrow \dfrac{1}{3}πr_1^2h + \dfrac{1}{3}πr_2^2h = \dfrac{4}{3}πR^3 \\[1em] \Rightarrow \dfrac{πh}{3}(r_1^2 + r_2^2) = \dfrac{π}{3} \times 4R^3 \\[1em] \Rightarrow h(r_1^2 + r_2^2) = 4R^3 \\[1em] \Rightarrow h = \dfrac{4R^3}{r_1^2 + r_2^2}.$

Hence, h = $\dfrac{4R^3}{r_1^2 + r_2^2}.$

A cone and a hemisphere have the same base and the same height. Find the ratio between their volumes.

Answer

Let radius and height of cone and hemisphere be a cm.

Volume of cone = $\dfrac{1}{3}π(\text{radius})^2\text{height} = \dfrac{1}{3}πa^3.$

Volume of hemisphere = $\dfrac{2}{3}π(\text{radius})^3 = \dfrac{4}{3}πa^3.$

$\text{Ratio} = \dfrac{\text{Vol. of cone}}{\text{Vol. of hemisphere}} \\[1em] = \dfrac{\dfrac{1}{3}πa^3}{\dfrac{2}{3}πa^3} \\[1em] = \dfrac{1}{2} \\[1em] = 1 : 2.$

Hence, ratio between volumes = 1 : 2.