##### Class 10^{th} Mathematics Tamilnadu Board Solution

**Exercise 8.1**- A solid right circular cylinder has radius of 14 cm and height of 8 cm. Find its…
- The total surface area of a solid right circular cylinder is 660 sq.cm. If its…
- Curved surface area and circumference at the base of a solid right circular…
- A mansion has 12 right cylindrical pillars each having radius 50 cm and height…
- The total surface area of a solid right circular cylinder is 231 cm^2 . Its…
- The total surface area of a solid right circular cylinder is 1540 cm^2 . If the…
- The radii of two right circular cylinders are in the ratio of 3 : 2 and their…
- The external surface area of a hollow cylinder is 540 πsq.cm. Its internal…
- The external diameter of a cylindrical shaped iron pipe is 25 cm and its length…
- The radius and height of a right circular solid cone are 7 cm and 24 cm…
- If the vertical angle and the radius of a right circular cone are 60° and 15 cm…
- If the circumference of the base of a solid right circular cone is 236 cm and…
- A heap of paddy is in the form of a cone whose diameter is 4.2 m and height is…
- The central angle and radius of a sector of a circular disc are 180° and 21 cm…
- Radius and slant height of a solid right circular cone are in the ratio 3 : 5.…
- If the curved surface area of a solid sphere is 98.56 cm^2 , then find the…
- If the curved surface area of a solid hemisphere is 2772 sq.cm, then find its…
- Radii of two solid hemispheres are in the ratio 3 : 5. Find the ratio of their…
- Find the curved surface area and total surface area of a hollow hemisphere…
- The inner curved surface area of a hemispherical dome of a building needs to be…

**Exercise 8.2**- Find the volume of a solid cylinder whose radius is 14 cm and height 30 cm.…
- A patient in a hospital is given soup daily in a cylindrical bowl of diameter 7…
- The sum of the base radius and the height of a solid right circular solid…
- Volume of a solid cylinder is 62.37 cu.cm. Find the radius if its height is 4.5…
- The radii of two right circular cylinders are in the ratio 2 : 3. Find the ratio…
- The radius and height of a cylinder are in the ratio 5 : 7. If its volume is…
- A rectangular sheet of metal foil with dimension 66 cm x 12 cm is rolled to form…
- A lead pencil is in the shape of right circular cylinder. The pencil is 28 cm…
- Radius and slant height of a cone are 20 cm and 29 cm respectively. Find its…
- The circumference of the base of a 12 m high wooden solid cone is 44 m. Find…
- A vessel is in the form of a frustum of a cone. Its radius at one end and the…
- The perimeter of the ends of a frustum of a cone are 44 cm and 8.4πcm. If the…
- A right angled ABC with sides 5 cm, 12 cm and 13 cm is revolved about the fixed…
- The radius and height of a right circular cone are in the ratio 2 : 3. Find the…
- The volume of a cone with circular base is 216 π cu.cm. If the base radius is 9…
- Find the mass of 200 steel spherical ball bearings, each of which has radius…
- The outer and the inner radii of a hollow sphere are 12 cm and 10 cm. Find its…
- The volume of a solid hemisphere is 1152 πcu.cm. Find its curved surface area.…
- Find the volume of the largest right circular cone that can be cut out of a…
- The radius of a spherical balloon increases from 7 cm to 14 cm as air is being…

**Exercise 8.3**- A play-top is in the form of a hemisphere surmounted on a cone. The diameter of…
- A solid is in the shape of a cylinder surmounted on a hemisphere. If the…
- A capsule is in the shape of a cylinder with two hemispheres stuck to each of…
- A tent is in the shape of a right circular cylinder surmounted by a cone. The…
- Using clay, a student made a right circular cone of height 48 cm and base radius…
- The radius of a solid sphere is 24 cm. It is melted and drawn into a long wire…
- A right circular conical vessel whose internal radius is 5 cm and height is 24…
- A solid sphere of diameter 6 cm is dropped into a right circular cylindrical…
- Through a cylindrical pipe of internal radius 7 cm, water flows out at the rate…
- Water in a cylindrical tank of diameter 4 m and height 10 m is released through…
- A spherical solid material of radius 18 cm is melted and recast into three…
- A hollow cylindrical pipe is of length 40 cm. Its internal and external radii…
- An iron right circular cone of diameter 8 cm and height 12 cm is melted and…
- A right circular cylinder having diameter 12 cm and height 15 cm is full of ice…
- A container with a rectangular base of length 4.4 m and breadth 2 m is used to…
- A cylindrical bucket of height 32 cm and radius 18 cm is filled with sand. The…
- A cylindrical shaped well of depth 20 m and diameter 14 m is dug. The dug out…

**Exercise 8.4**- The curved surface area of a right circular cylinder of radius 1 cm and height 1…
- The total surface area of a solid right circular cylinder whose radius is half…
- Base area of a right circular cylinder is 80 cm^2 . If its height is 5 cm, then…
- If the total surface area a solid right circular cylinder is 200 πcm^2 and its…
- The curved surface area of a right circular cylinder whose radius is a units and…
- Radius and height of a right circular cone and that of a right circular cylinder…
- If the diameter and height of a right circular cone are 12 cm and 8 cm…
- If the circumference at the base of a right circular cone and the slant height…
- If the volume and the base area of a right circular cone are 48πcm^3 and 12πcm^2…
- If the height and the base area of a right circular cone are 5 cm and 48 sq. cm…
- The ratios of the respective heights and the respective radii of two cylinders…
- If the radius of a sphere is 2 cm, then the curved surface area of the sphere…
- The total surface area of a solid hemisphere of diameter 2 cm is equal toA. 12…
- If the volume of a sphere is 9/6 pi cu. cm, then its radius isA. 4/3 cm B. 3/4…
- The surface areas of two spheres are in the ratio of 9 : 25. Then their volumes…
- The total surface area of a solid hemisphere whose radius is a units, is equal…
- If the surface area of a sphere is 100π cm^2 , then its radius is equal toA. 25…
- If the surface area of a sphere is 36π cm^2 , then the volume of the sphere is…
- If the total surface area of a solid hemisphere is 12π cm^2 then its curved…
- If the radius of a sphere is half of the radius of another sphere, then their…
- Curved surface area of solid sphere is 24 cm^2 . If the sphere is divided into…
- Two right circular cones have equal radii. If their slant heights are in the…

**Exercise 8.1**

- A solid right circular cylinder has radius of 14 cm and height of 8 cm. Find its…
- The total surface area of a solid right circular cylinder is 660 sq.cm. If its…
- Curved surface area and circumference at the base of a solid right circular…
- A mansion has 12 right cylindrical pillars each having radius 50 cm and height…
- The total surface area of a solid right circular cylinder is 231 cm^2 . Its…
- The total surface area of a solid right circular cylinder is 1540 cm^2 . If the…
- The radii of two right circular cylinders are in the ratio of 3 : 2 and their…
- The external surface area of a hollow cylinder is 540 πsq.cm. Its internal…
- The external diameter of a cylindrical shaped iron pipe is 25 cm and its length…
- The radius and height of a right circular solid cone are 7 cm and 24 cm…
- If the vertical angle and the radius of a right circular cone are 60° and 15 cm…
- If the circumference of the base of a solid right circular cone is 236 cm and…
- A heap of paddy is in the form of a cone whose diameter is 4.2 m and height is…
- The central angle and radius of a sector of a circular disc are 180° and 21 cm…
- Radius and slant height of a solid right circular cone are in the ratio 3 : 5.…
- If the curved surface area of a solid sphere is 98.56 cm^2 , then find the…
- If the curved surface area of a solid hemisphere is 2772 sq.cm, then find its…
- Radii of two solid hemispheres are in the ratio 3 : 5. Find the ratio of their…
- Find the curved surface area and total surface area of a hollow hemisphere…
- The inner curved surface area of a hemispherical dome of a building needs to be…

**Exercise 8.2**

- Find the volume of a solid cylinder whose radius is 14 cm and height 30 cm.…
- A patient in a hospital is given soup daily in a cylindrical bowl of diameter 7…
- The sum of the base radius and the height of a solid right circular solid…
- Volume of a solid cylinder is 62.37 cu.cm. Find the radius if its height is 4.5…
- The radii of two right circular cylinders are in the ratio 2 : 3. Find the ratio…
- The radius and height of a cylinder are in the ratio 5 : 7. If its volume is…
- A rectangular sheet of metal foil with dimension 66 cm x 12 cm is rolled to form…
- A lead pencil is in the shape of right circular cylinder. The pencil is 28 cm…
- Radius and slant height of a cone are 20 cm and 29 cm respectively. Find its…
- The circumference of the base of a 12 m high wooden solid cone is 44 m. Find…
- A vessel is in the form of a frustum of a cone. Its radius at one end and the…
- The perimeter of the ends of a frustum of a cone are 44 cm and 8.4πcm. If the…
- A right angled ABC with sides 5 cm, 12 cm and 13 cm is revolved about the fixed…
- The radius and height of a right circular cone are in the ratio 2 : 3. Find the…
- The volume of a cone with circular base is 216 π cu.cm. If the base radius is 9…
- Find the mass of 200 steel spherical ball bearings, each of which has radius…
- The outer and the inner radii of a hollow sphere are 12 cm and 10 cm. Find its…
- The volume of a solid hemisphere is 1152 πcu.cm. Find its curved surface area.…
- Find the volume of the largest right circular cone that can be cut out of a…
- The radius of a spherical balloon increases from 7 cm to 14 cm as air is being…

**Exercise 8.3**

- A play-top is in the form of a hemisphere surmounted on a cone. The diameter of…
- A solid is in the shape of a cylinder surmounted on a hemisphere. If the…
- A capsule is in the shape of a cylinder with two hemispheres stuck to each of…
- A tent is in the shape of a right circular cylinder surmounted by a cone. The…
- Using clay, a student made a right circular cone of height 48 cm and base radius…
- The radius of a solid sphere is 24 cm. It is melted and drawn into a long wire…
- A right circular conical vessel whose internal radius is 5 cm and height is 24…
- A solid sphere of diameter 6 cm is dropped into a right circular cylindrical…
- Through a cylindrical pipe of internal radius 7 cm, water flows out at the rate…
- Water in a cylindrical tank of diameter 4 m and height 10 m is released through…
- A spherical solid material of radius 18 cm is melted and recast into three…
- A hollow cylindrical pipe is of length 40 cm. Its internal and external radii…
- An iron right circular cone of diameter 8 cm and height 12 cm is melted and…
- A right circular cylinder having diameter 12 cm and height 15 cm is full of ice…
- A container with a rectangular base of length 4.4 m and breadth 2 m is used to…
- A cylindrical bucket of height 32 cm and radius 18 cm is filled with sand. The…
- A cylindrical shaped well of depth 20 m and diameter 14 m is dug. The dug out…

**Exercise 8.4**

- The curved surface area of a right circular cylinder of radius 1 cm and height 1…
- The total surface area of a solid right circular cylinder whose radius is half…
- Base area of a right circular cylinder is 80 cm^2 . If its height is 5 cm, then…
- If the total surface area a solid right circular cylinder is 200 πcm^2 and its…
- The curved surface area of a right circular cylinder whose radius is a units and…
- Radius and height of a right circular cone and that of a right circular cylinder…
- If the diameter and height of a right circular cone are 12 cm and 8 cm…
- If the circumference at the base of a right circular cone and the slant height…
- If the volume and the base area of a right circular cone are 48πcm^3 and 12πcm^2…
- If the height and the base area of a right circular cone are 5 cm and 48 sq. cm…
- The ratios of the respective heights and the respective radii of two cylinders…
- If the radius of a sphere is 2 cm, then the curved surface area of the sphere…
- The total surface area of a solid hemisphere of diameter 2 cm is equal toA. 12…
- If the volume of a sphere is 9/6 pi cu. cm, then its radius isA. 4/3 cm B. 3/4…
- The surface areas of two spheres are in the ratio of 9 : 25. Then their volumes…
- The total surface area of a solid hemisphere whose radius is a units, is equal…
- If the surface area of a sphere is 100π cm^2 , then its radius is equal toA. 25…
- If the surface area of a sphere is 36π cm^2 , then the volume of the sphere is…
- If the total surface area of a solid hemisphere is 12π cm^2 then its curved…
- If the radius of a sphere is half of the radius of another sphere, then their…
- Curved surface area of solid sphere is 24 cm^2 . If the sphere is divided into…
- Two right circular cones have equal radii. If their slant heights are in the…

###### Exercise 8.1

**Question 1.**A solid right circular cylinder has radius of 14 cm and height of 8 cm. Find its curved surface area and total surface area.

**Answer:**__Given__ : radius of cylinder = r = 14 cm

height of cylinder = h = 8 cm

__To find__ : Curved surface area = CSA of cylinder = ?

Total surface area = TSA of cylinder = ?

__Formula__ : CSA of cylinder = 2πrh

TSA of cylinder = 2πr (h + r)

__Solution__ : CSA of cylinder = 2rh = 2 × × 14 × 8

= 44 × 16 = 704 cm^{2}

TSA of cylinder = 2r (h + r) = 2 × × 14 (8 + 14)

= 88 × 22 = 1936 cm^{2}

∴ The curved surface area and the total surface area of the right circular cylinder is 704 cm^{2} and 1936 cm^{2} respectively.

**Question 2.**The total surface area of a solid right circular cylinder is 660 sq.cm. If its diameter of the base is 14 cm, find the height and curved surface area of the cylinder.

**Answer:**__Given__ : Total surface area (TSA) of cylinder = 660 sq.cm.

Diameter of the base of the cylinder = d = 14 cm

__To find__ : Height of cylinder = h = ?

Curved surface area (CSA) of cylinder = ?

__Formula__ : d = 2r

TSA of cylinder = 2πr (h + r)

CSA of cylinder = 2πrh

__Solution__ : diameter = d = 2r

⇒ Radius = r = = = 7 cm

TSA of cylinder = 2r (h + r)

⇒ 660 = 2 × × 7 (h + 7)

660 = 44 (h + 7)

44 (h + 7) = 660

⇒ h + 7 =

h + 7 = 15

⇒ h = 15 – 7

h = 8 cm

CSA of cylinder = 2rh = 2 × × 7 × 8

= 44 × 8 = 352 cm^{2}

∴ The height of the right circular cylinder is 8 cm and its curved surface area is 352 cm^{2}.

**Question 3.**Curved surface area and circumference at the base of a solid right circular cylinder are 4400 sq.cm and 110 cm respectively. Find its height and diameter.

**Answer:**__Given__ : CSA of cylinder = 4400 sq.cm

Circumference of the base of cylinder = 110 cm

__To find__ : Height of cylinder = h = ?

Diameter = d = ?

__Formula__ : circumference = 2πr

CSA of cylinder = 2πrh

__Solution__ : circumference of base = 2r

⇒ 110 = 2r (1)

CSA of cylinder = 2rh = 2r × h

4400 = 110 × h [from (1)]

⇒ h = = 40 cm

Circumference = d

⇒ 110 = × d

d = = 35 cm

∴ The height and diameter of the right circular cylinder are 40 cm and 35 cm respectively.

**Question 4.**A mansion has 12 right cylindrical pillars each having radius 50 cm and height 3.5 m. Find the cost to paint the lateral surface of the pillars at Rs 20 per square metre.

**Answer:**__Given__ : Radius of cylindrical pillar = r = 50 cm = 0.5 m

Height of cylindrical pillar = h = 3.5 m

Number of cylindrical pillars = 12

Cost of painting per square metre = Rs 20

__To find__ : The cost of painting the lateral surface of the pillars

__Formula__ : CSA of cylinder = 2πrh

__Solution__ : The lateral surface of a cylinder is the curved surface of the cylinder.

Area to be painted for one pillar = CSA of cylinder = 2πrh

= 2 × × 0.5 × 3.5

= 11 m^{2}

Total area to be painted for 12 cylinders = 11 × 12 = 132 m^{2}

Total cost of painting = Area to be painted × cost of painting per square meter

= 132 × 20 = Rs 2640

∴ The cost of painting 12 cylindrical pillars is Rs 2640.

**Question 5.**The total surface area of a solid right circular cylinder is 231 cm^{2}. Its curved surface area is two thirds of the total surface area. Find the radius and height of the cylinder.

**Answer:**__Given__ : TSA of cylinder = 231 cm^{2}

CSA of cylinder = TSA of cylinder = × 231 = 154 cm^{2}

__To find__ : Radius of cylinder = r = ?

Height of cylinder = h = ?

__Formula__ : CSA of cylinder = 2πrh

TSA of cylinder = 2πr (h + r)

__Solution__ : CSA of cylinder = 2πrh

⇒ 2 × × r × h = 154

r × h = = 24.5 (1)

TSA of cylinder = 2r (h + r)

⇒ 2rh + 2r^{2} = 231

154 + 2r^{2} = 231

⇒ 2r^{2} = 231 – 154 = 77

2 × × r^{2} = 77

⇒ r^{2} = =

⇒ r = = 3.5 cm

From (1),

r × h = 24.5

⇒ 3.5 × h = 24.5

⇒ h =

h = 7 cm

∴ The radius and height of the right circular cylinder is 3.5 cm and 7 cm respectively.

**Question 6.**The total surface area of a solid right circular cylinder is 1540 cm^{2}. If the height is four times the radius of the base, then find the height of the cylinder.

**Answer:**__Given__ : TSA of cylinder = 1540 cm^{2}

__To find__ : Height of cylinder = h = ?

__Formula__ : TSA of cylinder = 2πr (h + r)

__Solution__ : Let the radius and height of the cylinder be r and h respectively.

According to the given condition,

h = 4r

⇒ r =

TSA of cylinder = 2r (h + r)

⇒ 1540 = 2 × × (h + )

1540 =

h^{2} = = 28 × 28

⇒ h = 28 cm

∴ The height of the right circular cylinder is 8 cm.

**Question 7.**The radii of two right circular cylinders are in the ratio of 3 : 2 and their heights are in the ratio 5 : 3. Find the ratio of their curved surface areas.

**Answer:**__Given__ : Ratio of radii of two cylinders = r_{1} : r_{2} = 3 : 2

Ratio of heights of the two cylinders = h_{1} : h_{2} = 5 : 3

__To find__ : Ratio of CSA of the two cylinders = C_{1} : C_{2} = ?

__Formula__ : CSA of cylinder = 2rh

__Solution__ : CSA of first cylinder = C_{1} = 2r_{1}h_{1}

CSA of first cylinder = C_{2} = 2r_{2}h_{2}

∴

⇒

∴ C_{1} : C_{2} = 5 : 2

∴ The ratio of the curved surface areas of the two cylinders is 5 : 2.

**Question 8.**The external surface area of a hollow cylinder is 540 πsq.cm. Its internal diameter is 16 cm and height is 15 cm. Find the total surface area.

**Answer:**__Given__ : External surface area of hollow cylinder = CSA_{ext} = 540 cm^{2}

Internal diameter = d = 16 cm

Height of the hollow cylinder = h = 15 cm

__To find__ : Total surface area = TSA of hollow cylinder = ?

__Formula__ : r =

CSA of cylinder = 2rh

__Solution__ : Internal radius of cylinder = r = = = 8 cm

Internal surface area of hollow cylinder = CSA_{int} = 2πrh

= 2π × 8 × 15

= 16π × 15 = 240π cm^{2}

TSA of hollow cylinder = CSA_{ext} + CSA_{int} = 540 + 240

= 780 cm^{2}

∴ The total surface area of the hollow cylinder is 780 cm^{2}.

**Question 9.**The external diameter of a cylindrical shaped iron pipe is 25 cm and its length is 20 cm. If the thickness of the pipe is 1cm, find the total surface area of the pipe.

**Answer:**__Given__ : External diameter of pipe = d_{ext} = 25 cm

Thickness of pipe = 1 cm

Height of the hollow cylinder = h = 20 cm

__To find__ : Total surface area of pipe = TSA of hollow cylinder = ?

__Formula__ : r =

CSA of cylinder = 2rh

__Solution__ : External radius of pipe = r_{ext} = = = 12.5 cm

External surface area of pipe = CSA_{ext} = 2r_{ext}h

= 2 × 12.5 × 20

= 500 cm^{2}

Thickness of pipe = 1 cm

∴ Internal radius of pipe = r_{int} = r_{ext} – 1 = 12.5 – 1 = 11.5 cm

Internal surface area of pipe = CSA_{int} = 2r_{int}h

= 2 × 11.5 × 20

= 460 cm^{2}

TSA of hollow cylinder = CSA_{ext} + CSA_{int} = 500 + 460

= 960 cm^{2}

∴ The total surface area of the hollow cylinder is 960 cm^{2}.

**Question 10.**The radius and height of a right circular solid cone are 7 cm and 24 cm respectively. Find its curved surface area and total surface area.

**Answer:**__Given__ : Radius of cone = r = 7 cm

Height of cone = h = 24 cm

__To find__ : CSA of cone = ?

TSA of cone = ?

__Formula__ : l =

CSA of cone = rl

TSA of cone = r (r + l)

__Solution__ : l = = = = = 25 cm

CSA of cone = rl = × 7 × 25 = 22 × 25 = 550 cm^{2}

TSA of cone = r (r + l) = × 7 (25 + 7) = 22 × 32 = 704 cm^{2}

∴ The curved surface area and total surface area of the right circular cone is 550 cm^{2} and 704 cm^{2} respectively.

**Question 11.**If the vertical angle and the radius of a right circular cone are 60° and 15 cm respectively, then find its height and slant height.

**Answer:**__Given__ : Radius of cone = r = 15 cm

Vertical angle = θ = 60°

__To find__ : height of cone = h = ?

Slant height of cone = l = ?

__Formula__ : l =

__Solution__ :

Here, the vertical angle = ∠ A = 60°

AB = height

BC = radius = 15 cm

AC = slant height

In Δ ABC, AC is the hypotenuse.

tan θ =

⇒ tan 60° =

=

∴ h = = 5√3 cm

l = = = l = = 10√3 cm

∴ The height and slant height of the right circular cone are 5√3 cm and 10√3 cm respectively.

**Question 12.**If the circumference of the base of a solid right circular cone is 236 cm and its slant height is 12 cm, find its curved surface area.

**Answer:**__Given__ : Circumference of base of cone = 236 cm

Slant height of cone = l = 12 cm

__To find__ : CSA of cone = ?

__Formula__ : Circumference = 2r

CSA of cone = rl

__Solution__ : Circumference = 2r

⇒ 2 × r = 236

r = = 118

CSA of cone = rl = 118 × 12 = 1416 cm^{2}

∴ The curved surface area of the solid right circular cone is 1416 cm^{2}.

**Question 13.**A heap of paddy is in the form of a cone whose diameter is 4.2 m and height is 2.8 m. If the heap is to be covered exactly by a canvas to protect it from rain, then find the area of the canvas needed.

**Answer:**__Given__ : Diameter of paddy = d = 4.2 m

Height of paddy = h = 2.8 m

__To find__ : Area of canvas needed to cover paddy = ?

__Formula__ : l =

CSA of cone = rl

__Solution__ : radius of paddy = r = = = 2.1 m

l = = = = = 3.5 m

Area of canvas needed = CSA of cone = rl = × 2.1 × 3.5 = 23.1 m^{2}

∴ Area of canvas needed to cover the paddy is 23.1 m^{2}.

**Question 14.**The central angle and radius of a sector of a circular disc are 180° and 21 cm respectively. If the edges of the sector are joined together to make a hollow cone, then find the radius of the cone.

**Answer:**__Given__ : Radius of circular disc = r = 21 cm

Central angle of sector = θ = 180°

__To find__ : Radius of cone = R = ?

__Formula__ : Area of sector =

CSA of cone = rl

__Solution__ : When the edges of the sector are joined to form a cone, the radius of the sector becomes the slant height of the cone.

The area of the sector = CSA of the cone

⇒

21 × 21 = × R × 21

= 21R

⇒ R = = = 10.5 cm

∴ The radius of the hollow cone formed by the sector of the circular disc is 10.5 cm.

**Question 15.**Radius and slant height of a solid right circular cone are in the ratio 3 : 5. If the curved surface area is 60 πsq.cm, then find its total surface area.

**Answer:**__Given__ : Ratio of radius and slant height of cone = r : l = 3 : 5

CSA of cone = 60 sq.cm

__To find__ : TSA of cone = ?

__Formula__ : CSA of cone = rl

TSA of cone = r (r + l)

__Solution__ : Ratio of radius and slant height = 3 : 5

⇒

r =

CSA of cone = rl

60 = × × l

⇒ l^{2} = = 100

⇒ l = 10 cm

∴ r = = = 6 cm

TSA of cone = r (r + l) = × 6 (6 + 10) = = = cm^{2}

∴ The total surface area of the solid right circular cone is cm^{2}.

**Question 16.**If the curved surface area of a solid sphere is 98.56 cm^{2}, then find the radius of the sphere.

**Answer:**__Given__ : CSA of sphere = 98.56 cm^{2}

__To find__ : Radius of sphere = r = ?

__Formula__ : CSA of sphere = 4r^{2}

__Solution__ : CSA of sphere = 4r^{2}

⇒ 4r^{2} = 98.56

4 × × r^{2} = 98.56

r^{2} = = 7.84

⇒ r = 2.8 cm

∴ The radius of the solid sphere is 2.8 cm.

**Question 17.**If the curved surface area of a solid hemisphere is 2772 sq.cm, then find its total surface area.

**Answer:**__Given__ : CSA of hemisphere = 2772 sq.cm

__To find__ : TSA of hemisphere = ?

__Formula__ : CSA of hemisphere = 2r^{2}

TSA of hemisphere = r^{2}

__Solution__ : CSA of hemisphere = 2r^{2}

⇒ 2r^{2} = 2772

2 × × r^{2} = 2772

⇒ r^{2} = = 441

⇒ r = 21 cm

TSA of hemisphere = r^{2} = 3 × × 21 × 21 = 4158 cm^{2}

∴ The total surface area of the solid hemisphere is 4158 cm^{2}.

**Question 18.**Radii of two solid hemispheres are in the ratio 3 : 5. Find the ratio of their curved surface areas and the ratio of their total surface areas.

**Answer:**__Given__ : Ratio of radii of two hemispheres = r_{1} : r_{2} = 3 : 5

__To find__ : Ratio of curved surface areas = C_{1} : C_{2} = ?

Ratio of total surface areas = T_{1} : T_{2} = ?

__Formula__ : CSA of hemisphere = 2r^{2}

TSA of hemisphere = r^{2}

__Solution__ : CSA of first hemisphere = C_{1} = 2r_{1}^{2}

CSA of second hemisphere = C_{2} = 2r_{2}^{2}

⇒

TSA of first hemisphere = T_{1} = r_{1}^{2}

TSA of second hemisphere = T_{2} = r_{2}^{2}

⇒

∴ The ratio of the curved surface areas and total surface areas of the two solid hemispheres is 9 : 25 and 9 : 25 respectively.

**Question 19.**Find the curved surface area and total surface area of a hollow hemisphere whose outer and inner radii are 4.2 cm and 2.1 cm respectively.

**Answer:**__Given__ : Inner radius of hollow hemisphere = r_{int} = 2.1 cm

Outer radius of hollow hemisphere = r_{ext} = 4.2 cm

__To find__ : CSA of hollow hemisphere = ?

TSA of hollow hemisphere = ?

__Formula__ : CSA of hemisphere = 2r^{2}

TSA of hemisphere = r^{2}

__Solution__ : Outer surface area = CSA_{ext} = 2r_{ext}^{2} = 2 × 4.2 × 4.2 = 35.28 cm^{2}

Inner surface area = CSA_{int} = 2r_{int}^{2} = 2 × 2.1 × 2.1 = 8.82 cm^{2}

Area of the edges = (r_{ext}^{2} – r_{int}^{2}) = (4.2^{2} – 2.1^{2}) = 13.23 cm^{2}

Curved surface area of hollow hemisphere = CSA_{ext} + CSA_{int} = 35.28 + 8.82 = 44.1 cm^{2}

Total surface area of hollow hemisphere = CSA + Area of edges = 44.1 + 13.23 = 57.33 cm^{2}

∴ The curved surface area and total surface area of the hollow cylinder is 44.1 cm^{2} and 57.33 cm^{2} respectively.

**Question 20.**The inner curved surface area of a hemispherical dome of a building needs to be painted. If the circumference of the base is 17.6 m, find the cost of painting it at the rate of Rs 5 per sq. m.

**Answer:**__Given__ : Circumference of base of dome = 17.6 m

Rate of painting = Rs 5 per sq.m

__To find__ : Total cost of painting inside the dome = ?

__Formula__ : Circumference = 2r

CSA of hemisphere = 2r^{2}

__Solution__ : Circumference of the base = 2r

⇒ 2 × × r = 17.6

r = = 2.8 m

The area to be painted = CSA of hemisphere = 2r^{2} = 2 × × 2.8 × 2.8 = 49.28 m^{2}

Rate of painting = Rs 5 per sq.m

The total cost of painting = 49.28 × 5 = Rs 246.40

∴ The cost of painting the hemispherical dome on the inside is Rs 246.40.

**Question 1.**

A solid right circular cylinder has radius of 14 cm and height of 8 cm. Find its curved surface area and total surface area.

**Answer:**

__Given__ : radius of cylinder = r = 14 cm

height of cylinder = h = 8 cm

__To find__ : Curved surface area = CSA of cylinder = ?

Total surface area = TSA of cylinder = ?

__Formula__ : CSA of cylinder = 2πrh

TSA of cylinder = 2πr (h + r)

__Solution__ : CSA of cylinder = 2rh = 2 × × 14 × 8

= 44 × 16 = 704 cm^{2}

TSA of cylinder = 2r (h + r) = 2 × × 14 (8 + 14)

= 88 × 22 = 1936 cm^{2}

∴ The curved surface area and the total surface area of the right circular cylinder is 704 cm^{2} and 1936 cm^{2} respectively.

**Question 2.**

The total surface area of a solid right circular cylinder is 660 sq.cm. If its diameter of the base is 14 cm, find the height and curved surface area of the cylinder.

**Answer:**

__Given__ : Total surface area (TSA) of cylinder = 660 sq.cm.

Diameter of the base of the cylinder = d = 14 cm

__To find__ : Height of cylinder = h = ?

Curved surface area (CSA) of cylinder = ?

__Formula__ : d = 2r

TSA of cylinder = 2πr (h + r)

CSA of cylinder = 2πrh

__Solution__ : diameter = d = 2r

⇒ Radius = r = = = 7 cm

TSA of cylinder = 2r (h + r)

⇒ 660 = 2 × × 7 (h + 7)

660 = 44 (h + 7)

44 (h + 7) = 660

⇒ h + 7 =

h + 7 = 15

⇒ h = 15 – 7

h = 8 cm

CSA of cylinder = 2rh = 2 × × 7 × 8

= 44 × 8 = 352 cm^{2}

∴ The height of the right circular cylinder is 8 cm and its curved surface area is 352 cm^{2}.

**Question 3.**

Curved surface area and circumference at the base of a solid right circular cylinder are 4400 sq.cm and 110 cm respectively. Find its height and diameter.

**Answer:**

__Given__ : CSA of cylinder = 4400 sq.cm

Circumference of the base of cylinder = 110 cm

__To find__ : Height of cylinder = h = ?

Diameter = d = ?

__Formula__ : circumference = 2πr

CSA of cylinder = 2πrh

__Solution__ : circumference of base = 2r

⇒ 110 = 2r (1)

CSA of cylinder = 2rh = 2r × h

4400 = 110 × h [from (1)]

⇒ h = = 40 cm

Circumference = d

⇒ 110 = × d

d = = 35 cm

∴ The height and diameter of the right circular cylinder are 40 cm and 35 cm respectively.

**Question 4.**

A mansion has 12 right cylindrical pillars each having radius 50 cm and height 3.5 m. Find the cost to paint the lateral surface of the pillars at Rs 20 per square metre.

**Answer:**

__Given__ : Radius of cylindrical pillar = r = 50 cm = 0.5 m

Height of cylindrical pillar = h = 3.5 m

Number of cylindrical pillars = 12

Cost of painting per square metre = Rs 20

__To find__ : The cost of painting the lateral surface of the pillars

__Formula__ : CSA of cylinder = 2πrh

__Solution__ : The lateral surface of a cylinder is the curved surface of the cylinder.

Area to be painted for one pillar = CSA of cylinder = 2πrh

= 2 × × 0.5 × 3.5

= 11 m^{2}

Total area to be painted for 12 cylinders = 11 × 12 = 132 m^{2}

Total cost of painting = Area to be painted × cost of painting per square meter

= 132 × 20 = Rs 2640

∴ The cost of painting 12 cylindrical pillars is Rs 2640.

**Question 5.**

The total surface area of a solid right circular cylinder is 231 cm^{2}. Its curved surface area is two thirds of the total surface area. Find the radius and height of the cylinder.

**Answer:**

__Given__ : TSA of cylinder = 231 cm^{2}

CSA of cylinder = TSA of cylinder = × 231 = 154 cm^{2}

__To find__ : Radius of cylinder = r = ?

Height of cylinder = h = ?

__Formula__ : CSA of cylinder = 2πrh

TSA of cylinder = 2πr (h + r)

__Solution__ : CSA of cylinder = 2πrh

⇒ 2 × × r × h = 154

r × h = = 24.5 (1)

TSA of cylinder = 2r (h + r)

⇒ 2rh + 2r^{2} = 231

154 + 2r^{2} = 231

⇒ 2r^{2} = 231 – 154 = 77

2 × × r^{2} = 77

⇒ r^{2} = =

⇒ r = = 3.5 cm

From (1),

r × h = 24.5

⇒ 3.5 × h = 24.5

⇒ h =

h = 7 cm

∴ The radius and height of the right circular cylinder is 3.5 cm and 7 cm respectively.

**Question 6.**

The total surface area of a solid right circular cylinder is 1540 cm^{2}. If the height is four times the radius of the base, then find the height of the cylinder.

**Answer:**

__Given__ : TSA of cylinder = 1540 cm^{2}

__To find__ : Height of cylinder = h = ?

__Formula__ : TSA of cylinder = 2πr (h + r)

__Solution__ : Let the radius and height of the cylinder be r and h respectively.

According to the given condition,

h = 4r

⇒ r =

TSA of cylinder = 2r (h + r)

⇒ 1540 = 2 × × (h + )

1540 =

h^{2} = = 28 × 28

⇒ h = 28 cm

∴ The height of the right circular cylinder is 8 cm.

**Question 7.**

The radii of two right circular cylinders are in the ratio of 3 : 2 and their heights are in the ratio 5 : 3. Find the ratio of their curved surface areas.

**Answer:**

__Given__ : Ratio of radii of two cylinders = r_{1} : r_{2} = 3 : 2

Ratio of heights of the two cylinders = h_{1} : h_{2} = 5 : 3

__To find__ : Ratio of CSA of the two cylinders = C_{1} : C_{2} = ?

__Formula__ : CSA of cylinder = 2rh

__Solution__ : CSA of first cylinder = C_{1} = 2r_{1}h_{1}

CSA of first cylinder = C_{2} = 2r_{2}h_{2}

∴

⇒

∴ C_{1} : C_{2} = 5 : 2

∴ The ratio of the curved surface areas of the two cylinders is 5 : 2.

**Question 8.**

The external surface area of a hollow cylinder is 540 πsq.cm. Its internal diameter is 16 cm and height is 15 cm. Find the total surface area.

**Answer:**

__Given__ : External surface area of hollow cylinder = CSA_{ext} = 540 cm^{2}

Internal diameter = d = 16 cm

Height of the hollow cylinder = h = 15 cm

__To find__ : Total surface area = TSA of hollow cylinder = ?

__Formula__ : r =

CSA of cylinder = 2rh

__Solution__ : Internal radius of cylinder = r = = = 8 cm

Internal surface area of hollow cylinder = CSA_{int} = 2πrh

= 2π × 8 × 15

= 16π × 15 = 240π cm^{2}

TSA of hollow cylinder = CSA_{ext} + CSA_{int} = 540 + 240

= 780 cm^{2}

∴ The total surface area of the hollow cylinder is 780 cm^{2}.

**Question 9.**

The external diameter of a cylindrical shaped iron pipe is 25 cm and its length is 20 cm. If the thickness of the pipe is 1cm, find the total surface area of the pipe.

**Answer:**

__Given__ : External diameter of pipe = d_{ext} = 25 cm

Thickness of pipe = 1 cm

Height of the hollow cylinder = h = 20 cm

__To find__ : Total surface area of pipe = TSA of hollow cylinder = ?

__Formula__ : r =

CSA of cylinder = 2rh

__Solution__ : External radius of pipe = r_{ext} = = = 12.5 cm

External surface area of pipe = CSA_{ext} = 2r_{ext}h

= 2 × 12.5 × 20

= 500 cm^{2}

Thickness of pipe = 1 cm

∴ Internal radius of pipe = r_{int} = r_{ext} – 1 = 12.5 – 1 = 11.5 cm

Internal surface area of pipe = CSA_{int} = 2r_{int}h

= 2 × 11.5 × 20

= 460 cm^{2}

TSA of hollow cylinder = CSA_{ext} + CSA_{int} = 500 + 460

= 960 cm^{2}

∴ The total surface area of the hollow cylinder is 960 cm^{2}.

**Question 10.**

The radius and height of a right circular solid cone are 7 cm and 24 cm respectively. Find its curved surface area and total surface area.

**Answer:**

__Given__ : Radius of cone = r = 7 cm

Height of cone = h = 24 cm

__To find__ : CSA of cone = ?

TSA of cone = ?

__Formula__ : l =

CSA of cone = rl

TSA of cone = r (r + l)

__Solution__ : l = = = = = 25 cm

CSA of cone = rl = × 7 × 25 = 22 × 25 = 550 cm^{2}

TSA of cone = r (r + l) = × 7 (25 + 7) = 22 × 32 = 704 cm^{2}

∴ The curved surface area and total surface area of the right circular cone is 550 cm^{2} and 704 cm^{2} respectively.

**Question 11.**

If the vertical angle and the radius of a right circular cone are 60° and 15 cm respectively, then find its height and slant height.

**Answer:**

__Given__ : Radius of cone = r = 15 cm

Vertical angle = θ = 60°

__To find__ : height of cone = h = ?

Slant height of cone = l = ?

__Formula__ : l =

__Solution__ :

Here, the vertical angle = ∠ A = 60°

AB = height

BC = radius = 15 cm

AC = slant height

In Δ ABC, AC is the hypotenuse.

tan θ =

⇒ tan 60° =

=

∴ h = = 5√3 cm

l = = = l = = 10√3 cm

∴ The height and slant height of the right circular cone are 5√3 cm and 10√3 cm respectively.

**Question 12.**

If the circumference of the base of a solid right circular cone is 236 cm and its slant height is 12 cm, find its curved surface area.

**Answer:**

__Given__ : Circumference of base of cone = 236 cm

Slant height of cone = l = 12 cm

__To find__ : CSA of cone = ?

__Formula__ : Circumference = 2r

CSA of cone = rl

__Solution__ : Circumference = 2r

⇒ 2 × r = 236

r = = 118

CSA of cone = rl = 118 × 12 = 1416 cm^{2}

∴ The curved surface area of the solid right circular cone is 1416 cm^{2}.

**Question 13.**

A heap of paddy is in the form of a cone whose diameter is 4.2 m and height is 2.8 m. If the heap is to be covered exactly by a canvas to protect it from rain, then find the area of the canvas needed.

**Answer:**

__Given__ : Diameter of paddy = d = 4.2 m

Height of paddy = h = 2.8 m

__To find__ : Area of canvas needed to cover paddy = ?

__Formula__ : l =

CSA of cone = rl

__Solution__ : radius of paddy = r = = = 2.1 m

l = = = = = 3.5 m

Area of canvas needed = CSA of cone = rl = × 2.1 × 3.5 = 23.1 m^{2}

∴ Area of canvas needed to cover the paddy is 23.1 m^{2}.

**Question 14.**

The central angle and radius of a sector of a circular disc are 180° and 21 cm respectively. If the edges of the sector are joined together to make a hollow cone, then find the radius of the cone.

**Answer:**

__Given__ : Radius of circular disc = r = 21 cm

Central angle of sector = θ = 180°

__To find__ : Radius of cone = R = ?

__Formula__ : Area of sector =

CSA of cone = rl

__Solution__ : When the edges of the sector are joined to form a cone, the radius of the sector becomes the slant height of the cone.

The area of the sector = CSA of the cone

⇒

21 × 21 = × R × 21

= 21R

⇒ R = = = 10.5 cm

∴ The radius of the hollow cone formed by the sector of the circular disc is 10.5 cm.

**Question 15.**

Radius and slant height of a solid right circular cone are in the ratio 3 : 5. If the curved surface area is 60 πsq.cm, then find its total surface area.

**Answer:**

__Given__ : Ratio of radius and slant height of cone = r : l = 3 : 5

CSA of cone = 60 sq.cm

__To find__ : TSA of cone = ?

__Formula__ : CSA of cone = rl

TSA of cone = r (r + l)

__Solution__ : Ratio of radius and slant height = 3 : 5

⇒

r =

CSA of cone = rl

60 = × × l

⇒ l^{2} = = 100

⇒ l = 10 cm

∴ r = = = 6 cm

TSA of cone = r (r + l) = × 6 (6 + 10) = = = cm^{2}

∴ The total surface area of the solid right circular cone is cm^{2}.

**Question 16.**

If the curved surface area of a solid sphere is 98.56 cm^{2}, then find the radius of the sphere.

**Answer:**

__Given__ : CSA of sphere = 98.56 cm^{2}

__To find__ : Radius of sphere = r = ?

__Formula__ : CSA of sphere = 4r^{2}

__Solution__ : CSA of sphere = 4r^{2}

⇒ 4r^{2} = 98.56

4 × × r^{2} = 98.56

r^{2} = = 7.84

⇒ r = 2.8 cm

∴ The radius of the solid sphere is 2.8 cm.

**Question 17.**

If the curved surface area of a solid hemisphere is 2772 sq.cm, then find its total surface area.

**Answer:**

__Given__ : CSA of hemisphere = 2772 sq.cm

__To find__ : TSA of hemisphere = ?

__Formula__ : CSA of hemisphere = 2r^{2}

TSA of hemisphere = r^{2}

__Solution__ : CSA of hemisphere = 2r^{2}

⇒ 2r^{2} = 2772

2 × × r^{2} = 2772

⇒ r^{2} = = 441

⇒ r = 21 cm

TSA of hemisphere = r^{2} = 3 × × 21 × 21 = 4158 cm^{2}

∴ The total surface area of the solid hemisphere is 4158 cm^{2}.

**Question 18.**

Radii of two solid hemispheres are in the ratio 3 : 5. Find the ratio of their curved surface areas and the ratio of their total surface areas.

**Answer:**

__Given__ : Ratio of radii of two hemispheres = r_{1} : r_{2} = 3 : 5

__To find__ : Ratio of curved surface areas = C_{1} : C_{2} = ?

Ratio of total surface areas = T_{1} : T_{2} = ?

__Formula__ : CSA of hemisphere = 2r^{2}

TSA of hemisphere = r^{2}

__Solution__ : CSA of first hemisphere = C_{1} = 2r_{1}^{2}

CSA of second hemisphere = C_{2} = 2r_{2}^{2}

⇒

TSA of first hemisphere = T_{1} = r_{1}^{2}

TSA of second hemisphere = T_{2} = r_{2}^{2}

⇒

∴ The ratio of the curved surface areas and total surface areas of the two solid hemispheres is 9 : 25 and 9 : 25 respectively.

**Question 19.**

Find the curved surface area and total surface area of a hollow hemisphere whose outer and inner radii are 4.2 cm and 2.1 cm respectively.

**Answer:**

__Given__ : Inner radius of hollow hemisphere = r_{int} = 2.1 cm

Outer radius of hollow hemisphere = r_{ext} = 4.2 cm

__To find__ : CSA of hollow hemisphere = ?

TSA of hollow hemisphere = ?

__Formula__ : CSA of hemisphere = 2r^{2}

TSA of hemisphere = r^{2}

__Solution__ : Outer surface area = CSA_{ext} = 2r_{ext}^{2} = 2 × 4.2 × 4.2 = 35.28 cm^{2}

Inner surface area = CSA_{int} = 2r_{int}^{2} = 2 × 2.1 × 2.1 = 8.82 cm^{2}

Area of the edges = (r_{ext}^{2} – r_{int}^{2}) = (4.2^{2} – 2.1^{2}) = 13.23 cm^{2}

Curved surface area of hollow hemisphere = CSA_{ext} + CSA_{int} = 35.28 + 8.82 = 44.1 cm^{2}

Total surface area of hollow hemisphere = CSA + Area of edges = 44.1 + 13.23 = 57.33 cm^{2}

∴ The curved surface area and total surface area of the hollow cylinder is 44.1 cm^{2} and 57.33 cm^{2} respectively.

**Question 20.**

The inner curved surface area of a hemispherical dome of a building needs to be painted. If the circumference of the base is 17.6 m, find the cost of painting it at the rate of Rs 5 per sq. m.

**Answer:**

__Given__ : Circumference of base of dome = 17.6 m

Rate of painting = Rs 5 per sq.m

__To find__ : Total cost of painting inside the dome = ?

__Formula__ : Circumference = 2r

CSA of hemisphere = 2r^{2}

__Solution__ : Circumference of the base = 2r

⇒ 2 × × r = 17.6

r = = 2.8 m

The area to be painted = CSA of hemisphere = 2r^{2} = 2 × × 2.8 × 2.8 = 49.28 m^{2}

Rate of painting = Rs 5 per sq.m

The total cost of painting = 49.28 × 5 = Rs 246.40

∴ The cost of painting the hemispherical dome on the inside is Rs 246.40.

###### Exercise 8.2

**Question 1.**Find the volume of a solid cylinder whose radius is 14 cm and height 30 cm.

**Answer:**GIVEN : radius of solid cylinder “r” = 14cm

height of solid cylinder “h” = 30cm

we take π =

TO FIND : volume of a solid cylinder = ?

PROCEDURE :

As we know, that

Volume of a cylinder “V” = π r^{2}h

So we now put the values of r and h in the above formula

So we get, V = π ×(14)^{2}×(30)

= π ×196×30

= ×196×30

= 22 × 28 × 30

= 18480 cm^{3}

∴ The volume of the given cylinder is 18480 cm^{3}.

**Question 2.**A patient in a hospital is given soup daily in a cylindrical bowl of diameter 7 cm. If the bowl is filled with soup to a height of 4 cm, then find the quantity of soup to be prepared daily in the hospital to serve 250 patients?

**Answer:**GIVEN : diameter of cylindrical bowl = 7cm

Radius of cylinder “r” = 3.5cm

Height upto which the soup is filled “h” = 4cm

No. of patients daily = 250

we take π =

TO FIND : quantity of soup to be prepared daily in the hospital to serve 250 patients = ?

PROCEDURE :

Volume of a cylindrical bowl “V” = π r^{2}h

Putting the values in the above formula, we get

V = × (3.5)^{2} × 4

V = ×3.5× 3.5× 4

= ×3.5×4

= 22× 3.5× 2

= 154cm^{3}

∴ volume of 1 bowl = 154cm^{3}

⇒ volume of 250 bowls = 250 × 154cm^{3} = 38500cm^{3}

Quantity of soup in the 250 bowls = 38500 × litres

{∵ 1litre = 1000cm^{3})

Quantity of soup in the 250 bowls = litres = 38.5 litres.

∴ Quantity of soup in the 250 bowls is 38.5 litres.

**Question 3.**The sum of the base radius and the height of a solid right circular solid cylinder is 37 cm. If the total surface area of the cylinder is 1628 sq.cm, then find the volume of the cylinder.

**Answer:**GIVEN : radius of cylinder be “r” and its height be “h”

r + h = 37cm

TSA (total surface area) of the cylinder = 1628 sq.cm

we take π =

TO FIND : volume of a solid cylinder = ?

PROCEDURE :

As we know that, TSA of a cylinder = 2π r(r + h)

Now putting the values of “r + h” and “TSA” in the above formula,

We get,

1628 = 2π r(37)

= 2π r

44 = 2π r

r = = = = 44 × = 7cm

∴ r = 7cm

Now we have, r + h = 37

So, we have, 7 + h = 37

So, h = 37-7 = 30cm

∴ h = 30cm

Now that we have values of radius and height of the cylinder, we can find the volume of the cylinder.

Volume of the cylinder “V” = π r^{2}h

V = π ×(7)^{2}×(30)

= × 7 × 7 × 30

= 22 × 7 × 30

= 4620 cm^{3}

∴ The volume of the given cylinder is 4620 cm^{3}.

**Question 4.**Volume of a solid cylinder is 62.37 cu.cm. Find the radius if its height is 4.5 cm.

**Answer:**GIVEN : radius of solid cylinder = “r”

height of solid cylinder “h” = 4.5cm

volume of the cylinder “V” = 62.37cm^{2}

we take π =

TO FIND : radius of a solid cylinder = ?

PROCEDURE :

As we know, that

Volume of a cylinder “V” = π r^{2}h

So we now put the values of V and h in the above formula

So we get, 62.37 = π ×(r)^{2}×(4.5)

= π ×(r)^{2}

13.86 = ×(r)^{2}

r^{2} = = × 7 = 0.63 × 7 = 4.41cm^{2}

∴ r = √ 4.41 = 2.1cm

∴ radius of the given cylinder = 2.1cm.

**Question 5.**The radii of two right circular cylinders are in the ratio 2 : 3. Find the ratio of their volumes if their heights are in the ratio 5 : 3.

**Answer:**GIVEN :

For cylinder C1 : radius is r1

Height is h1

Volume is V1

For cylinder C2 : radius is r2

Height is h2

Volume is V2

So we have, =

=

we take π =

TO FIND : = ?

PROCEDURE :

As we know, that

Volume of a cylinder “V” = π r^{2}h

So, = = ×

= ×

= ×

= ×

=

∴ the ratio between the 2 cylinders ie. =

ie. V_{1} : V_{2} = 20:27

**Question 6.**The radius and height of a cylinder are in the ratio 5 : 7. If its volume is 4400 cu.cm, find the radius of the cylinder.

**Answer:**GIVEN : radius of solid cylinder = “r”

height of solid cylinder = “h”

=

volume of the cylinder “V” = 4400cm^{2}

we take π =

TO FIND : radius of a solid cylinder = ?

PROCEDURE :

As we know, that

Volume of a cylinder “V” = π r^{2}h

Also, we have, =

⇒ = ⇒ h = r ×

So we now put the values of V and h in the above formula for V,

So we get, 4400 = π ×(r)^{2}×( r × )

4400 = × r^{3}×

4400 × × = r^{3}

200 × 5 = r^{3}

1000 = r^{3}

So, r = ∛ 1000 = 10

∴ radius of the given cylinder = 10cm.

**Question 7.**A rectangular sheet of metal foil with dimension 66 cm x 12 cm is rolled to form a cylinder of height 12 cm. Find the volume of the cylinder.

**Answer:**GIVEN : rectangular sheet of metal foil with dimension :

66 cm x 12cm ie. l × b

Height of the cylinder = 12cm

we take π =

TO FIND : volume of the cylinder = ?

PROCEDURE :

As the cylinder if formed from the rectangular foil, we can say that length of the rectangle is now the circumference of the base of the cylinder.

⇒ 66 = 2π r

So, r = = = = = = 10.5cm

∴ r = 10.5cm

Now, As we know, that

Volume of a cylinder “V” = π r^{2}h

We put the values of r and h in the above formula, we get,

V = × (10.5)^{2} × 12

= × 110.25 × 12

= 4158 cm^{3}

∴ volume of the cylinder is 4158 cm^{3}

**Question 8.**A lead pencil is in the shape of right circular cylinder. The pencil is 28 cm long and its radius is 3 mm. If the lead is of radius 1 mm, then find the volume of the wood used in the pencil.

**Answer:**GIVEN :

inner diameter of the cylinder of lead “r” = 1mm = 0.1cm

height of the pencil “h” = 28cm

outer radius of the pencil “R” = 3mm = 0.3cm

we take π =

TO FIND : volume of the wood used in the pencil “V” = ?

PROCEDURE :

Here, we will use the concept of hollow cylinder.

So, volume of the wood used in the pencil,

“V” = π × h× (R + r )× (R – r)

So, we put all the values in the above formula, and get

V = × 28 × (0.3 + 0.1)× (0.3 - 0.1)

= 22 × 4 × (0.4)× (0.2)

= 22 × 4 × 4 × 2 × 10^{-2}

= 704 × 10^{-2}

= 7.04 cm^{3}

∴ volume of the wood used in the pencil is 7.04 cm^{3}

**Question 9.**Radius and slant height of a cone are 20 cm and 29 cm respectively. Find its volume.

**Answer:**GIVEN : radius of cone “r” = 20cm

Slant height of cone “l” = 29cm

we take π =

TO FIND : volume of the cylinder “V” = ?

PROCEDURE :

As we know that, volume of cone “V” = × π r^{2}h

As we are not given the value of h, we have to find it.

As it is a right cone, l^{2} = r^{2} + h^{2}

h^{2} = l^{2} – r^{2}

= (29)^{2} – (20)^{2}

= 841 – 400

= 441

h^{2} = 441

h = √ 441 = 21cm

So now, we put the values of r and h in the above formula

We get, V = × π r^{2}h

= × π × (20)^{2}× (21)

= × × 400 × 21

= 8800

∴ volume of the given cone = 8800cm^{3}

**Question 10.**The circumference of the base of a 12 m high wooden solid cone is 44 m. Find the volume.

**Answer:**GIVEN:

Height of the cone = 12m

circumference of the base = 44m

we take π =

TO FIND : volume of the cone = ?

PROCEDURE :

As we know the circumference of the cone = 2π r

2π r = 44

r = = = = = 7cm

now, as we know that, volume of cone “V” = × π r^{2}h

putting all the values in the above formula, we get

V = × π × (7)^{2}× (12)

= × × 49 × 12

= × 22 × 7 × 12

= 22 × 7× 4

= 616 cm^{3}

∴ volume of the given cone is 616 cm^{3}

**Question 11.**A vessel is in the form of a frustum of a cone. Its radius at one end and the height are 8 cm and 14 cm respectively. If its volume is , then find the radius at the other end.

**Answer:**GIVEN :

Volume of the frustum cone “V” =

Radius of another end = “r”

Radius (R) = 8 cm

height (h) = 14 cm

we take π =

TO FIND : Radius of another end = ?

PRODEDURE :

As we know, the volume of frustum of cone,

×π ×h×(R^{2} + r^{2} + R×r) =

××(14)×(8^{2} + r^{2} + 8×r) =

(8^{2} + r^{2} + 8×r) = × 3 × ×

(64 + r^{2} + 8r) = 258× 7×

(64 + r^{2} + 8r) =

r^{2} + 8r + 64 = 129

r^{2} + 8r + 64 -129 = 0

r^{2} + 8r - 65 = 0

(r + 13) (r - 5) = 0

r + 13 = 0 or r - 5 = 0

r = -13 or r = 5 cm

as the radius cannot be negative,

∴ the required radius = 5cm.

**Question 12.**The perimeter of the ends of a frustum of a cone are 44 cm and 8.4πcm. If the depth is 14 cm., then find its volume.

**Answer:**GIVEN :

Perimeter of the upper end of frustum of a cone = 44cm

Perimeter of the lower end of frustum of a cone = 8.4cm

Height of the frustum of the cone is 14 cm

we take π =

TO FIND : volume of the frustum “V” = ?

PROCEDURE :

As we know, the volume of frustum of cone,

V = ×π ×h×(R^{2} + r^{2} + R×r)

Now first, for upper end,

2π R = 44

R = = = = 7cm

For the lower end,

2π r = 8.4π

2r = 8.4

r = 4.2cm

now that we have all the values, we will find the volumeof frustum.

V = ×π ×h×(R^{2} + r^{2} + R×r)

= ×× 14× (7^{2} + (4.2)^{2} + 7× 4.2)

= ×22 × 2 × (49 + 29.4 + 17.64)

= ×96.04

= 44 × 32.013

= 1408.57cm^{3}

∴ volume of frustum is 1408.57cm^{3}

**Question 13.**A right angled ΔABC with sides 5 cm, 12 cm and 13 cm is revolved about the fixed side of 12 cm. Find the volume of the solid generated.

**Answer:**GIVEN : Δ ABC with right angle at B

AC will be the hypotenuse = 13cm

AB = 12cm

BC = 5cm

we take π =

TO FIND : volume of the solid generated = ?

PROCEDURE :

The solid formed on rotating the ΔABC about AB is a cone with radius “r” = 5cm

And height “h” = 12cm

∴ volume of the cone “V” = × π r^{2}h

= × × (5)^{2} × 12

= × 25 × 4

= cm^{3}

∴ volume of the solid generated ie. Cone is cm^{3}

**Question 14.**The radius and height of a right circular cone are in the ratio 2 : 3. Find the slant height if its volume is 100.48 cu.cm. ( Take π = 3.14)

**Answer:**GIVEN : radius of cone = ”r”

Height of cone = “h”

=

Volume of the cone = 100.48 cu.cm

We take π = 3.14

TO FIND : slant height “l” = ?

PROCEDURE :

We know that volume of the cone “V” = × π r^{2}h

Now =

⇒ r =

Now putting the values in the formula of volume, we get

V = × π r^{2}h

100.48 = × 3.14 × ()^{2} × h

100.48 = × 3.14 × × h^{3}

h^{3} = 100.48 × 3 × ×

h^{3} = 32 × 3 × = 8× 27

h = ∛ 8× 27 = ∛ 2× 2× 2× 3× 3× 3

h = 2× 3 = 6cm

now, r = = = 4cm

now, slant height “l” can be found using :

l^{2} = r^{2} + h^{2}

= 4^{2} + 6^{2} = 16 + 36 = 52

L = √ 52 = 2√13cm

∴ slant height of the given cone is 2√13cm.

**Question 15.**The volume of a cone with circular base is 216 π cu.cm. If the base radius is 9 cm, then find the height of the cone.

**Answer:**GIVEN : volume of the cone “V” = 216cu.cm

Base radius of cone “r” = 9cm

we take π =

TO FIND : height of the cone “h” = ?

PROCEDURE :

We know that volume of the cone “V” = × π r^{2}h

Now putting all the values in the above formula, we get

V = × π r^{2}h

216 = × π × (9)^{2} × h

h = 216 × 3 × = 216 × = 8cm

∴ height of the cone “h” is 8cm.

**Question 16.**Find the mass of 200 steel spherical ball bearings, each of which has radius 0.7 cm, given that the density of steel is 7.95 g/cm^{3}. (Mass = Volume × Density)

**Answer:**GIVEN : no. of steel balls “n” = 200

Radius of each ball “r” = 0.7cm

we take π =

Density of steel = 7.95 g/cm^{3}

TO FIND : Mass of 200 balls = ?

PROCEDURE :

We know that, volume of sphere "V” = × π r^{3}

Putting the values in the above formula, we get

V = × × (0.7)^{3}

= × × 0.343

= 1.437cm^{3}

This was the volume of one ball.

Now volume of 200 balls “V’ ” = n× V = 200 × 1.437cm^{3}

V’ = 287.466 cm^{3}

Now, it is given that, Density of steel = 7.95 g/cm^{3}

So, mass of the balls = volume × density

Mass of 200 balls = volume of 200 balls × density

= 287.466 cm^{3} × 7.95 g/cm^{3}

= 2285.46g

= 2.285 Kg

∴ Mass of 200 balls bearings is 2.285 Kg.

**Question 17.**The outer and the inner radii of a hollow sphere are 12 cm and 10 cm. Find its volume.

**Answer:**GIVEN : outer radius of hollow the sphere “R” = 12cm

inner radius of hollow the sphere “r” = 10cm

we take π =

TO FIND : volume of the hollow sphere = ?

PROCEDURE :

We know that volume of the hollow sphere “V” = × π × (R^{3} – r^{3})

Putting all the values in the above formula, we get

V = ×× (12^{3} – 10^{3})

= ××(1728-1000)

= ××728

= 3050.66 cm^{3}

∴ volume of the hollow sphere is 3050.66 cm^{3}

**Question 18.**The volume of a solid hemisphere is 1152 πcu.cm. Find its curved surface area.

**Answer:**GIVEN : volume of a solid hemisphere “V” = 1152cu.cm

we take π =

TO FIND : curved surface area (CSA) = ?

PROCEDURE :

As we know that volume of a hemisphere “V” = × × π ×r^{3}

So, putting all the values in the above formula, we get

V = × × π ×r^{3}

1152 = × × π ×r^{3}

1152 = × × r^{3}

r^{3} = 1152 × 2 × = 1152 × = 576×3 = 1728

r = ∛ 1728 = 12cm

so, now that we have radius, we can find the curved surface area of the hemisphere.

Now, CSA = 2× π ×r^{2}

= 2×π × (12)^{2}

= 2×π ×144

= 288π cm^{2}

∴ curved surface area of the hemisphere is 288π cm^{2}

**Question 19.**Find the volume of the largest right circular cone that can be cut out of a cube whose edge is 14 cm.

**Answer:**GIVEN : side of cube ”s” = 14cm

we take π =

TO FIND : volume of the largest right circular cone that can be cut out of a cube = ?

PROCEDURE :

For the cone to largest, its diameter should be equal to side of the cube ie. d = 14cm

And, height of the cone should also be equal to the side of the cube ie. h = 14cm

Now, radius of the cone “r” = 7cm

Now, volume of cone “V” = × π r^{2}h

Now putting all the values in the above formula, we get

V = × π × (7)^{2} × 14

= × × 7 × 7 × 14

= × 22 × 7× 14

= 718.666cm^{3}

∴ volume of the largest right circular cone that can be cut out of a cube is 718.666cm^{3}.

**Question 20.**The radius of a spherical balloon increases from 7 cm to 14 cm as air is being pumped into it. Find the ratio of volumes of the balloon in the two cases.

**Answer:**GIVEN : radius of balloon before the increase “r” = 7cm

radius of balloon after the increase “R” = 14cm

TO FIND : ratio of volumes of the balloon in both cases

PROCEDURE :

Volume of balloon before the increase “V1” = ×π ×r^{3}

Volume of balloon after the increase “V2” = ×π ×R^{3}

Ratio of the volumes in the 2 cases =

= = = = = =

∴ the ratio of volumes of the balloon in the two cases is = ie. 1 : 8.

**Question 1.**

Find the volume of a solid cylinder whose radius is 14 cm and height 30 cm.

**Answer:**

GIVEN : radius of solid cylinder “r” = 14cm

height of solid cylinder “h” = 30cm

we take π =

TO FIND : volume of a solid cylinder = ?

PROCEDURE :

As we know, that

Volume of a cylinder “V” = π r^{2}h

So we now put the values of r and h in the above formula

So we get, V = π ×(14)^{2}×(30)

= π ×196×30

= ×196×30

= 22 × 28 × 30

= 18480 cm^{3}

∴ The volume of the given cylinder is 18480 cm^{3}.

**Question 2.**

A patient in a hospital is given soup daily in a cylindrical bowl of diameter 7 cm. If the bowl is filled with soup to a height of 4 cm, then find the quantity of soup to be prepared daily in the hospital to serve 250 patients?

**Answer:**

GIVEN : diameter of cylindrical bowl = 7cm

Radius of cylinder “r” = 3.5cm

Height upto which the soup is filled “h” = 4cm

No. of patients daily = 250

we take π =

TO FIND : quantity of soup to be prepared daily in the hospital to serve 250 patients = ?

PROCEDURE :

Volume of a cylindrical bowl “V” = π r^{2}h

Putting the values in the above formula, we get

V = × (3.5)^{2} × 4

V = ×3.5× 3.5× 4

= ×3.5×4

= 22× 3.5× 2

= 154cm^{3}

∴ volume of 1 bowl = 154cm^{3}

⇒ volume of 250 bowls = 250 × 154cm^{3} = 38500cm^{3}

Quantity of soup in the 250 bowls = 38500 × litres

{∵ 1litre = 1000cm^{3})

Quantity of soup in the 250 bowls = litres = 38.5 litres.

∴ Quantity of soup in the 250 bowls is 38.5 litres.

**Question 3.**

The sum of the base radius and the height of a solid right circular solid cylinder is 37 cm. If the total surface area of the cylinder is 1628 sq.cm, then find the volume of the cylinder.

**Answer:**

GIVEN : radius of cylinder be “r” and its height be “h”

r + h = 37cm

TSA (total surface area) of the cylinder = 1628 sq.cm

we take π =

TO FIND : volume of a solid cylinder = ?

PROCEDURE :

As we know that, TSA of a cylinder = 2π r(r + h)

Now putting the values of “r + h” and “TSA” in the above formula,

We get,

1628 = 2π r(37)

= 2π r

44 = 2π r

r = = = = 44 × = 7cm

∴ r = 7cm

Now we have, r + h = 37

So, we have, 7 + h = 37

So, h = 37-7 = 30cm

∴ h = 30cm

Now that we have values of radius and height of the cylinder, we can find the volume of the cylinder.

Volume of the cylinder “V” = π r^{2}h

V = π ×(7)^{2}×(30)

= × 7 × 7 × 30

= 22 × 7 × 30

= 4620 cm^{3}

∴ The volume of the given cylinder is 4620 cm^{3}.

**Question 4.**

Volume of a solid cylinder is 62.37 cu.cm. Find the radius if its height is 4.5 cm.

**Answer:**

GIVEN : radius of solid cylinder = “r”

height of solid cylinder “h” = 4.5cm

volume of the cylinder “V” = 62.37cm^{2}

we take π =

TO FIND : radius of a solid cylinder = ?

PROCEDURE :

As we know, that

Volume of a cylinder “V” = π r^{2}h

So we now put the values of V and h in the above formula

So we get, 62.37 = π ×(r)^{2}×(4.5)

= π ×(r)^{2}

13.86 = ×(r)^{2}

r^{2} = = × 7 = 0.63 × 7 = 4.41cm^{2}

∴ r = √ 4.41 = 2.1cm

∴ radius of the given cylinder = 2.1cm.

**Question 5.**

The radii of two right circular cylinders are in the ratio 2 : 3. Find the ratio of their volumes if their heights are in the ratio 5 : 3.

**Answer:**

GIVEN :

For cylinder C1 : radius is r1

Height is h1

Volume is V1

For cylinder C2 : radius is r2

Height is h2

Volume is V2

So we have, =

=

we take π =

TO FIND : = ?

PROCEDURE :

As we know, that

Volume of a cylinder “V” = π r^{2}h

So, = = ×

= ×

= ×

= ×

=

∴ the ratio between the 2 cylinders ie. =

ie. V_{1} : V_{2} = 20:27

**Question 6.**

The radius and height of a cylinder are in the ratio 5 : 7. If its volume is 4400 cu.cm, find the radius of the cylinder.

**Answer:**

GIVEN : radius of solid cylinder = “r”

height of solid cylinder = “h”

=

volume of the cylinder “V” = 4400cm^{2}

we take π =

TO FIND : radius of a solid cylinder = ?

PROCEDURE :

As we know, that

Volume of a cylinder “V” = π r^{2}h

Also, we have, =

⇒ = ⇒ h = r ×

So we now put the values of V and h in the above formula for V,

So we get, 4400 = π ×(r)^{2}×( r × )

4400 = × r^{3}×

4400 × × = r^{3}

200 × 5 = r^{3}

1000 = r^{3}

So, r = ∛ 1000 = 10

∴ radius of the given cylinder = 10cm.

**Question 7.**

A rectangular sheet of metal foil with dimension 66 cm x 12 cm is rolled to form a cylinder of height 12 cm. Find the volume of the cylinder.

**Answer:**

GIVEN : rectangular sheet of metal foil with dimension :

66 cm x 12cm ie. l × b

Height of the cylinder = 12cm

we take π =

TO FIND : volume of the cylinder = ?

PROCEDURE :

As the cylinder if formed from the rectangular foil, we can say that length of the rectangle is now the circumference of the base of the cylinder.

⇒ 66 = 2π r

So, r = = = = = = 10.5cm

∴ r = 10.5cm

Now, As we know, that

Volume of a cylinder “V” = π r^{2}h

We put the values of r and h in the above formula, we get,

V = × (10.5)^{2} × 12

= × 110.25 × 12

= 4158 cm^{3}

∴ volume of the cylinder is 4158 cm^{3}

**Question 8.**

A lead pencil is in the shape of right circular cylinder. The pencil is 28 cm long and its radius is 3 mm. If the lead is of radius 1 mm, then find the volume of the wood used in the pencil.

**Answer:**

GIVEN :

inner diameter of the cylinder of lead “r” = 1mm = 0.1cm

height of the pencil “h” = 28cm

outer radius of the pencil “R” = 3mm = 0.3cm

we take π =

TO FIND : volume of the wood used in the pencil “V” = ?

PROCEDURE :

Here, we will use the concept of hollow cylinder.

So, volume of the wood used in the pencil,

“V” = π × h× (R + r )× (R – r)

So, we put all the values in the above formula, and get

V = × 28 × (0.3 + 0.1)× (0.3 - 0.1)

= 22 × 4 × (0.4)× (0.2)

= 22 × 4 × 4 × 2 × 10^{-2}

= 704 × 10^{-2}

= 7.04 cm^{3}

∴ volume of the wood used in the pencil is 7.04 cm^{3}

**Question 9.**

Radius and slant height of a cone are 20 cm and 29 cm respectively. Find its volume.

**Answer:**

GIVEN : radius of cone “r” = 20cm

Slant height of cone “l” = 29cm

we take π =

TO FIND : volume of the cylinder “V” = ?

PROCEDURE :

As we know that, volume of cone “V” = × π r^{2}h

As we are not given the value of h, we have to find it.

As it is a right cone, l^{2} = r^{2} + h^{2}

h^{2} = l^{2} – r^{2}

= (29)^{2} – (20)^{2}

= 841 – 400

= 441

h^{2} = 441

h = √ 441 = 21cm

So now, we put the values of r and h in the above formula

We get, V = × π r^{2}h

= × π × (20)^{2}× (21)

= × × 400 × 21

= 8800

∴ volume of the given cone = 8800cm^{3}

**Question 10.**

The circumference of the base of a 12 m high wooden solid cone is 44 m. Find the volume.

**Answer:**

GIVEN:

Height of the cone = 12m

circumference of the base = 44m

we take π =

TO FIND : volume of the cone = ?

PROCEDURE :

As we know the circumference of the cone = 2π r

2π r = 44

r = = = = = 7cm

now, as we know that, volume of cone “V” = × π r^{2}h

putting all the values in the above formula, we get

V = × π × (7)^{2}× (12)

= × × 49 × 12

= × 22 × 7 × 12

= 22 × 7× 4

= 616 cm^{3}

∴ volume of the given cone is 616 cm^{3}

**Question 11.**

A vessel is in the form of a frustum of a cone. Its radius at one end and the height are 8 cm and 14 cm respectively. If its volume is , then find the radius at the other end.

**Answer:**

GIVEN :

Volume of the frustum cone “V” =

Radius of another end = “r”

Radius (R) = 8 cm

height (h) = 14 cm

we take π =

TO FIND : Radius of another end = ?

PRODEDURE :

As we know, the volume of frustum of cone,

×π ×h×(R^{2} + r^{2} + R×r) =

××(14)×(8^{2} + r^{2} + 8×r) =

(8^{2} + r^{2} + 8×r) = × 3 × ×

(64 + r^{2} + 8r) = 258× 7×

(64 + r^{2} + 8r) =

r^{2} + 8r + 64 = 129

r^{2} + 8r + 64 -129 = 0

r^{2} + 8r - 65 = 0

(r + 13) (r - 5) = 0

r + 13 = 0 or r - 5 = 0

r = -13 or r = 5 cm

as the radius cannot be negative,

∴ the required radius = 5cm.

**Question 12.**

The perimeter of the ends of a frustum of a cone are 44 cm and 8.4πcm. If the depth is 14 cm., then find its volume.

**Answer:**

GIVEN :

Perimeter of the upper end of frustum of a cone = 44cm

Perimeter of the lower end of frustum of a cone = 8.4cm

Height of the frustum of the cone is 14 cm

we take π =

TO FIND : volume of the frustum “V” = ?

PROCEDURE :

As we know, the volume of frustum of cone,

V = ×π ×h×(R^{2} + r^{2} + R×r)

Now first, for upper end,

2π R = 44

R = = = = 7cm

For the lower end,

2π r = 8.4π

2r = 8.4

r = 4.2cm

now that we have all the values, we will find the volumeof frustum.

V = ×π ×h×(R^{2} + r^{2} + R×r)

= ×× 14× (7^{2} + (4.2)^{2} + 7× 4.2)

= ×22 × 2 × (49 + 29.4 + 17.64)

= ×96.04

= 44 × 32.013

= 1408.57cm^{3}

∴ volume of frustum is 1408.57cm^{3}

**Question 13.**

A right angled ΔABC with sides 5 cm, 12 cm and 13 cm is revolved about the fixed side of 12 cm. Find the volume of the solid generated.

**Answer:**

GIVEN : Δ ABC with right angle at B

AC will be the hypotenuse = 13cm

AB = 12cm

BC = 5cm

we take π =

TO FIND : volume of the solid generated = ?

PROCEDURE :

The solid formed on rotating the ΔABC about AB is a cone with radius “r” = 5cm

And height “h” = 12cm

∴ volume of the cone “V” = × π r^{2}h

= × × (5)^{2} × 12

= × 25 × 4

= cm^{3}

∴ volume of the solid generated ie. Cone is cm^{3}

**Question 14.**

The radius and height of a right circular cone are in the ratio 2 : 3. Find the slant height if its volume is 100.48 cu.cm. ( Take π = 3.14)

**Answer:**

GIVEN : radius of cone = ”r”

Height of cone = “h”

=

Volume of the cone = 100.48 cu.cm

We take π = 3.14

TO FIND : slant height “l” = ?

PROCEDURE :

We know that volume of the cone “V” = × π r^{2}h

Now =

⇒ r =

Now putting the values in the formula of volume, we get

V = × π r^{2}h

100.48 = × 3.14 × ()^{2} × h

100.48 = × 3.14 × × h^{3}

h^{3} = 100.48 × 3 × ×

h^{3} = 32 × 3 × = 8× 27

h = ∛ 8× 27 = ∛ 2× 2× 2× 3× 3× 3

h = 2× 3 = 6cm

now, r = = = 4cm

now, slant height “l” can be found using :

l^{2} = r^{2} + h^{2}

= 4^{2} + 6^{2} = 16 + 36 = 52

L = √ 52 = 2√13cm

∴ slant height of the given cone is 2√13cm.

**Question 15.**

The volume of a cone with circular base is 216 π cu.cm. If the base radius is 9 cm, then find the height of the cone.

**Answer:**

GIVEN : volume of the cone “V” = 216cu.cm

Base radius of cone “r” = 9cm

we take π =

TO FIND : height of the cone “h” = ?

PROCEDURE :

We know that volume of the cone “V” = × π r^{2}h

Now putting all the values in the above formula, we get

V = × π r^{2}h

216 = × π × (9)^{2} × h

h = 216 × 3 × = 216 × = 8cm

∴ height of the cone “h” is 8cm.

**Question 16.**

Find the mass of 200 steel spherical ball bearings, each of which has radius 0.7 cm, given that the density of steel is 7.95 g/cm^{3}. (Mass = Volume × Density)

**Answer:**

GIVEN : no. of steel balls “n” = 200

Radius of each ball “r” = 0.7cm

we take π =

Density of steel = 7.95 g/cm^{3}

TO FIND : Mass of 200 balls = ?

PROCEDURE :

We know that, volume of sphere "V” = × π r^{3}

Putting the values in the above formula, we get

V = × × (0.7)^{3}

= × × 0.343

= 1.437cm^{3}

This was the volume of one ball.

Now volume of 200 balls “V’ ” = n× V = 200 × 1.437cm^{3}

V’ = 287.466 cm^{3}

Now, it is given that, Density of steel = 7.95 g/cm^{3}

So, mass of the balls = volume × density

Mass of 200 balls = volume of 200 balls × density

= 287.466 cm^{3} × 7.95 g/cm^{3}

= 2285.46g

= 2.285 Kg

∴ Mass of 200 balls bearings is 2.285 Kg.

**Question 17.**

The outer and the inner radii of a hollow sphere are 12 cm and 10 cm. Find its volume.

**Answer:**

GIVEN : outer radius of hollow the sphere “R” = 12cm

inner radius of hollow the sphere “r” = 10cm

we take π =

TO FIND : volume of the hollow sphere = ?

PROCEDURE :

We know that volume of the hollow sphere “V” = × π × (R^{3} – r^{3})

Putting all the values in the above formula, we get

V = ×× (12^{3} – 10^{3})

= ××(1728-1000)

= ××728

= 3050.66 cm^{3}

∴ volume of the hollow sphere is 3050.66 cm^{3}

**Question 18.**

The volume of a solid hemisphere is 1152 πcu.cm. Find its curved surface area.

**Answer:**

GIVEN : volume of a solid hemisphere “V” = 1152cu.cm

we take π =

TO FIND : curved surface area (CSA) = ?

PROCEDURE :

As we know that volume of a hemisphere “V” = × × π ×r^{3}

So, putting all the values in the above formula, we get

V = × × π ×r^{3}

1152 = × × π ×r^{3}

1152 = × × r^{3}

r^{3} = 1152 × 2 × = 1152 × = 576×3 = 1728

r = ∛ 1728 = 12cm

so, now that we have radius, we can find the curved surface area of the hemisphere.

Now, CSA = 2× π ×r^{2}

= 2×π × (12)^{2}

= 2×π ×144

= 288π cm^{2}

∴ curved surface area of the hemisphere is 288π cm^{2}

**Question 19.**

Find the volume of the largest right circular cone that can be cut out of a cube whose edge is 14 cm.

**Answer:**

GIVEN : side of cube ”s” = 14cm

we take π =

TO FIND : volume of the largest right circular cone that can be cut out of a cube = ?

PROCEDURE :

For the cone to largest, its diameter should be equal to side of the cube ie. d = 14cm

And, height of the cone should also be equal to the side of the cube ie. h = 14cm

Now, radius of the cone “r” = 7cm

Now, volume of cone “V” = × π r^{2}h

Now putting all the values in the above formula, we get

V = × π × (7)^{2} × 14

= × × 7 × 7 × 14

= × 22 × 7× 14

= 718.666cm^{3}

∴ volume of the largest right circular cone that can be cut out of a cube is 718.666cm^{3}.

**Question 20.**

The radius of a spherical balloon increases from 7 cm to 14 cm as air is being pumped into it. Find the ratio of volumes of the balloon in the two cases.

**Answer:**

GIVEN : radius of balloon before the increase “r” = 7cm

radius of balloon after the increase “R” = 14cm

TO FIND : ratio of volumes of the balloon in both cases

PROCEDURE :

Volume of balloon before the increase “V1” = ×π ×r^{3}

Volume of balloon after the increase “V2” = ×π ×R^{3}

Ratio of the volumes in the 2 cases =

= = = = = =

∴ the ratio of volumes of the balloon in the two cases is = ie. 1 : 8.

###### Exercise 8.3

**Question 1.**A play-top is in the form of a hemisphere surmounted on a cone. The diameter of the hemisphere is 3.6 cm. The total height of the play-top is 4.2 cm. Find its total surface area.

**Answer:**A play-top is made up of a hemisphere (at the top) and cone(at the bottom).

Given, Diameter of the hemisphere = 3.6 cm

Radius of the hemisphere =

= 1.8 cm

Total height of the play-top = 4.2 cm

Height of cone = (Total height of the play-top)-( Radius of the hemisphere)

⇒ Height of cone = 4.2-1.8

= 2.4 cm

Now, Total surface area of play-top = Lateral surface area of hemisphere+ Lateral surface area of cone

Lateral surface area of hemisphere = 2r^{2}

= 21.81.8

= 6.48

Lateral surface area of cone = rl ,

where l =

⇒ l =

⇒ l =

⇒ l =

⇒ l = 3

So, Lateral surface area of cone = 1.83

= 5.4

Total surface area of play-top = 6.48+5.4

= 11.88 cm^{2}

Total surface area is 11.88 cm^{2}.

**Question 2.**A solid is in the shape of a cylinder surmounted on a hemisphere. If the diameter and the total height of the solid are 21 cm, 25.5 cm respectively, then find its volume.

**Answer:**Given, Diameter = 21 cm

Total height of the solid = 25.5 cm

Radius =

= 10.5 cm

Height of cylinder = 25.5-10.5

= 15 cm

Volume of cylinder = r^{2}h

= 10.510.515

= 1653.75

Volume of hemisphere = r^{3}

=

= 771.75

Volume of solid = Volume of cylinder+ Volume of hemisphere

= 1653.75 +771.75

= 2425.5

= 7616.07 cm^{3}

Volume of solid is 7616.07 cm^{3}

**Question 3.**A capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends. If the length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm, find its surface area.

**Answer:**Given, Diameter of the capsule = 5 mm

Length of the entire capsule = 14 mm

Radius of the capsule = mm

= 2.5 mm

Height of cylinder = 14-5

= 9 mm

Lateral surface area of cylinder = 2rh

= 22.59

= 45 mm^{2}

Lateral surface area of hemisphere = 2r^{2}

= 22.52.5

= 12.5 mm^{2}

Surface area of the solid = 12.5 +45 +12.5

= 70

= 70

= 220 mm^{2}

Surface area of capsule is 220 mm^{2}.

**Question 4.**A tent is in the shape of a right circular cylinder surmounted by a cone. The total height and the diameter of the base are 13.5 m and 28 m. If the height of the cylindrical portion is 3 m, find the total surface area of the tent.

**Answer:**Given, Total height = 13.5 m

Diameter of the base = 28 m

Height of the cylindrical portion = 3 m

Height of cone = 13.5-3

= 10.5 m

Radius of the base =

= 14 m

Lateral surface area of cylinder = 2rh

= 2143

= 84 m^{2}

Lateral surface area of cone = rl ,

where l =

⇒ l =

⇒ l =

⇒ l =

⇒ l = 17.5

So, Lateral surface area of cone = 1417.5

= 245 m^{2}

Total surface area = Lateral surface area of cylinder+ Lateral surface area of cone

Total surface area = 84 +245

= 329

= 329

= 1034 m^{2}

Total surface area of the tent is 1034 m^{2}.

**Question 5.**Using clay, a student made a right circular cone of height 48 cm and base radius 12 cm. Another student reshapes it in the form of a sphere. Find the radius of the sphere.

**Answer:**Given, height of cone(h) = 48 cm

Base radius(r) = 12 cm

Volume of cone = r^{2}h

= 121248

= 2304

We know,

Volume of sphere = r^{3}

To find radius of sphere,

r^{3} = 2304

⇒ r^{3} = 2304

⇒ r^{3} = 5763

⇒ r^{3} = 1728

⇒ r = 12 cm

Radius of the sphere is 12 cm.

**Question 6.**The radius of a solid sphere is 24 cm. It is melted and drawn into a long wire of uniform cross section. Find the length of the wire if its radius is 1.2 mm.

**Answer:**Given, Radius of a solid sphere(r) = 24 cm

Radius of wire (R) = 1.2 mm = 0.12 cm

length of the wire(l) = ?

Solid sphere is melted and drawn into a long wire of uniform cross section. Thus,

Volume of sphere = Volume of wire(cylinder)

⇒ r^{3} = r^{2}l

⇒ (24)^{3} = (0.12)^{2}l

⇒ (24)^{3} = (0.12)^{2}l

⇒ l =

⇒ l =

⇒ l = 128100100

⇒ l = 12.8 km

The length of the wire is 12.8 km.

**Question 7.**A right circular conical vessel whose internal radius is 5 cm and height is 24 cm is full of water. The water is emptied into an empty cylindrical vessel with internal radius 10 cm. Find the height of the water level in the cylindrical vessel.

**Answer:**Given, Radius of right circular conical vessel(r) = 5 cm

Height of right circular conical vessel(h) = 24 cm

Radius of cylinder(R) = 10 cm

Height of cylinder(H) = ?

As water is emptied into an empty cylindrical vessel,

Volume of right circular cone = Volume of cylinder

⇒ r^{2}h = R^{2}H

⇒ (5)^{2}24 = (10)^{2}H

⇒ H =

⇒ H = 2 cm

Height of the water level in the cylindrical vessel is 2 cm.

**Question 8.**A solid sphere of diameter 6 cm is dropped into a right circular cylindrical vessel with diameter 12 cm, which is partly filled with water. If the sphere is completely submerged in water, how much does the water level in the cylindrical vessel increase?

**Answer:**Given, Diameter of the sphere(d) = 6cm

Radius of the sphere (r) = = 3cm

Diameter of the cylindrical vessel(D) = 12 cm

radius of the cylindrical vessel (R) = = 6 cm

Let level of water raised be h cm.

Volume of the water raised in the vessel = Volume of the sphere

⇒ R^{2}h = r^{3}

⇒ h =

⇒ h =

⇒ h = 1 cm

Level of water raised in the vessel is 1 cm

**Question 9.**Through a cylindrical pipe of internal radius 7 cm, water flows out at the rate of 5 cm/sec. Calculate the volume of water (in litres) discharged through the pipe in half an hour.

**Answer:**We need to find volume of water (in liters) discharged through the pipe in half an hour.

Half an hour = 30 minutes

Let us change minutes into seconds. For that we have to multiply 30 by 60. Because 60 seconds = 1 minute.

⇒ 30 60 = 1800 seconds.

Volume of water discharged in half an hour = r^{2} 5 1800

= (7)^{2} 5 1800

= 7 7 5 1800

= 22 7 5 1800

= 1386000 cubic cm ()

= (1000 cm^{3} = 1 liter)

= 1386 liters

Volume of water discharged through the pipe in half an hour is 1386 liters.

**Question 10.**Water in a cylindrical tank of diameter 4 m and height 10 m is released through a cylindrical pipe of diameter 10 cm at the rate of 2.5 Km/hr. How much time will it take to empty the half of the tank? Assume that the tank is full of water to begin with.

**Answer:**Given,

Diameter of the cylindrical tank = 4 m

Radius of the cylindrical tank = 2 m

Height of the tank = 10 m

Diameter of the cylindrical pipe = 10 cm

Radius of the cylindrical pipe = = 5 cm

= 0.05 m

speed of water = 2.5 km/hr

= 2.5 x 1000 (1000 m = 1 km)

= 2500 m/hr

⇒ Volume of water discharged from the cylindrical pipe = Volume of cylindrical tank

⇒ Area of cross section time speed = r^{2} h

⇒ r^{2} time speed = r^{2} h

⇒ ()^{2} time 2500 = (2)^{2} (10)

⇒ () () time 2500 = (2) 2 (10)

⇒ Time = 2 10 () () ()

⇒ Time =

= 3.2 hour

= 3 hours 0.260 mins(1 hour = 60 minute)

= 3 hour 12 minute

Time taken to empty the half of the tank is 3 hour 12 minutes.

**Question 11.**A spherical solid material of radius 18 cm is melted and recast into three small solid spherical spheres of different sizes. If the radii of two spheres are 2cm and 12 cm, find the radius of the third sphere.

**Answer:**Volume of sphere = r^{3}

⇒ (18)^{3} = (2)^{3} + (12)^{3} + r^{3}

⇒ (18)^{3} = [(2)^{3} + (12)^{3} + r^{3} ]

⇒ (18)^{3} = [(2)^{3} + (12)^{3} + r^{3} ]

⇒ 5832 = 8 + 1728 + r^{3}

⇒ 5832 = 1736 + r^{3}

⇒ 5832-1736 = r^{3}

⇒ 4096 = r^{3}

⇒ r =

⇒ r = 16 cm

Radius of the third sphere is 16 cm.

**Question 12.**A hollow cylindrical pipe is of length 40 cm. Its internal and external radii are 4 cm and 12 cm respectively. It is melted and cast into a solid cylinder of length 20 cm. Find the radius of the new solid.

**Answer:**Given, Length of cylindrical pipe (h) = 40 cm

Internal radius of the pipe (r) = 4 cm

External radius of the pipe (R) = 12 cm

Height of cylinder = 20 cm

Volume of hollow cylindrical pipe = Volume of cylinder

⇒ h (R^{2} - r^{2}) = r^{2} h

⇒ h (R^{2} - r^{2}) = r^{2} h

⇒ (40) (12^{2} - 4^{2}) = r^{2} (20)

⇒ (40) (144 - 16) = r^{2} (20)

⇒ (40) = r^{2}

⇒ r^{2} = (40)

⇒ r^{2} = 2 128

⇒ r^{2} = 256

⇒ r =

⇒ r = 16 cm

Therefore radius of cylinder = 16 cm

**Question 13.**An iron right circular cone of diameter 8 cm and height 12 cm is melted and recast into spherical lead shots each of radius 4 mm. How many lead shots can be made?

**Answer:**Let the number of lead shots be n.

Given,

Diameter of right circular cone = 8 cm

Radius of right circular cone (r) = 4 cm

Height of right circular cone (h) = 12 cm

Radius of spherical lead shot (r) = 4 mm

= cm (10 mm = 1 cm)

Volume of right circular cone = nNumber of spherical lead shots

n =

⇒ n =

⇒ n = (4)^{2} (12) ()^{3}

⇒ n = 4 4 12

⇒ n = 3 5 5 10

⇒ n = 750 lead shots

Therefore 750 lead shots can be made.

**Question 14.**A right circular cylinder having diameter 12 cm and height 15 cm is full of ice cream. The ice cream is to be filled in cones of height 12 cm and diameter 6 cm, having a hemispherical shape on top. Find the number of such cones which can be filled with the ice cream available.

**Answer:**Given, Diameter of right circular cylinder = 12 cm

Radius of right circular cylinder (r_{1}) =

= 6 cm

Height of right circular cylinder (h_{1}) = 15 cm

Volume of Cylindrical ice-cream container = r_{1}^{2}h_{1}

= 6 6 15

= cm^{3}

Diameter of cone = 6 cm

Radius of cone (r_{2}) =

= 3 cm

Height of cone (h_{2}) = 12 cm

Radius of hemisphere = radius of cone = 3 cm

⇒ Volume of cone full of ice-cream = volume of cone + volume of hemisphere

= r_{2}^{2}h_{2} + r_{2}^{3}

= ( r_{2}^{2}h_{2} + 2r_{2}^{3})

= (3^{2}× 12 + 2× 3^{3})

= (9 ×12 + 2 × 27)

= (108 +54)

= 162

= cm^{3}

Let n be the number of cones full of ice cream.

⇒ Volume of Cylindrical ice-cream container = nVolume of one cone full with ice cream

= n

⇒ N =

⇒ n = 10

Hence, the required number of cones = 10

**Question 15.**A container with a rectangular base of length 4.4 m and breadth 2 m is used to collect rain water. The height of the water level in the container is 4 cm and the water is transferred into a cylindrical vessel with radius 40 cm. What will be the height of the water level in the cylinder?

**Answer:**Given,

Length of container(l) = 4.4 m

Breadth of container(b) = 2 m

Height of container(h) = 4 m

Radius of cylindrical vessel(r) = 40 cm

⇒ Volume of container = Volume of cylindrical vessel

⇒ lbh = r^{2}h

⇒ 4.424 = (40)^{2}h

⇒ H =

⇒ H = 70 cm

The height of the water level in the cylinder is 70 cm.

**Question 16.**A cylindrical bucket of height 32 cm and radius 18 cm is filled with sand. The bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 24 cm, find the radius and slant height of the heap.

**Answer:**Given, Height of cylindrical bucket(h) = 32 cm

Radius of cylindrical bucket(r) = 18 cm

Height of conical heap (H) = 24 cm

Radius of conical heap (R) = ?

⇒ Volume of cylindrical bucket = Volume of conical heap

⇒ r^{2}h = R^{2}H

⇒ 181832 = R^{2}24

⇒ 181832 = R^{2}8

⇒ R^{2} =

⇒ R^{2} = 18184

⇒ R =

⇒ R = 182

⇒ R = 36 cm

Slant height (L) =

L =

L =

L =

L = 12 cm

Slant height = 12 cm

Radius of conical heap = 36 cm

**Question 17.**A cylindrical shaped well of depth 20 m and diameter 14 m is dug. The dug out soil is evenly spread to form a cuboid-platform with base dimension 20 m x 14 m. Find the height of the platform.

**Answer:**Given, Depth of well = Height of well = 20 m

Diameter of well = 14 m

Radius of well =

= 7 m

Volume of well = r^{2}h

= 7720

= 3080 m^{3}

⇒ Volume of platform = Volume of well

⇒ lbh = 3080

⇒ 2014h = 3080

⇒ H =

⇒ H = 11 m

The height of the platform is 11 m.

**Question 1.**

A play-top is in the form of a hemisphere surmounted on a cone. The diameter of the hemisphere is 3.6 cm. The total height of the play-top is 4.2 cm. Find its total surface area.

**Answer:**

A play-top is made up of a hemisphere (at the top) and cone(at the bottom).

Given, Diameter of the hemisphere = 3.6 cm

Radius of the hemisphere =

= 1.8 cm

Total height of the play-top = 4.2 cm

Height of cone = (Total height of the play-top)-( Radius of the hemisphere)

⇒ Height of cone = 4.2-1.8

= 2.4 cm

Now, Total surface area of play-top = Lateral surface area of hemisphere+ Lateral surface area of cone

Lateral surface area of hemisphere = 2r^{2}

= 21.81.8

= 6.48

Lateral surface area of cone = rl ,

where l =

⇒ l =

⇒ l =

⇒ l =

⇒ l = 3

So, Lateral surface area of cone = 1.83

= 5.4

Total surface area of play-top = 6.48+5.4

= 11.88 cm^{2}

Total surface area is 11.88 cm^{2}.

**Question 2.**

A solid is in the shape of a cylinder surmounted on a hemisphere. If the diameter and the total height of the solid are 21 cm, 25.5 cm respectively, then find its volume.

**Answer:**

Given, Diameter = 21 cm

Total height of the solid = 25.5 cm

Radius =

= 10.5 cm

Height of cylinder = 25.5-10.5

= 15 cm

Volume of cylinder = r^{2}h

= 10.510.515

= 1653.75

Volume of hemisphere = r^{3}

=

= 771.75

Volume of solid = Volume of cylinder+ Volume of hemisphere

= 1653.75 +771.75

= 2425.5

= 7616.07 cm^{3}

Volume of solid is 7616.07 cm^{3}

**Question 3.**

A capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends. If the length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm, find its surface area.

**Answer:**

Given, Diameter of the capsule = 5 mm

Length of the entire capsule = 14 mm

Radius of the capsule = mm

= 2.5 mm

Height of cylinder = 14-5

= 9 mm

Lateral surface area of cylinder = 2rh

= 22.59

= 45 mm^{2}

Lateral surface area of hemisphere = 2r^{2}

= 22.52.5

= 12.5 mm^{2}

Surface area of the solid = 12.5 +45 +12.5

= 70

= 70

= 220 mm^{2}

Surface area of capsule is 220 mm^{2}.

**Question 4.**

A tent is in the shape of a right circular cylinder surmounted by a cone. The total height and the diameter of the base are 13.5 m and 28 m. If the height of the cylindrical portion is 3 m, find the total surface area of the tent.

**Answer:**

Given, Total height = 13.5 m

Diameter of the base = 28 m

Height of the cylindrical portion = 3 m

Height of cone = 13.5-3

= 10.5 m

Radius of the base =

= 14 m

Lateral surface area of cylinder = 2rh

= 2143

= 84 m^{2}

Lateral surface area of cone = rl ,

where l =

⇒ l =

⇒ l =

⇒ l =

⇒ l = 17.5

So, Lateral surface area of cone = 1417.5

= 245 m^{2}

Total surface area = Lateral surface area of cylinder+ Lateral surface area of cone

Total surface area = 84 +245

= 329

= 329

= 1034 m^{2}

Total surface area of the tent is 1034 m^{2}.

**Question 5.**

Using clay, a student made a right circular cone of height 48 cm and base radius 12 cm. Another student reshapes it in the form of a sphere. Find the radius of the sphere.

**Answer:**

Given, height of cone(h) = 48 cm

Base radius(r) = 12 cm

Volume of cone = r^{2}h

= 121248

= 2304

We know,

Volume of sphere = r^{3}

To find radius of sphere,

r^{3} = 2304

⇒ r^{3} = 2304

⇒ r^{3} = 5763

⇒ r^{3} = 1728

⇒ r = 12 cm

Radius of the sphere is 12 cm.

**Question 6.**

The radius of a solid sphere is 24 cm. It is melted and drawn into a long wire of uniform cross section. Find the length of the wire if its radius is 1.2 mm.

**Answer:**

Given, Radius of a solid sphere(r) = 24 cm

Radius of wire (R) = 1.2 mm = 0.12 cm

length of the wire(l) = ?

Solid sphere is melted and drawn into a long wire of uniform cross section. Thus,

Volume of sphere = Volume of wire(cylinder)

⇒ r^{3} = r^{2}l

⇒ (24)^{3} = (0.12)^{2}l

⇒ (24)^{3} = (0.12)^{2}l

⇒ l =

⇒ l =

⇒ l = 128100100

⇒ l = 12.8 km

The length of the wire is 12.8 km.

**Question 7.**

A right circular conical vessel whose internal radius is 5 cm and height is 24 cm is full of water. The water is emptied into an empty cylindrical vessel with internal radius 10 cm. Find the height of the water level in the cylindrical vessel.

**Answer:**

Given, Radius of right circular conical vessel(r) = 5 cm

Height of right circular conical vessel(h) = 24 cm

Radius of cylinder(R) = 10 cm

Height of cylinder(H) = ?

As water is emptied into an empty cylindrical vessel,

Volume of right circular cone = Volume of cylinder

⇒ r^{2}h = R^{2}H

⇒ (5)^{2}24 = (10)^{2}H

⇒ H =

⇒ H = 2 cm

Height of the water level in the cylindrical vessel is 2 cm.

**Question 8.**

A solid sphere of diameter 6 cm is dropped into a right circular cylindrical vessel with diameter 12 cm, which is partly filled with water. If the sphere is completely submerged in water, how much does the water level in the cylindrical vessel increase?

**Answer:**

Given, Diameter of the sphere(d) = 6cm

Radius of the sphere (r) = = 3cm

Diameter of the cylindrical vessel(D) = 12 cm

radius of the cylindrical vessel (R) = = 6 cm

Let level of water raised be h cm.

Volume of the water raised in the vessel = Volume of the sphere

⇒ R^{2}h = r^{3}

⇒ h =

⇒ h =

⇒ h = 1 cm

Level of water raised in the vessel is 1 cm

**Question 9.**

Through a cylindrical pipe of internal radius 7 cm, water flows out at the rate of 5 cm/sec. Calculate the volume of water (in litres) discharged through the pipe in half an hour.

**Answer:**

We need to find volume of water (in liters) discharged through the pipe in half an hour.

Half an hour = 30 minutes

Let us change minutes into seconds. For that we have to multiply 30 by 60. Because 60 seconds = 1 minute.

⇒ 30 60 = 1800 seconds.

Volume of water discharged in half an hour = r^{2} 5 1800

= (7)^{2} 5 1800

= 7 7 5 1800

= 22 7 5 1800

= 1386000 cubic cm ()

= (1000 cm^{3} = 1 liter)

= 1386 liters

Volume of water discharged through the pipe in half an hour is 1386 liters.

**Question 10.**

Water in a cylindrical tank of diameter 4 m and height 10 m is released through a cylindrical pipe of diameter 10 cm at the rate of 2.5 Km/hr. How much time will it take to empty the half of the tank? Assume that the tank is full of water to begin with.

**Answer:**

Given,

Diameter of the cylindrical tank = 4 m

Radius of the cylindrical tank = 2 m

Height of the tank = 10 m

Diameter of the cylindrical pipe = 10 cm

Radius of the cylindrical pipe = = 5 cm

= 0.05 m

speed of water = 2.5 km/hr

= 2.5 x 1000 (1000 m = 1 km)

= 2500 m/hr

⇒ Volume of water discharged from the cylindrical pipe = Volume of cylindrical tank

⇒ Area of cross section time speed = r^{2} h

⇒ r^{2} time speed = r^{2} h

⇒ ()^{2} time 2500 = (2)^{2} (10)

⇒ () () time 2500 = (2) 2 (10)

⇒ Time = 2 10 () () ()

⇒ Time =

= 3.2 hour

= 3 hours 0.260 mins(1 hour = 60 minute)

= 3 hour 12 minute

Time taken to empty the half of the tank is 3 hour 12 minutes.

**Question 11.**

A spherical solid material of radius 18 cm is melted and recast into three small solid spherical spheres of different sizes. If the radii of two spheres are 2cm and 12 cm, find the radius of the third sphere.

**Answer:**

Volume of sphere = r^{3}

⇒ (18)^{3} = (2)^{3} + (12)^{3} + r^{3}

⇒ (18)^{3} = [(2)^{3} + (12)^{3} + r^{3} ]

⇒ (18)^{3} = [(2)^{3} + (12)^{3} + r^{3} ]

⇒ 5832 = 8 + 1728 + r^{3}

⇒ 5832 = 1736 + r^{3}

⇒ 5832-1736 = r^{3}

⇒ 4096 = r^{3}

⇒ r =

⇒ r = 16 cm

Radius of the third sphere is 16 cm.

**Question 12.**

A hollow cylindrical pipe is of length 40 cm. Its internal and external radii are 4 cm and 12 cm respectively. It is melted and cast into a solid cylinder of length 20 cm. Find the radius of the new solid.

**Answer:**

Given, Length of cylindrical pipe (h) = 40 cm

Internal radius of the pipe (r) = 4 cm

External radius of the pipe (R) = 12 cm

Height of cylinder = 20 cm

Volume of hollow cylindrical pipe = Volume of cylinder

⇒ h (R^{2} - r^{2}) = r^{2} h

⇒ h (R^{2} - r^{2}) = r^{2} h

⇒ (40) (12^{2} - 4^{2}) = r^{2} (20)

⇒ (40) (144 - 16) = r^{2} (20)

⇒ (40) = r^{2}

⇒ r^{2} = (40)

⇒ r^{2} = 2 128

⇒ r^{2} = 256

⇒ r =

⇒ r = 16 cm

Therefore radius of cylinder = 16 cm

**Question 13.**

An iron right circular cone of diameter 8 cm and height 12 cm is melted and recast into spherical lead shots each of radius 4 mm. How many lead shots can be made?

**Answer:**

Let the number of lead shots be n.

Given,

Diameter of right circular cone = 8 cm

Radius of right circular cone (r) = 4 cm

Height of right circular cone (h) = 12 cm

Radius of spherical lead shot (r) = 4 mm

= cm (10 mm = 1 cm)

Volume of right circular cone = nNumber of spherical lead shots

n =

⇒ n =

⇒ n = (4)^{2} (12) ()^{3}

⇒ n = 4 4 12

⇒ n = 3 5 5 10

⇒ n = 750 lead shots

Therefore 750 lead shots can be made.

**Question 14.**

A right circular cylinder having diameter 12 cm and height 15 cm is full of ice cream. The ice cream is to be filled in cones of height 12 cm and diameter 6 cm, having a hemispherical shape on top. Find the number of such cones which can be filled with the ice cream available.

**Answer:**

Given, Diameter of right circular cylinder = 12 cm

Radius of right circular cylinder (r_{1}) =

= 6 cm

Height of right circular cylinder (h_{1}) = 15 cm

Volume of Cylindrical ice-cream container = r_{1}^{2}h_{1}

= 6 6 15

= cm^{3}

Diameter of cone = 6 cm

Radius of cone (r_{2}) =

= 3 cm

Height of cone (h_{2}) = 12 cm

Radius of hemisphere = radius of cone = 3 cm

⇒ Volume of cone full of ice-cream = volume of cone + volume of hemisphere

= r_{2}^{2}h_{2} + r_{2}^{3}

= ( r_{2}^{2}h_{2} + 2r_{2}^{3})

= (3^{2}× 12 + 2× 3^{3})

= (9 ×12 + 2 × 27)

= (108 +54)

= 162

= cm^{3}

Let n be the number of cones full of ice cream.

⇒ Volume of Cylindrical ice-cream container = nVolume of one cone full with ice cream

= n

⇒ N =

⇒ n = 10

Hence, the required number of cones = 10

**Question 15.**

A container with a rectangular base of length 4.4 m and breadth 2 m is used to collect rain water. The height of the water level in the container is 4 cm and the water is transferred into a cylindrical vessel with radius 40 cm. What will be the height of the water level in the cylinder?

**Answer:**

Given,

Length of container(l) = 4.4 m

Breadth of container(b) = 2 m

Height of container(h) = 4 m

Radius of cylindrical vessel(r) = 40 cm

⇒ Volume of container = Volume of cylindrical vessel

⇒ lbh = r^{2}h

⇒ 4.424 = (40)^{2}h

⇒ H =

⇒ H = 70 cm

The height of the water level in the cylinder is 70 cm.

**Question 16.**

A cylindrical bucket of height 32 cm and radius 18 cm is filled with sand. The bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 24 cm, find the radius and slant height of the heap.

**Answer:**

Given, Height of cylindrical bucket(h) = 32 cm

Radius of cylindrical bucket(r) = 18 cm

Height of conical heap (H) = 24 cm

Radius of conical heap (R) = ?

⇒ Volume of cylindrical bucket = Volume of conical heap

⇒ r^{2}h = R^{2}H

⇒ 181832 = R^{2}24

⇒ 181832 = R^{2}8

⇒ R^{2} =

⇒ R^{2} = 18184

⇒ R =

⇒ R = 182

⇒ R = 36 cm

Slant height (L) =

L =

L =

L =

L = 12 cm

Slant height = 12 cm

Radius of conical heap = 36 cm

**Question 17.**

A cylindrical shaped well of depth 20 m and diameter 14 m is dug. The dug out soil is evenly spread to form a cuboid-platform with base dimension 20 m x 14 m. Find the height of the platform.

**Answer:**

Given, Depth of well = Height of well = 20 m

Diameter of well = 14 m

Radius of well =

= 7 m

Volume of well = r^{2}h

= 7720

= 3080 m^{3}

⇒ Volume of platform = Volume of well

⇒ lbh = 3080

⇒ 2014h = 3080

⇒ H =

⇒ H = 11 m

The height of the platform is 11 m.

###### Exercise 8.4

**Question 1.**The curved surface area of a right circular cylinder of radius 1 cm and height 1 cm is equal to

A. πcm^{2}

B. 2πcm^{2}

C. 3π cm^{3}

D. 2 cm^{2}

**Answer:**Radius (R) = 1 cm

Height (H) = 1 cm

Curved Surface Area = (2πRH) = 2π cm^{2}

Out of all the four Options,

Correct Option: B

**Question 2.**The total surface area of a solid right circular cylinder whose radius is half of its height h is equal to

A. sq. units

B. sq. units

C. sq. units

D. sq. units

**Answer:**Total Surface Area = 2πrh + 2πr^{2}

where r is the radius and h is the height

According to the problem r

Total surface area

⇒ Total surface area

⇒ Total surface area

Out of all the four Options,

Correct Option : C

**Question 3.**Base area of a right circular cylinder is 80 cm^{2}. If its height is 5 cm, then the volume is equal to

A. 400 cm^{3}

B. 16 cm^{3}

C. 200 cm^{3}

D.

**Answer:**Base Area (A) = 80 cm^{2}

Height (H) = 5 cm

Volume = (A×H) = 400 cm^{3}

Out of all the four Options,

Correct Option : A

**Question 4.**If the total surface area a solid right circular cylinder is 200 πcm^{2}and its radius is 5 cm, then the sum of its height and radius is

A. 20 cm

B. 25 cm

C. 30 cm

D. 15 cm

**Answer:**Total Surface Area = 2πrh + 2πr^{2}

where r is the radius and h is the height

Total Surface Area = 200π cm^{2} (Given)

Radius (r) = 5 cm

Equating the Total Surface Area we can say

2πrh + 2πr^{2} = 200π

⇒ rh + r^{2} = 100

⇒ r(r + h) = 100

⇒ r + h = 20 cm

Out of all the four Options,

Correct Option : A

**Question 5.**The curved surface area of a right circular cylinder whose radius is a units and height is b units, is equal to

A. πa^{2}b sq.cm

B. 2πab

C. 2π sq.cm

D. 2 sq.cm

**Answer:**Radius (R) = a units

Height (H) = b units

Curved Surface Area = (2πRH) = 2πab sq. units

Out of all the four Options,

Correct Option: B

**Question 6.**Radius and height of a right circular cone and that of a right circular cylinder are respectively, equal. If the volume of the cylinder is 120 cm^{3}, then the volume of the cone is equal to

A. 1200 cm^{3}

B. 360 cm^{3}

C. 40 cm^{3}

D. 90 cm^{3}

**Answer:**Volume of Cylinder = 3 × Volume of Cone

Volume of the cylinder = 120 cm^{3} (Given)

⇒ Volume of Cone

⇒ Volume of Cone = 40 cm^{3}

Out of all the four Options,

Correct Option : C

**Question 7.**If the diameter and height of a right circular cone are 12 cm and 8 cm respectively, then the slant height is

A. 10 cm

B. 20 cm

C. 30 cm

D. 96 cm

**Answer:**Diameter = 12 cm.

⇒ Radius (R) = 6 cm.

Height (H) = 8 cm

Slant height (l) = √(R^{2} + H^{2}) = √(6^{2} + 8^{2})

⇒ Slant height (l) = √100 = 10 cm

Out of all the four Options,

Correct Option : A

**Question 8.**If the circumference at the base of a right circular cone and the slant height are 120 πcm and 10 cm respectively, then the curved surface area of the cone is equal to

A. 1200πcm^{2}

B. 600πcm^{2}

C. 300πcm^{2}

D. 600 cm^{2}

**Answer:**Base Circumference = 120π cm

Let radius be r

Base Circumference = 2πr

2πr = 120π

⇒ r = 60 cm

Slant Height (l) = 10 cm

Curved Surface Area = π×r×l

Curved Surface Area = 600π cm^{2}

Out of all the four Options,

Correct Option :B

**Question 9.**If the volume and the base area of a right circular cone are 48πcm^{3} and 12πcm^{2}respectively, then the height of the cone is equal to

A. 6 cm

B. 8 cm

C. 10 cm

D. 12 cm

**Answer:**Volume of cone = 48π cm^{3}

Base Area = 12π cm^{2}

Volume of cone

Height

⇒ Height

⇒ Height = 12 cm.

Out of all the four Options,

Correct Option :D

**Question 10.**If the height and the base area of a right circular cone are 5 cm and 48 sq. cm respectively, then the volume of the cone is equal to

A. 240 cm^{3}

B. 120 cm^{3}

C. 80 cm^{3}

D. 480 cm^{3}

**Answer:**Base Area (A) = 48 cm^{2}

Height (H) = 5 cm

Volume

⇒ Volume

⇒ Volume = 80 cm^{3}

Out of all the four Options,

Correct Option : C

**Question 11.**The ratios of the respective heights and the respective radii of two cylinders are 1:2 and 2:1 respectively. Then their respective volumes are in the ratio

A. 4 : 1

B. 1 : 4

C. 2 : 1

D. 1 : 2

**Answer:**Let R_{1},R_{2} be the radius and H_{1},H_{2} be the height V_{1},V_{2} of the two cylinders

Volume Of Cylinder = πR^{2}H

So we can say,

Out of all the four Options,

Correct Option : C

**Question 12.**If the radius of a sphere is 2 cm, then the curved surface area of the sphere is equal to

A. 8π cm^{2}

B. 16 cm^{2}

C. 12π cm^{2}

D. 16π cm^{2}

**Answer:**Radius of sphere (r) = 2 cm

Surface Area of Sphere = 4πr^{2}

⇒ Surface Area of Sphere = 4π × 2^{2}

⇒ Surface Area of Sphere = 16π cm^{2}

Out of all the four Options,

Correct Option :D

**Question 13.**The total surface area of a solid hemisphere of diameter 2 cm is equal to

A. 12 cm^{2}

B. 12π cm^{2}

C. 4π cm^{2}

D. 3π cm^{2}

**Answer:**Diameter = 2 cm

Radius (r) = 1 cm

Total surface Area = Curved surface Area + Base Area

⇒ Total surface Area = 2πr^{2} + πr^{2} = 3πr^{2}

⇒ Total surface Area = 3π × 1^{2} = 3π cm^{2}

Out of all the four Options,

Correct Option : D

**Question 14.**If the volume of a sphere is cu. cm, then its radius is

A. cm

B. cm

C. cm

D. cm

**Answer:**Let radius of sphere be r cm

Volume of Sphere

Volume of Sphere

So we can say that

Out of all the four Options,

Correct Option : B

**Question 15.**The surface areas of two spheres are in the ratio of 9 : 25. Then their volumes are in the ratio

A. 81 : 625

B. 729 : 15625

C. 27 : 75

D. 27 : 125

**Answer:**Let R_{1},R_{2} be the radius of the two spheres and V_{1},V_{2} be the Volume of this two spheres, A_{1},A_{2} be the area of two spheres

Surface Area of Sphere = 4πr^{2}

So we can say,

…Equation (i)

Volume of Sphere

Putting the values from Equation (i) we get

Out of all the four Options,

Correct Option : D

**Question 16.**The total surface area of a solid hemisphere whose radius is a units, is equal to

A. 2πa^{2}sq. units

B. 3πa^{2} sq. units

C. 3πa sq. units

D. 3a^{2} sq.units

**Answer:**Radius = a units

Total surface Area = Curved surface Area + Base Area

⇒ Total surface Area = 2πa^{2} + πa^{2} = 3πa^{2} sq. units

Out of all the four Options,

Correct Option : B

**Question 17.**If the surface area of a sphere is 100π cm^{2}, then its radius is equal to

A. 25 cm

B. 100 cm

C. 5 cm

D. 10 cm

**Answer:**Let radius of sphere be r cm

Surface Area of Sphere = 4πr^{2}

Surface Area of Sphere = 100π cm^{2}

So we can say,

4πr^{2} = 100π

⇒ r^{2} = 25

⇒ r = 5 cm

Out of all the four Options,

Correct Option: C

**Question 18.**If the surface area of a sphere is 36π cm^{2}, then the volume of the sphere is equal to

A. 12π cm^{3}

B. 36π cm^{3}

C. 72π cm^{3}

D. 108π cm^{3}

**Answer:**Let radius of sphere be r cm

Surface Area of Sphere = 4πr^{2}

Surface Area of Sphere = 36π cm^{2}

So we can say,

4πr^{2} = 36π

⇒ r^{2} = 9

⇒ r = 3 cm

Volume of Sphere

⇒ Volume of Sphere

⇒ Volume of Sphere = 4π × 3^{2}

⇒ Volume of Sphere = 36π cm^{3}

Out of all the four Options,

Correct Option : B

**Question 19.**If the total surface area of a solid hemisphere is 12π cm^{2} then its curved surface area is equal to

A. 6π cm^{2}

B. 24π cm^{2}

C. 36π cm^{2}

D. 8π cm^{2}

**Answer:**Total surface Area = Curved surface Area + Base Area

⇒ Total surface Area = 2πr^{2} + πr^{2} = 3πr^{2}

Total surface area of a solid hemisphere = 12π cm^{2}(Given)

3πr^{2} = 12π

⇒ r^{2} = 4

⇒ r = 2cm

Curved surface Area = 2πr^{2}

Curved surface Area = 2π×2^{2}

Curved surface Area = 8π cm^{3}

Out of all the four Options,

Correct Option : D

**Question 20.**If the radius of a sphere is half of the radius of another sphere, then their respective volumes are in the ratio

A. 1 : 8

B. 2: 1

C. 1 : 2

D. 8 : 1

**Answer:**Let R_{1},R_{2} be the radius of the two spheres and V_{1},V_{2} be the Volume of this two spheres

Volume of Sphere

R_{2} = 2R_{1}

So we can say,

Out of all the four Options,

Correct Option :A

**Question 21.**Curved surface area of solid sphere is 24 cm^{2}. If the sphere is divided into two hemispheres, then the total surface area of one of the hemispheres is

A. 12 cm^{2}

B. 8 cm^{2}

C. 16 cm^{2}

D. 18 cm^{2}

**Answer:**Curved Surface Area = 24 cm^{2}

Let radius be r

Curved Surface Area of solid sphere = 4πr^{2}

4πr^{2} = 24

Total surface Area of hemisphere = 2πr^{2} + πr^{2} = 3πr^{2}

Total surface Area of hemisphere

Total surface Area of hemisphere = 18 cm^{2}

Out of all the four Options,

Correct Option :D

**Question 22.**Two right circular cones have equal radii. If their slant heights are in the ratio 4 : 3, then their respective curved surface areas are in the ratio

A. 16 : 9

B. 2 : 3

C. 4 : 3

D. 3 : 4

**Answer:**Curved Surface Area = πRL

where R is the radius and L is the slant height

Let L_{1},L_{2} be the slant heights of the two cone and A_{1},A_{2} be the area of this two cones

So we can say,

Out of all the four Options,

Correct Option : C

**Question 1.**

The curved surface area of a right circular cylinder of radius 1 cm and height 1 cm is equal to

A. πcm^{2}

B. 2πcm^{2}

C. 3π cm^{3}

D. 2 cm^{2}

**Answer:**

Radius (R) = 1 cm

Height (H) = 1 cm

Curved Surface Area = (2πRH) = 2π cm^{2}

Out of all the four Options,

Correct Option: B

**Question 2.**

The total surface area of a solid right circular cylinder whose radius is half of its height h is equal to

A. sq. units

B. sq. units

C. sq. units

D. sq. units

**Answer:**

Total Surface Area = 2πrh + 2πr^{2}

where r is the radius and h is the height

According to the problem r

Total surface area

⇒ Total surface area

⇒ Total surface area

Out of all the four Options,

Correct Option : C

**Question 3.**

Base area of a right circular cylinder is 80 cm^{2}. If its height is 5 cm, then the volume is equal to

A. 400 cm^{3}

B. 16 cm^{3}

C. 200 cm^{3}

D.

**Answer:**

Base Area (A) = 80 cm^{2}

Height (H) = 5 cm

Volume = (A×H) = 400 cm^{3}

Out of all the four Options,

Correct Option : A

**Question 4.**

If the total surface area a solid right circular cylinder is 200 πcm^{2}and its radius is 5 cm, then the sum of its height and radius is

A. 20 cm

B. 25 cm

C. 30 cm

D. 15 cm

**Answer:**

Total Surface Area = 2πrh + 2πr^{2}

where r is the radius and h is the height

Total Surface Area = 200π cm^{2} (Given)

Radius (r) = 5 cm

Equating the Total Surface Area we can say

2πrh + 2πr^{2} = 200π

⇒ rh + r^{2} = 100

⇒ r(r + h) = 100

⇒ r + h = 20 cm

Out of all the four Options,

Correct Option : A

**Question 5.**

The curved surface area of a right circular cylinder whose radius is a units and height is b units, is equal to

A. πa^{2}b sq.cm

B. 2πab

C. 2π sq.cm

D. 2 sq.cm

**Answer:**

Radius (R) = a units

Height (H) = b units

Curved Surface Area = (2πRH) = 2πab sq. units

Out of all the four Options,

Correct Option: B

**Question 6.**

Radius and height of a right circular cone and that of a right circular cylinder are respectively, equal. If the volume of the cylinder is 120 cm^{3}, then the volume of the cone is equal to

A. 1200 cm^{3}

B. 360 cm^{3}

C. 40 cm^{3}

D. 90 cm^{3}

**Answer:**

Volume of Cylinder = 3 × Volume of Cone

Volume of the cylinder = 120 cm^{3} (Given)

⇒ Volume of Cone

⇒ Volume of Cone = 40 cm^{3}

Out of all the four Options,

Correct Option : C

**Question 7.**

If the diameter and height of a right circular cone are 12 cm and 8 cm respectively, then the slant height is

A. 10 cm

B. 20 cm

C. 30 cm

D. 96 cm

**Answer:**

Diameter = 12 cm.

⇒ Radius (R) = 6 cm.

Height (H) = 8 cm

Slant height (l) = √(R^{2} + H^{2}) = √(6^{2} + 8^{2})

⇒ Slant height (l) = √100 = 10 cm

Out of all the four Options,

Correct Option : A

**Question 8.**

If the circumference at the base of a right circular cone and the slant height are 120 πcm and 10 cm respectively, then the curved surface area of the cone is equal to

A. 1200πcm^{2}

B. 600πcm^{2}

C. 300πcm^{2}

D. 600 cm^{2}

**Answer:**

Base Circumference = 120π cm

Let radius be r

Base Circumference = 2πr

2πr = 120π

⇒ r = 60 cm

Slant Height (l) = 10 cm

Curved Surface Area = π×r×l

Curved Surface Area = 600π cm^{2}

Out of all the four Options,

Correct Option :B

**Question 9.**

If the volume and the base area of a right circular cone are 48πcm^{3} and 12πcm^{2}respectively, then the height of the cone is equal to

A. 6 cm

B. 8 cm

C. 10 cm

D. 12 cm

**Answer:**

Volume of cone = 48π cm^{3}

Base Area = 12π cm^{2}

Volume of cone

Height

⇒ Height

⇒ Height = 12 cm.

Out of all the four Options,

Correct Option :D

**Question 10.**

If the height and the base area of a right circular cone are 5 cm and 48 sq. cm respectively, then the volume of the cone is equal to

A. 240 cm^{3}

B. 120 cm^{3}

C. 80 cm^{3}

D. 480 cm^{3}

**Answer:**

Base Area (A) = 48 cm^{2}

Height (H) = 5 cm

Volume

⇒ Volume

⇒ Volume = 80 cm^{3}

Out of all the four Options,

Correct Option : C

**Question 11.**

The ratios of the respective heights and the respective radii of two cylinders are 1:2 and 2:1 respectively. Then their respective volumes are in the ratio

A. 4 : 1

B. 1 : 4

C. 2 : 1

D. 1 : 2

**Answer:**

Let R_{1},R_{2} be the radius and H_{1},H_{2} be the height V_{1},V_{2} of the two cylinders

Volume Of Cylinder = πR^{2}H

So we can say,

Out of all the four Options,

Correct Option : C

**Question 12.**

If the radius of a sphere is 2 cm, then the curved surface area of the sphere is equal to

A. 8π cm^{2}

B. 16 cm^{2}

C. 12π cm^{2}

D. 16π cm^{2}

**Answer:**

Radius of sphere (r) = 2 cm

Surface Area of Sphere = 4πr^{2}

⇒ Surface Area of Sphere = 4π × 2^{2}

⇒ Surface Area of Sphere = 16π cm^{2}

Out of all the four Options,

Correct Option :D

**Question 13.**

The total surface area of a solid hemisphere of diameter 2 cm is equal to

A. 12 cm^{2}

B. 12π cm^{2}

C. 4π cm^{2}

D. 3π cm^{2}

**Answer:**

Diameter = 2 cm

Radius (r) = 1 cm

Total surface Area = Curved surface Area + Base Area

⇒ Total surface Area = 2πr^{2} + πr^{2} = 3πr^{2}

⇒ Total surface Area = 3π × 1^{2} = 3π cm^{2}

Out of all the four Options,

Correct Option : D

**Question 14.**

If the volume of a sphere is cu. cm, then its radius is

A. cm

B. cm

C. cm

D. cm

**Answer:**

Let radius of sphere be r cm

Volume of Sphere

Volume of Sphere

So we can say that

Out of all the four Options,

Correct Option : B

**Question 15.**

The surface areas of two spheres are in the ratio of 9 : 25. Then their volumes are in the ratio

A. 81 : 625

B. 729 : 15625

C. 27 : 75

D. 27 : 125

**Answer:**

Let R_{1},R_{2} be the radius of the two spheres and V_{1},V_{2} be the Volume of this two spheres, A_{1},A_{2} be the area of two spheres

Surface Area of Sphere = 4πr^{2}

So we can say,

…Equation (i)

Volume of Sphere

Putting the values from Equation (i) we get

Out of all the four Options,

Correct Option : D

**Question 16.**

The total surface area of a solid hemisphere whose radius is a units, is equal to

A. 2πa^{2}sq. units

B. 3πa^{2} sq. units

C. 3πa sq. units

D. 3a^{2} sq.units

**Answer:**

Radius = a units

Total surface Area = Curved surface Area + Base Area

⇒ Total surface Area = 2πa^{2} + πa^{2} = 3πa^{2} sq. units

Out of all the four Options,

Correct Option : B

**Question 17.**

If the surface area of a sphere is 100π cm^{2}, then its radius is equal to

A. 25 cm

B. 100 cm

C. 5 cm

D. 10 cm

**Answer:**

Let radius of sphere be r cm

Surface Area of Sphere = 4πr^{2}

Surface Area of Sphere = 100π cm^{2}

So we can say,

4πr^{2} = 100π

⇒ r^{2} = 25

⇒ r = 5 cm

Out of all the four Options,

Correct Option: C

**Question 18.**

If the surface area of a sphere is 36π cm^{2}, then the volume of the sphere is equal to

A. 12π cm^{3}

B. 36π cm^{3}

C. 72π cm^{3}

D. 108π cm^{3}

**Answer:**

Let radius of sphere be r cm

Surface Area of Sphere = 4πr^{2}

Surface Area of Sphere = 36π cm^{2}

So we can say,

4πr^{2} = 36π

⇒ r^{2} = 9

⇒ r = 3 cm

Volume of Sphere

⇒ Volume of Sphere

⇒ Volume of Sphere = 4π × 3^{2}

⇒ Volume of Sphere = 36π cm^{3}

Out of all the four Options,

Correct Option : B

**Question 19.**

If the total surface area of a solid hemisphere is 12π cm^{2} then its curved surface area is equal to

A. 6π cm^{2}

B. 24π cm^{2}

C. 36π cm^{2}

D. 8π cm^{2}

**Answer:**

Total surface Area = Curved surface Area + Base Area

⇒ Total surface Area = 2πr^{2} + πr^{2} = 3πr^{2}

Total surface area of a solid hemisphere = 12π cm^{2}(Given)

3πr^{2} = 12π

⇒ r^{2} = 4

⇒ r = 2cm

Curved surface Area = 2πr^{2}

Curved surface Area = 2π×2^{2}

Curved surface Area = 8π cm^{3}

Out of all the four Options,

Correct Option : D

**Question 20.**

If the radius of a sphere is half of the radius of another sphere, then their respective volumes are in the ratio

A. 1 : 8

B. 2: 1

C. 1 : 2

D. 8 : 1

**Answer:**

Let R_{1},R_{2} be the radius of the two spheres and V_{1},V_{2} be the Volume of this two spheres

Volume of Sphere

R_{2} = 2R_{1}

So we can say,

Out of all the four Options,

Correct Option :A

**Question 21.**

Curved surface area of solid sphere is 24 cm^{2}. If the sphere is divided into two hemispheres, then the total surface area of one of the hemispheres is

A. 12 cm^{2}

B. 8 cm^{2}

C. 16 cm^{2}

D. 18 cm^{2}

**Answer:**

Curved Surface Area = 24 cm^{2}

Let radius be r

Curved Surface Area of solid sphere = 4πr^{2}

4πr^{2} = 24

Total surface Area of hemisphere = 2πr^{2} + πr^{2} = 3πr^{2}

Total surface Area of hemisphere

Total surface Area of hemisphere = 18 cm^{2}

Out of all the four Options,

Correct Option :D

**Question 22.**

Two right circular cones have equal radii. If their slant heights are in the ratio 4 : 3, then their respective curved surface areas are in the ratio

A. 16 : 9

B. 2 : 3

C. 4 : 3

D. 3 : 4

**Answer:**

Curved Surface Area = πRL

where R is the radius and L is the slant height

Let L_{1},L_{2} be the slant heights of the two cone and A_{1},A_{2} be the area of this two cones

So we can say,

Out of all the four Options,

Correct Option : C