##### Class 10^{th} Mathematics CBSE Solution

**Exercise 13.1**- 2 cubes each of volume 64 cm^3 are joined end to end. Find the surface area of…
- A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder.…
- A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same…
- A cubical block of side 7 cm is surmounted by a hemisphere. What is the…
- A hemispherical depression is cut out from one face of a cubical wooden block…
- A medicine capsule is in the shape of a cylinder with two hemispheres stuck to…
- A tent is in the shape of a cylinder surmounted by a conical top. If the height…
- From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical…
- A wooden article was made by scooping out a hemisphere from each end of a solid…

**Exercise 13.2**- A solid is in the shape of a cone standing on a hemisphere with both their…
- Rachel, an engineering student, was asked to make a model shaped like a…
- A gulabjamun contains sugar syrup up to about30% of its volume. Find…
- A pen stand made of wood is in the shape of a cuboid with four conical…
- A vessel is in the form of an inverted cone. Its height is 8 cm and the radius…
- A solid iron pole consists of a cylinder of height 220 cm and base diameter 24…
- A solid consisting of a right circular cone of height 120 cm and radius 60 cm…
- A spherical glass vessel has a cylindrical neck 8 cm long, 2 cm in diameter;…

**Exercise 13.3**- A metallic sphere of radius 4.2 cm is melted and recast into the shape of a…
- Metallic spheres of radii 6 cm, 8 cm and 10 cm, respectively, are melted to…
- A 20 m deep well with diameter 7 m is dug and the earth from digging is evenly…
- A well of diameter 3 m is dug 14 m deep. The earth taken out of it has been…
- A container shaped like a right circular cylinder having diameter 12 cm and…
- How many silver coins, 1.75 cm in diameter and of thickness 2 mm, must be…
- A cylindrical bucket, 32 cm high and with radius of base 18 cm, is filled with…
- Water in a canal, 6 m wide and 1.5 m deep, is flowing with a speed of 10 km/h.…
- A farmer connects a pipe of internal diameter 20 cm from a canal into a…

**Exercise 13.4**- A drinking glass is in the shape of a frustum of acone of height 14 cm. The…
- The slant height of a frustum of a cone is 4 cm and the perimeters…
- A fez, the cap used by the Turks, is shaped like the frustum of a cone (see…
- A container, opened from the top and made up of a metal sheet, is in the form…
- A metallic right circular cone 20 cm high and whose vertical angle is 60 is cut…

**Exercise 13.5**- A copper wire, 3 mm in diameter, is wound about a cylinder whose length is 12…
- A right triangle, whose sides are 3 cm and 4 cm (other than hypotenuse) is made…
- A cistern, internally measuring 150 cm 120 cm 110 cm, has 129600 cm3 of water…
- In one fortnight of a given month, there was a rainfall of 10 cm in a river…
- An oil funnel made of tin sheet consists of a10 cm long cylindrical portion…
- Derive the formula for the curved surface area and total surface area of the…
- Derive the formula for the volume of the frustum of a cone, given to you in…

**Exercise 13.1**

- 2 cubes each of volume 64 cm^3 are joined end to end. Find the surface area of…
- A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder.…
- A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same…
- A cubical block of side 7 cm is surmounted by a hemisphere. What is the…
- A hemispherical depression is cut out from one face of a cubical wooden block…
- A medicine capsule is in the shape of a cylinder with two hemispheres stuck to…
- A tent is in the shape of a cylinder surmounted by a conical top. If the height…
- From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical…
- A wooden article was made by scooping out a hemisphere from each end of a solid…

**Exercise 13.2**

- A solid is in the shape of a cone standing on a hemisphere with both their…
- Rachel, an engineering student, was asked to make a model shaped like a…
- A gulabjamun contains sugar syrup up to about30% of its volume. Find…
- A pen stand made of wood is in the shape of a cuboid with four conical…
- A vessel is in the form of an inverted cone. Its height is 8 cm and the radius…
- A solid iron pole consists of a cylinder of height 220 cm and base diameter 24…
- A solid consisting of a right circular cone of height 120 cm and radius 60 cm…
- A spherical glass vessel has a cylindrical neck 8 cm long, 2 cm in diameter;…

**Exercise 13.3**

- A metallic sphere of radius 4.2 cm is melted and recast into the shape of a…
- Metallic spheres of radii 6 cm, 8 cm and 10 cm, respectively, are melted to…
- A 20 m deep well with diameter 7 m is dug and the earth from digging is evenly…
- A well of diameter 3 m is dug 14 m deep. The earth taken out of it has been…
- A container shaped like a right circular cylinder having diameter 12 cm and…
- How many silver coins, 1.75 cm in diameter and of thickness 2 mm, must be…
- A cylindrical bucket, 32 cm high and with radius of base 18 cm, is filled with…
- Water in a canal, 6 m wide and 1.5 m deep, is flowing with a speed of 10 km/h.…
- A farmer connects a pipe of internal diameter 20 cm from a canal into a…

**Exercise 13.4**

- A drinking glass is in the shape of a frustum of acone of height 14 cm. The…
- The slant height of a frustum of a cone is 4 cm and the perimeters…
- A fez, the cap used by the Turks, is shaped like the frustum of a cone (see…
- A container, opened from the top and made up of a metal sheet, is in the form…
- A metallic right circular cone 20 cm high and whose vertical angle is 60 is cut…

**Exercise 13.5**

- A copper wire, 3 mm in diameter, is wound about a cylinder whose length is 12…
- A right triangle, whose sides are 3 cm and 4 cm (other than hypotenuse) is made…
- A cistern, internally measuring 150 cm 120 cm 110 cm, has 129600 cm3 of water…
- In one fortnight of a given month, there was a rainfall of 10 cm in a river…
- An oil funnel made of tin sheet consists of a10 cm long cylindrical portion…
- Derive the formula for the curved surface area and total surface area of the…
- Derive the formula for the volume of the frustum of a cone, given to you in…

###### Exercise 13.1

**Question 1.**2 cubes each of volume 64 cm^{3} are joined end to end. Find the surface area of the resulting cuboid.

**Answer:**

In the figure above two cubes are joined end to end such that a cuboid is formed.

Volume of cube = 64 cm^{3}

We know that,

Side of cube =

=

= 4 cm

Side of cube = 4 cm

Now these two cubes when joined, we can see from the figure that, their lengths are added and breadth and height remains the same.

Length of new cuboid = 8 cm

Height of new cuboid = 4 cm

Width of new cuboid = 4 cm

We know that:

Surface Area = 2(lb + lh + bh)

Surface Area = 2(8 x 4 + 8 x 4 + 4 x 4)

= 2 x 80

= 160 cm^{2}

**Question 2.**A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm. Find the inner surface area of the vessel.

**Answer:**

According to the question,

Radius (*r*) of the cylindrical part = 7cm

And

Radius(r) of the hemispherical part = 7 cm

Height of hemispherical part = Radius = 7 cm

Height of cylindrical part (*h*) = 13 - 7

= 6 cm

Now,

Inner surface area of the vessel = CSA of cylindrical part + CSA of hemispherical part

= 2rh + 2 Ï€r^{2}

Inner surface area of vessel = (2 × × 7 × 6) + (2 × × 7 × 7)

= (44 × 6) + ( 44 × 7)

= 44 (6 + 7)

= 44 × 13

= 572 cm^{2}

**Question 3.**A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius. The total height of the toy is 15.5 cm. Find the total surface area of the toy.

**Answer:**As per the question:

The radius of the conical part = 3.5 cm

And,

Radius of the hemispherical part = 3.5 cm

Height of hemispherical part = Radius (*r*)

= 3.5 cm

= cm

Height of conical part (*h*) = 15.5 - 3.5 = 12 cm

for cone we know that,

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

where r is the radius of base, l is the slant height and h is the height of cone.

Slant height of the conical part

=

=

Total surface area of toy = CSA of conical part + CSA of hemispherical part

= Ï€rl + 2Ï€r^{2}

= 137.5 + 77

= 214.5 cm^{2}

**Question 4.**A cubical block of side 7 cm is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have? Find the surface area of the solid.

**Answer:**From the figure:

The greatest diameter possible for the hemisphere is equal to the cube's edge= 7cm

Radius (*r*) of hemispherical part = 7/2cm

The total surface area of solid = Surface area of cubical part + CSA of hemispherical part - Area of the base of hemispherical part

= 6 (Edge)^{2} + 2Ï€r^{2} – Ï€r^{2}

= 6 (Edge)^{2} + Ï€r^{2}

Total surface area of solid = 6 (7)^{2} +( × × )

= 294 + 38.5

= 332.5 cm^{2}

**Question 5.**A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter l of the hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid.

**Answer:**

According to the question,

Diameter of hemisphere = Edge of cube

= *l*

Radius of hemisphere =

**Curved Surface Area of hemisphere = 2 Ï€ r**^{2}

**Surface Area of cube = 6(Edge)2**

Total surface area of solid = Surface area of cubical part + CSA of hemispherical part - Area of base of hemispherical part

= 6 (Edge)^{2} + 2Ï€r^{2} – Ï€r^{2}

= 6 (Edge)^{2} + Ï€r^{2}

TSA of Solid = 6l^{2} + Ï€ x ()^{2}

**Question 6.**A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends (see Fig. 13.10). The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area.

**Answer:**According to the question,

Radius (r) of cylindrical part = Radius (r) of hemispherical part

Length of cylindrical part (h) = Length of the entire capsule - 2 × r

= 14 - 5

= 9 cm

Surface area of capsule = 2 × C S A of hemispherical part + C S A of cylindrical part

( Curved Surface Area of cylinder = 2Î rh, Curved Surface Area of Hemisphere = 2Î r^{2})

= 2 x 2Ï€r^{2} + 2Ï€rh

= 70 x

= 220 mm^{2}**Therefore, Surface Area of capsule = 220 mm**^{2}

**Question 7.**A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are 2.1 m and 4 m respectively, and the slant height of the top is 2.8 m, find the area of the canvas used for making the tent. Also, find the cost of the canvas of the tent at the rate of Rs 500 per m^{2}. (Note that the base of the tent will not be covered with canvas)

**Answer:****Given:**

Height (*h*) of the cylindrical part = 2.1 m

Diameter of the cylindrical part = 4 m

Radius of the cylindrical part = 2 m

Slant height (*l*) = 2.8 m

**Solution:**

We know for a cone,

Area of canvas used = CSA of conical part + CSA of cylindrical part

= Ï€rl + 2Ï€rh

=( Ï€ × 2 × 2.8 )+( 2Ï€ × 2 × 2.1)

= 2Ï€ [2.8 + (2 × 2.1)]

= 2Ï€ [2.8 + 4.2]

= 2 × × 7

= 44 m^{2}

Cost of 1 m^{2} canvas = Rs 500

Cost of 44 m^{2} canvas = 44 × 500

= Rs 22000

**Question 8.**From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest cm^{2}

**Answer:****Given:**

Height (*h*) of the conical part = Height (*h*) of the cylindrical part = 2.4 cm

Diameter of the cylindrical part = 1.4 cm

Radius (*r*) of the cylindrical part = 0.7 cm

we know, slant height of cone, l = √(r^{2} + h^{2})

Slant height of the cylindrical part (l) =

=

=

= 2.5

Total surface area of the remaining solid will be

= CSA of cylindrical part + CSA of conical part + Area of cylindrical base

= 2Ï€rh + Ï€rl + Ï€r^{2}

=( 2 × × 0.7 × 2.4 )+( × 0.7 × 2.5) +( × 0.7 × 0.7)

= (4.4 × 2.4) +( 2.2 × 2.5) +( 2.2 × 0.7)

= 10.56 + 5.50 + 1.54

= 12.10 + 5.50 cm^{2}

= 17.60 cm^{2} (approx)

**Hence, the total surface area of remaining solid is 17.60 cm**^{2}.

**Question 9.**A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in Fig. 13.11. If the height of the cylinder is 10 cm, and its base is of radius 3.5 cm, find the total surface area of the article.

**Answer:****Given:**

Radius (*r*) of cylindrical part = 3.5 cm

Radius (*r*) of hemispherical part = 3.5 cm

Height of cylindrical part (*h*) = 10 cm

Surface area of article = CSA of cylindrical part + 2 × CSA of hemispherical part

= 2 Ï€rh + 2 × 2 Ï€r^{2}

= (2 Ï€ × 3.5 × 10) + (2 × 2 Ï€ × 3.5 × 3.5)

= 70 Ï€ + 49 Ï€

= 119 Ï€

= 17 × 22

= 374 cm^{2}

Hence, the total surface of the article = 374 cm^{2}

**Question 1.**

2 cubes each of volume 64 cm^{3} are joined end to end. Find the surface area of the resulting cuboid.

**Answer:**

In the figure above two cubes are joined end to end such that a cuboid is formed.

Volume of cube = 64 cm^{3}

We know that,

Side of cube =

=

= 4 cm

Side of cube = 4 cmNow these two cubes when joined, we can see from the figure that, their lengths are added and breadth and height remains the same.

Length of new cuboid = 8 cm

Height of new cuboid = 4 cm

Width of new cuboid = 4 cm

We know that:

Surface Area = 2(lb + lh + bh)

Surface Area = 2(8 x 4 + 8 x 4 + 4 x 4)

= 2 x 80

= 160 cm^{2}

**Question 2.**

A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm. Find the inner surface area of the vessel.

**Answer:**

According to the question,

Radius (*r*) of the cylindrical part = 7cm

And

Radius(r) of the hemispherical part = 7 cm

Height of hemispherical part = Radius = 7 cm

Height of cylindrical part (*h*) = 13 - 7

= 6 cm

Now,

Inner surface area of the vessel = CSA of cylindrical part + CSA of hemispherical part

= 2rh + 2 Ï€r^{2}

Inner surface area of vessel = (2 × × 7 × 6) + (2 × × 7 × 7)

= (44 × 6) + ( 44 × 7)

= 44 (6 + 7)

= 44 × 13

= 572 cm^{2}

**Question 3.**

A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius. The total height of the toy is 15.5 cm. Find the total surface area of the toy.

**Answer:**

As per the question:

The radius of the conical part = 3.5 cm

And,

Radius of the hemispherical part = 3.5 cm

Height of hemispherical part = Radius (*r*)

= 3.5 cm

= cm

Height of conical part (*h*) = 15.5 - 3.5 = 12 cm

r

^{2}+ h

^{2}= l

^{2}

where r is the radius of base, l is the slant height and h is the height of cone.

Slant height of the conical part

=

=

Total surface area of toy = CSA of conical part + CSA of hemispherical part

= Ï€rl + 2Ï€r^{2}

= 137.5 + 77

= 214.5 cm^{2}

**Question 4.**

A cubical block of side 7 cm is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have? Find the surface area of the solid.

**Answer:**

From the figure:

The greatest diameter possible for the hemisphere is equal to the cube's edge= 7cm

Radius (*r*) of hemispherical part = 7/2cm

The total surface area of solid = Surface area of cubical part + CSA of hemispherical part - Area of the base of hemispherical part

= 6 (Edge)^{2} + 2Ï€r^{2} – Ï€r^{2}

= 6 (Edge)^{2} + Ï€r^{2}

Total surface area of solid = 6 (7)^{2} +( × × )

= 294 + 38.5

= 332.5 cm^{2}

**Question 5.**

A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter l of the hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid.

**Answer:**

According to the question,

Diameter of hemisphere = Edge of cube

= *l*

Radius of hemisphere =

**Curved Surface Area of hemisphere = 2 Ï€ r**

^{2}**Surface Area of cube = 6(Edge)2**

Total surface area of solid = Surface area of cubical part + CSA of hemispherical part - Area of base of hemispherical part

= 6 (Edge)^{2} + 2Ï€r^{2} – Ï€r^{2}

= 6 (Edge)^{2} + Ï€r^{2}

TSA of Solid = 6l^{2} + Ï€ x ()^{2}

**Question 6.**

A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends (see Fig. 13.10). The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area.

**Answer:**

According to the question,

Radius (r) of cylindrical part = Radius (r) of hemispherical part

Length of cylindrical part (h) = Length of the entire capsule - 2 × r

= 14 - 5

= 9 cm

Surface area of capsule = 2 × C S A of hemispherical part + C S A of cylindrical part

( Curved Surface Area of cylinder = 2Î rh, Curved Surface Area of Hemisphere = 2Î r

^{2})

= 2 x 2Ï€r^{2} + 2Ï€rh

= 70 x

= 220 mm^{2}**Therefore, Surface Area of capsule = 220 mm ^{2}**

**Question 7.**

A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are 2.1 m and 4 m respectively, and the slant height of the top is 2.8 m, find the area of the canvas used for making the tent. Also, find the cost of the canvas of the tent at the rate of Rs 500 per m^{2}. (Note that the base of the tent will not be covered with canvas)

**Answer:**

**Given:**

Height (*h*) of the cylindrical part = 2.1 m

Diameter of the cylindrical part = 4 m

Radius of the cylindrical part = 2 m

Slant height (*l*) = 2.8 m**Solution:**

We know for a cone,

Area of canvas used = CSA of conical part + CSA of cylindrical part

= Ï€rl + 2Ï€rh

=( Ï€ × 2 × 2.8 )+( 2Ï€ × 2 × 2.1)

= 2Ï€ [2.8 + (2 × 2.1)]

= 2Ï€ [2.8 + 4.2]

= 2 × × 7

= 44 m^{2}

Cost of 1 m^{2} canvas = Rs 500

Cost of 44 m^{2} canvas = 44 × 500

= Rs 22000

**Question 8.**

From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest cm^{2}

**Answer:**

**Given:**

Height (*h*) of the conical part = Height (*h*) of the cylindrical part = 2.4 cm

Diameter of the cylindrical part = 1.4 cm

Radius (*r*) of the cylindrical part = 0.7 cm

we know, slant height of cone, l = √(r^{2} + h^{2})

Slant height of the cylindrical part (l) =

=

=

= 2.5

Total surface area of the remaining solid will be

= CSA of cylindrical part + CSA of conical part + Area of cylindrical base

= 2Ï€rh + Ï€rl + Ï€r^{2}

=( 2 × × 0.7 × 2.4 )+( × 0.7 × 2.5) +( × 0.7 × 0.7)

= (4.4 × 2.4) +( 2.2 × 2.5) +( 2.2 × 0.7)

= 10.56 + 5.50 + 1.54

= 12.10 + 5.50 cm^{2}

= 17.60 cm^{2} (approx)**Hence, the total surface area of remaining solid is 17.60 cm ^{2}.**

**Question 9.**

A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in Fig. 13.11. If the height of the cylinder is 10 cm, and its base is of radius 3.5 cm, find the total surface area of the article.

**Answer:**

**Given:**

Radius (*r*) of cylindrical part = 3.5 cm

Radius (*r*) of hemispherical part = 3.5 cm

Height of cylindrical part (*h*) = 10 cm

Surface area of article = CSA of cylindrical part + 2 × CSA of hemispherical part

= 2 Ï€rh + 2 × 2 Ï€r^{2}

= (2 Ï€ × 3.5 × 10) + (2 × 2 Ï€ × 3.5 × 3.5)

= 70 Ï€ + 49 Ï€

= 119 Ï€

= 17 × 22

= 374 cm^{2}

Hence, the total surface of the article = 374 cm^{2}

###### Exercise 13.2

**Question 1.**A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 1 cm and the height of the cone is equal to its radius. Find the volume of the solid in terms of Ï€.

**Answer:**

**Given:**

Height (*h*) of conical part = Radius(*r*) of conical part = 1 cm

Radius(*r*) of hemispherical part = Radius of conical part (*r*) = 1 cm

Volume of solid = Volume of conical part + Volume of hemispherical part

= Ï€r^{2}h + Ï€r^{3}

= (1)^{2} (1) + (1)^{3}

= Ï€ cm^{3}

Hence, Volume of the solid = Ï€ cm^{3}

**Question 2.**Rachel, an engineering student, was asked to make a model shaped like a cylinder with two cones attached at its two ends by using a thin aluminum sheet. The diameter of the model is 3 cm and its length is 12 cm. If each cone has a height of 2 cm, find the volume of air contained in the model that Rachel made. (Assume the outer and inner dimensions of the model to be nearly the same)

**Answer:**

Height (*h*_{1}) of each conical part = 2 cm

Height (*h*_{2}) of cylindrical part = 12 - 2 × Height of conical part

= 12 - 2 ×2

= 8 cm

Radius = Diameter/ 2

Radius (*r*) of cylindrical part = cm

Radius of conical part = cm

Volume of air present in the model = Volume of cylinder + 2 × Volume of cones

**volume of cylinder = Î r**^{2}h

Volume of air present in the model = Ï€r^{2}h_{2} + 2 x r^{2}h_{1}

Volume of air present in the model = Ï€ ()^{2} (8) 2 x ()^{2} (2)

= 18Ï€ + 3Ï€

= 21Ï€

= 21 × 22/7

= 3 × 22

= 66cm^{3}

Hence, volume of air contained in the model that Rachel made = 66cm^{3}

**Question 3.**A gulabjamun contains sugar syrup up to about30% of its volume. Find approximately how much syrup would be found in 45 gulabjamun each shaped like a cylinder with two hemispherical ends with length 5 cm and diameter 2.8 cm (see Fig. 13.15)

**Answer:**

Radius (*r*) of cylindrical part == 1.4 cm

Radius (*R*) of hemispherical part == 1.4 cm

Length of each hemispherical part = 1.4 cm

The radius of hemispherical part = 1.4 cm

Length (*h*) of cylindrical part = 5 - 2 × Length of hemispherical part

= 5 -( 2 × 1.4 )

= 5 - 2.8

= 2.2 cm

The volume of one gulabjamun = Vol. of cylindrical part + 2 × Vol. of hemispherical part

= (Ï€r^{2}h )+ (2 × Ï€R^{3})

= (Ï€r^{2}h) + (R^{3})

= [Ï€ × (1.4)^{2} × 2.2] +[ (1.4)^{3}]

=

=

= 13.552 + 11.498

= 25.05 cm^{3}

Volume of 45 gulabjamun = 45 × 25.05

= 1,127.25 cm^{3}

The volume of sugar syrup = 30% of the volume

= × 1127.25

= 338.17 cm^{3}

**= 338 cm**^{3}(approx)

**Question 4.**A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold pens. The dimensions of the cuboid are 15 cm by 10 cm by3.5 cm. The radius of each of the depressions is 0.5cm and the depth is 1.4 cm. Find the volume of wood in the entire stand (see Fig. 13.16)

**Answer:**Given:

Depth (*h*) of each conical depression = 1.4 cm

Radius (*r*) of each conical depression = 0.5 cm

Volume of wood = Volume of cuboid - 4 × Volume of cones

= lbh – 4 × r^{2}h

= 15 × 10 × 3.5 – 4 × × × ()^{2} × 1.4

= 525 – 1.47

= 523.53 cm^{3}

Hence,

The volume of wood in the entire stand= 523.53 cm^{3}

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

**Answer:**

.................LEAD SHOTS EACH OF DIAMETER = 0.5 CM

When lead shots are dropped in the Vessel, the same amount of water will come out of the vessel as the Total Volume Of Lead Shots.

Height (*h*) of conical vessel = 8 cm

Radius (*r*_{1}) of conical vessel = 5 cm

Radius (*r*_{2}) of lead shots = 0.5 cm

Volume of water = Volume of Vessel

Let *n* number of lead shots were dropped in the vessel

Volume of water spilled = Volume of dropped lead shots

^{}

**Hence, the number of lead shots dropped in the vessel is 100.**

**Question 6.**A solid iron pole consists of a cylinder of height 220 cm and base diameter 24 cm, which is surmounted by another cylinder of height 60 cm and radius 8 cm. Find the mass of the pole, given that 1 cm^{3} of iron has approximately 8gm mass. (Use Ï€ = 3.14)

**Answer:**

**Given:**

Height (*H*) of larger cylinder = 220 cm

Radius (*R*) of larger cylinder = = 12 cm

Height (*h*) of smaller cylinder = 60 cm

Radius (*r*) of smaller cylinder = 8 cm

Total volume = Volume (Larger cylinder) + Volume (Smaller cylinder)

= Ï€(R)^{2}H + Ï€(r)^{2}h

= Ï€ (12)^{2} × 220 + Ï€ (8)^{2} × 60

= Ï€ [(144 × 220 )+ (64 × 60)]

= Ï€ [31680+ 3840]

= 35520 × 3.14

= 111532.8 cm^{3}

Mass of 1 cm^{3} iron = 8 g

Mass of 111532.8 cm^{3} iron = 111532.8 × 8

= 892262.4 g

= 892.262 kg

**Question 7.**A solid consisting of a right circular cone of height 120 cm and radius 60 cm standing on a hemisphere of radius 60 cm is placed upright in a right circular cylinder full of water such that it touches the bottom. Find the volume of water left in the cylinder, if the radius of the cylinder is 60 cm and its height is 180 cm.

**Answer:**

Radius (*r*) of hemispherical part = Radius (*r*) of conical part = 60 cm

Height (*h*) of conical part of solid = 120 cm

Height (H) of cylinder = 180 cm

Radius (*r*) of cylinder = 60 cm

Volume of water left = Volume of cylinder - Volume of solid

= Volume of cylinder – (Volume of cone + Volume of hemisphere)

= Ï€r^{2}H– (r^{2}h + r^{3})

= [Ï€ (60)^{2} (180) ]– [ (60)^{2} × 120 + (60)^{3}]

= 648000 Ï€ - ( 144000 Ï€ + 144000 Ï€ )

= 648000 Ï€ - 288000 Ï€

= 360000Ï€

= 1131428.57 cm^{3}

= 1.131 m^{3}(approx)

**Question 8.**A spherical glass vessel has a cylindrical neck 8 cm long, 2 cm in diameter; the diameter of the spherical part is 8.5 cm. By measuring the amount of water it holds, a child finds its volume to be 345 cm^{3}. Check whether she is correct, taking the above as the inside measurements, and Ï€ = 3.14

**Answer:**

Given:

Height (*h*) of cylindrical part = 8 cm

Radius (*r2*) of cylindrical part = = 1 cm

Radius (*r1*) spherical part = = 4.25 cm

Volume of vessel = Volume of sphere + Volume of cylinder

= (r1)^{3} + Ï€(r2)^{2}h

= (4.25)^{3} + Ï€ (1)^{2} (8)

=( × 3.14 × 76.765625 )+ (8 × 3.14)

= 321.392 + 25.12

= 346.512

= 346.51 cm^{3}

Hence,

From the result we can conclude that she is wrong

**Question 1.**

A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 1 cm and the height of the cone is equal to its radius. Find the volume of the solid in terms of Ï€.

**Answer:**

**Given:**

Height (*h*) of conical part = Radius(*r*) of conical part = 1 cm

Radius(*r*) of hemispherical part = Radius of conical part (*r*) = 1 cm

Volume of solid = Volume of conical part + Volume of hemispherical part

= Ï€r^{2}h + Ï€r^{3}

= (1)^{2} (1) + (1)^{3}

= Ï€ cm^{3}

Hence, Volume of the solid = Ï€ cm^{3}

**Question 2.**

Rachel, an engineering student, was asked to make a model shaped like a cylinder with two cones attached at its two ends by using a thin aluminum sheet. The diameter of the model is 3 cm and its length is 12 cm. If each cone has a height of 2 cm, find the volume of air contained in the model that Rachel made. (Assume the outer and inner dimensions of the model to be nearly the same)

**Answer:**

Height (*h _{1}*) of each conical part = 2 cm

Height (*h _{2}*) of cylindrical part = 12 - 2 × Height of conical part

= 12 - 2 ×2

= 8 cm

Radius (*r*) of cylindrical part = cm

Radius of conical part = cm

Volume of air present in the model = Volume of cylinder + 2 × Volume of cones

**volume of cylinder = Î r**

^{2}hVolume of air present in the model = Ï€r^{2}h_{2} + 2 x r^{2}h_{1}

Volume of air present in the model = Ï€ ()^{2} (8) 2 x ()^{2} (2)

= 18Ï€ + 3Ï€

= 21Ï€

= 21 × 22/7= 3 × 22

= 66cm^{3}

Hence, volume of air contained in the model that Rachel made = 66cm^{3}

**Question 3.**

A gulabjamun contains sugar syrup up to about30% of its volume. Find approximately how much syrup would be found in 45 gulabjamun each shaped like a cylinder with two hemispherical ends with length 5 cm and diameter 2.8 cm (see Fig. 13.15)

**Answer:**

Radius (*r*) of cylindrical part == 1.4 cm

Radius (*R*) of hemispherical part == 1.4 cm

Length of each hemispherical part = 1.4 cm

The radius of hemispherical part = 1.4 cm

Length (*h*) of cylindrical part = 5 - 2 × Length of hemispherical part

= 5 -( 2 × 1.4 )

= 5 - 2.8

= 2.2 cm

The volume of one gulabjamun = Vol. of cylindrical part + 2 × Vol. of hemispherical part

= (Ï€r^{2}h )+ (2 × Ï€R^{3})

= (Ï€r^{2}h) + (R^{3})

= [Ï€ × (1.4)^{2} × 2.2] +[ (1.4)^{3}]

=

=

= 13.552 + 11.498

= 25.05 cm^{3}

Volume of 45 gulabjamun = 45 × 25.05

= 1,127.25 cm^{3}

The volume of sugar syrup = 30% of the volume

= × 1127.25

= 338.17 cm^{3}

**= 338 cm ^{3}(approx)**

**Question 4.**

A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold pens. The dimensions of the cuboid are 15 cm by 10 cm by3.5 cm. The radius of each of the depressions is 0.5cm and the depth is 1.4 cm. Find the volume of wood in the entire stand (see Fig. 13.16)

**Answer:**

Given:

Depth (*h*) of each conical depression = 1.4 cm

Radius (*r*) of each conical depression = 0.5 cm

Volume of wood = Volume of cuboid - 4 × Volume of cones

= lbh – 4 × r^{2}h

= 15 × 10 × 3.5 – 4 × × × ()^{2} × 1.4

= 525 – 1.47

= 523.53 cm^{3}

Hence,

The volume of wood in the entire stand= 523.53 cm^{3}

**Question 5.**

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

**Answer:**

.................LEAD SHOTS EACH OF DIAMETER = 0.5 CM

When lead shots are dropped in the Vessel, the same amount of water will come out of the vessel as the Total Volume Of Lead Shots.

Height (*h*) of conical vessel = 8 cm

Radius (*r _{1}*) of conical vessel = 5 cm

Radius (*r _{2}*) of lead shots = 0.5 cm

Volume of water = Volume of Vessel

Let

*n*number of lead shots were dropped in the vessel

Volume of water spilled = Volume of dropped lead shots

^{}

**Hence, the number of lead shots dropped in the vessel is 100.**

**Question 6.**

A solid iron pole consists of a cylinder of height 220 cm and base diameter 24 cm, which is surmounted by another cylinder of height 60 cm and radius 8 cm. Find the mass of the pole, given that 1 cm^{3} of iron has approximately 8gm mass. (Use Ï€ = 3.14)

**Answer:**

**Given:**

Height (*H*) of larger cylinder = 220 cm

Radius (*R*) of larger cylinder = = 12 cm

Height (*h*) of smaller cylinder = 60 cm

Radius (*r*) of smaller cylinder = 8 cm

Total volume = Volume (Larger cylinder) + Volume (Smaller cylinder)

= Ï€(R)^{2}H + Ï€(r)^{2}h

= Ï€ (12)^{2} × 220 + Ï€ (8)^{2} × 60

= Ï€ [(144 × 220 )+ (64 × 60)]

= Ï€ [31680+ 3840]

= 35520 × 3.14

= 111532.8 cm^{3}

Mass of 1 cm^{3} iron = 8 g

Mass of 111532.8 cm^{3} iron = 111532.8 × 8

= 892262.4 g

= 892.262 kg

**Question 7.**

A solid consisting of a right circular cone of height 120 cm and radius 60 cm standing on a hemisphere of radius 60 cm is placed upright in a right circular cylinder full of water such that it touches the bottom. Find the volume of water left in the cylinder, if the radius of the cylinder is 60 cm and its height is 180 cm.

**Answer:**

Radius (*r*) of hemispherical part = Radius (*r*) of conical part = 60 cm

Height (*h*) of conical part of solid = 120 cm

Height (H) of cylinder = 180 cm

Radius (*r*) of cylinder = 60 cm

Volume of water left = Volume of cylinder - Volume of solid

= Volume of cylinder – (Volume of cone + Volume of hemisphere)

= Ï€r^{2}H– (r^{2}h + r^{3})

= [Ï€ (60)^{2} (180) ]– [ (60)^{2} × 120 + (60)^{3}]

= 648000 Ï€ - ( 144000 Ï€ + 144000 Ï€ )

= 648000 Ï€ - 288000 Ï€

= 360000Ï€

= 1131428.57 cm^{3}

= 1.131 m^{3}(approx)

**Question 8.**

A spherical glass vessel has a cylindrical neck 8 cm long, 2 cm in diameter; the diameter of the spherical part is 8.5 cm. By measuring the amount of water it holds, a child finds its volume to be 345 cm^{3}. Check whether she is correct, taking the above as the inside measurements, and Ï€ = 3.14

**Answer:**

Given:

Height (*h*) of cylindrical part = 8 cm

Radius (*r2*) of cylindrical part = = 1 cm

Radius (*r1*) spherical part = = 4.25 cm

Volume of vessel = Volume of sphere + Volume of cylinder

= (r1)^{3} + Ï€(r2)^{2}h

= (4.25)^{3} + Ï€ (1)^{2} (8)

=( × 3.14 × 76.765625 )+ (8 × 3.14)

= 321.392 + 25.12

= 346.512

= 346.51 cm^{3}

Hence,

From the result we can conclude that she is wrong

###### Exercise 13.3

**Question 1.**A metallic sphere of radius 4.2 cm is melted and recast into the shape of a cylinder of radius 6 cm. Find the height of the cylinder

**Answer:****To find: Height of the cylinder (h)**

**Given:**

**Radius (***r*_{1}) of sphere = 4.2 cm

**Radius (***r*_{2}) of cylinder = 6 cm

Let the height of the cylinder be *h*.

The object formed by recasting the sphere will be the same in volume.

Volume of sphere = Volume of cylinder

Hence, the height of the cylinder = 2.74 cm

**Question 2.**Metallic spheres of radii 6 cm, 8 cm and 10 cm, respectively, are melted to form a single solid sphere. Find the radius of the resulting sphere.

**Answer:**

Radius (*r*1) of 1st sphere = 6 cm

Radius (*r*2) of 2nd sphere = 8 cm

Radius (*r*3) of 3rd sphere = 10 cm

Let the radius of the resulting sphere be *r*.

The object formed by recasting these spheres will be the same in volume as the sum of the volumes of these spheres.

The volume of 3 spheres = Volume of resulting sphere

[r1^{3} + r2^{3} + r3^{3}] = r^{3}

[6^{3} + 8^{3} + 10^{3}] = r^{3}

r^{3} = 216 + 512 + 1000

r^{3} = 1728

r = 12 cm

**Hence, the radius of the resulting sphere, r = 12 cm.**

**Question 3.**A 20 m deep well with diameter 7 m is dug and the earth from digging is evenly spread out to form a platform 22 m by 14 m. Find the height of the platform.

**Answer:**The diagram for the situation is as follows:

As per the question,

The shape of the well will be cylindrical

Depth (h) of well = 20 m

Radius (r) of circular end of well =

Area of platform = Length × Breadth = 22 × 14 m^{2}

Let height of the platform = H

Volume of soil dug from the well would be equal to the volume of soil that is scattered on the platform.

Volume of soil from well = Volume of soil used to make such platform

Ï€r^{2}h = Area × Height

Ï€ × ()^{2} × 20 = 22 × 14 × H

H = × ×

= 2.5 m

Hence, the height of the platform = 2.5 m

**Question 4.**A well of diameter 3 m is dug 14 m deep. The earth taken out of it has been spread evenly all around it in the shape of a circular ring of width 4 m to form an embankment. Find the height of the embankment.

**Answer:**

The shape of the well will be cylindrical.

Depth (*h*_{1}) of well = 14 m

Radius (*r*) of the circular end of well = 3/2 m

Width of embankment = 4 m

As per the question,

Our embankment will be in a cylindrical shape having outer radius (*R*)as 11/2 m and inner radius (r) = 3/2 m

Let the height of embankment be *h*_{2}

**Volume of soil dug from well = Volume of earth used to form embankment**

Ï€r^{2} × h_{1} = Ï€ × (R^{2} – r^{2}) × h_{2}⇒ 31.5 = 28 h

_{}

⇒ h = 1.125 m

**Question 5.**A container shaped like a right circular cylinder having diameter 12 cm and height 15 cm is full of ice cream. The ice cream is to be filled into cones of height 12 cm and diameter 6 cm, having a hemispherical shape on the top. Find the number of such cones which can be filled with ice cream.

**Answer:**

Height (*H*) of cylindrical container = 15 cm

Radius (*r*_{1}) of circular end of container = = 6 cm

Radius (*r*_{2}) of circular end of ice-cream cone = = 3 cm

Height (*h*) of conical part of ice-cream cone = 12 cm

Let *n* cones be filled with ice-cream of the container

Volume of ice-cream in cylinder = *n* × (Volume of 1 ice-cream cone + Volume of hemispherical shape on the top)

**Question 1.**

A metallic sphere of radius 4.2 cm is melted and recast into the shape of a cylinder of radius 6 cm. Find the height of the cylinder

**Answer:**

**To find: Height of the cylinder (h)****Given:****Radius ( r_{1}) of sphere = 4.2 cm**

**Radius ( r_{2}) of cylinder = 6 cm**

Let the height of the cylinder be *h*.

The object formed by recasting the sphere will be the same in volume.

Volume of sphere = Volume of cylinder

Hence, the height of the cylinder = 2.74 cm

**Question 2.**

Metallic spheres of radii 6 cm, 8 cm and 10 cm, respectively, are melted to form a single solid sphere. Find the radius of the resulting sphere.

**Answer:**

Radius (*r*1) of 1st sphere = 6 cm

Radius (*r*2) of 2nd sphere = 8 cm

Radius (*r*3) of 3rd sphere = 10 cm

Let the radius of the resulting sphere be *r*.

The object formed by recasting these spheres will be the same in volume as the sum of the volumes of these spheres.

The volume of 3 spheres = Volume of resulting sphere

[r1^{3} + r2^{3} + r3^{3}] = r^{3}

[6^{3} + 8^{3} + 10^{3}] = r^{3}

r^{3} = 216 + 512 + 1000

r^{3} = 1728

r = 12 cm

**Hence, the radius of the resulting sphere, r = 12 cm.**

**Question 3.**

A 20 m deep well with diameter 7 m is dug and the earth from digging is evenly spread out to form a platform 22 m by 14 m. Find the height of the platform.

**Answer:**

The diagram for the situation is as follows:

As per the question,

The shape of the well will be cylindrical

Depth (h) of well = 20 m

Radius (r) of circular end of well =

Area of platform = Length × Breadth = 22 × 14 m^{2}

Let height of the platform = H

Volume of soil dug from the well would be equal to the volume of soil that is scattered on the platform.

Volume of soil from well = Volume of soil used to make such platform

Ï€r^{2}h = Area × Height

Ï€ × ()^{2} × 20 = 22 × 14 × H

H = × ×

= 2.5 m

Hence, the height of the platform = 2.5 m

**Question 4.**

A well of diameter 3 m is dug 14 m deep. The earth taken out of it has been spread evenly all around it in the shape of a circular ring of width 4 m to form an embankment. Find the height of the embankment.

**Answer:**

The shape of the well will be cylindrical.

Depth (*h*_{1}) of well = 14 m

Radius (*r*) of the circular end of well = 3/2 m

Width of embankment = 4 m

As per the question,

Our embankment will be in a cylindrical shape having outer radius (*R*)as 11/2 m and inner radius (r) = 3/2 m

Let the height of embankment be *h*_{2}

**Volume of soil dug from well = Volume of earth used to form embankment**

Ï€r^{2} × h_{1} = Ï€ × (R^{2} – r^{2}) × h_{2}⇒ 31.5 = 28 h_{}

⇒ h = 1.125 m

**Question 5.**

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

**Answer:**

Height (*H*) of cylindrical container = 15 cm

Radius (*r*_{1}) of circular end of container = = 6 cm

Radius (*r*_{2}) of circular end of ice-cream cone = = 3 cm

Height (*h*) of conical part of ice-cream cone = 12 cm

Let *n* cones be filled with ice-cream of the container

Volume of ice-cream in cylinder = *n* × (Volume of 1 ice-cream cone + Volume of hemispherical shape on the top)