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#### Page No 138:

#### Question 1:

Choose the correct answer from the given four options:

A cylindrical pencil sharpened at one edge is the combination of

(A) a cone and a cylinder

(B) frustum of a cone and a cylinder

(C) a hemisphere and a cylinder

(D) two cylinders.

#### Answer:

The nip of a sharpened pencil is in conical shape and the rest of the part is cylindrical therefore pencil is a combination of cylinder and a cone.

Hence, the correct answer is option (A).

#### Page No 138:

#### Question 2:

Choose the correct answer from the given four options:

A *surahi *is the combination of

(A) a sphere and a cylinder

(B) a hemisphere and a cylinder

(C) two hemispheres

(D) a cylinder and a cone.

#### Answer:

The top part of surahi is in cylindrical shape and bottom part is in spherical shape therefore surahi is a combination of sphere and a cylinder.

Hence, the correct answer is option (A).

#### Page No 138:

#### Question 3:

Choose the correct answer from the given four options:

A plumbline (*sahul*) is the combination of (see Fig. 12.2)

(A) a cone and a cylinder

(B) a hemisphere and a cone

(C) frustum of a cone and a cylinder

(D) sphere and cylinder

#### Answer:

The upper part of plumbline is hemispherical, and the bottom part is conical therefore, it is a combination of hemisphere and cone.

Hence, the correct answer is option (B).

#### Page No 138:

#### Question 4:

Choose the correct answer from the given four options:

The shape of a glass (tumbler) (see Fig. 12.3) is usually in the form of

(A) a cone

(B) frustum of a cone

(C) a cylinder

(D) a sphere

#### Answer:

We know that, the shape of frustum of a cone is

So, the shape of glass is a frustum [ an inverted frustum].

Hence, the correct answer is option (B).

#### Page No 139:

#### Question 5:

Choose the correct answer from the given four options:

The shape of a *gilli*, in the *gilli-danda *game (see Fig. 12.4), is a combination of

(A) two cylinders

(B) a cone and a cylinder

(C) two cones and a cylinder

(D) two cylinders and a cone

#### Answer:

As the left and right part of a gilli are conical and the central part is cylindrical.

Therefore, it is a combination of a cylinder and two cones.

Hence, the correct answer is option (C).

#### Page No 139:

#### Question 6:

Choose the correct answer from the given four options:

A shuttle cock used for playing badminton has the shape of the combination of

(A) a cylinder and a sphere

(B) a cylinder and a hemisphere

(C) a sphere and a cone

(D) frustum of a cone and a hemisphere

#### Answer:

The cork of a shuttle is in hemispherical shapes and the upper part is in the shape of frustum of a cone. Therefore, it is a combination of frustum of a cone and a hemisphere.

Hence, the correct answer is Option (D).

#### Page No 139:

#### Question 7:

Choose the correct answer from the given four options:

A cone is cut through a plane parallel to its base and then the cone that is formed on one side of that plane is removed. The new part that is left over on the other side of the plane is called

(A) a frustum of a cone

(B) cone

(C) cylinder

(D) sphere

#### Answer:

When a cone is divided into two parts by a plane through any point on its axis parallel to its base, the upper and lower parts obtained are cone and a frustum respectively.

Hence, the correct answer is option (A).

#### Page No 139:

#### Question 8:

Choose the correct answer from the given four options:

A hollow cube of internal edge 22cm is filled with spherical marbles of diameter 0.5 cm and it is assumed that $\frac{1}{8}$ space of the cube remains unfilled. Then the number of marbles that the cube can accomodate is

(A) 142244

(B) 142344

(C) 142444

(D) 142544

#### Answer:

Given, edge of cube = 22 cm

As we know,

Volume of given hollow cube = *a*^{3} = (22)^{3} = 10648 cm^{3}

Filled space of cube = $\frac{7}{8}$ of total volume of cube = $\frac{7}{8}$ × 10648 = 9317 cm^{3}

Now,

Diameter of marble, *D* = 0.5 cm

Radius of marble, *r* = $\frac{D}{2}$ = $\frac{0.5}{2}$ = 0.25 cm

Volume of one marble = Volume of sphere of radius *r*

$\therefore $Volume of one marble = $\frac{4}{3}\mathrm{\pi}{r}^{3}$ = $\frac{4}{3}\times \frac{22}{7}\times {\left(0.25\right)}^{3}$

$=\frac{1.375}{21}\phantom{\rule{0ex}{0ex}}=0.0655{\mathrm{cm}}^{3}$

$\therefore \mathrm{Required}\mathrm{number}\mathrm{of}\mathrm{marbles}=\frac{\mathrm{Total}\mathrm{space}\mathrm{filled}\mathrm{by}\mathrm{marbles}\mathrm{in}\mathrm{cube}}{\mathrm{Volume}\mathrm{of}\mathrm{one}\mathrm{marble}}$

$=\frac{9317}{0.0655}=142244\left(\mathrm{approx}\right)$

$\therefore $ The number of marbles that the cube can accomodate is 142244 (approx).

Hence, the correct answer is option (A).

#### Page No 139:

#### Question 9:

Choose the correct answer from the given four options:

A metallic spherical shell of internal and external diameters 4 cm and 8 cm, respectively is melted and recast into the form a cone of base diameter 8cm. The height of the cone is

(A) 12cm

(B) 14cm

(C) 15cm

(D) 18cm

#### Answer:

Volume of spherical shell = Volume of cone recast by melting .....(1)

For Spherical Shell,

Internal diameter, *d*_{1} = 4 cm

Internal radius, *r*_{1} = 2 cm [ as radius = $\frac{1}{2}$ diameter]

External diameter, *d*_{2} = 8 cm

External radius, *r*_{2} = 4 cm

Now,

Volume of spherical shell = $\frac{4}{3}\mathrm{\pi}\left[{\left({\mathrm{r}}_{2}\right)}^{3}-{\left({\mathrm{r}}_{1}\right)}^{3}\right]$

= $\frac{4}{3}\mathrm{\pi}\left[{\left(4\right)}^{3}-{\left(2\right)}^{3}\right]$

= $\frac{4}{3}\mathrm{\pi}\left[64-8\right]$

= $\frac{224}{3}\mathrm{\pi}$ cm^{3}

Volume of cone = $\frac{224\mathrm{\pi}}{3}$ cm^{3} [Using(1)]

For cone,

Base diameter = 8 cm

Base radius, *r* = 4 cm [ as radius = 1/2 diameter]

Let Height of cone be *h *cm

Volume of given cone = $\frac{1}{3}\mathrm{\pi}{\left(4\right)}^{2}h$

$\Rightarrow \frac{224\mathrm{\pi}}{3}=\frac{1}{3}\mathrm{\pi}{\left(4\right)}^{2}h$

⇒ 16*h *= 224

⇒ h = 14 cm

So, Height of cone is 14 cm.

Hence, the correct answer is Option(B).

#### Page No 139:

#### Question 10:

Choose the correct answer from the given four options:

A solid piece of iron in the form of a cuboid of dimensions 49cm × 33cm × 24cm, is moulded to form a solid sphere. The radius of the sphere is

(A) 21cm

(B) 23cm

(C) 25cm

(D) 19cm

#### Answer:

For given cuboid,

Length, *l* = 49 cm, Breadth, *b* = 33 cm and Height, *h* = 24 cm

Volume of cuboid = 49×(33) × (24) cm^{3} [Volume of cuboid = *lbh*]

Now,

Let the radius of sphere be *r*.

Volume of cuboid = volume of sphere

$\Rightarrow 49\times 33\times 24=\frac{4}{3}\pi {r}^{3}$

$\Rightarrow 29106\times \frac{22}{7}={r}^{3}$

$\Rightarrow {r}^{3}=9261\phantom{\rule{0ex}{0ex}}\Rightarrow r=21\mathrm{cm}$

Hence, the correct answer is Option(A).

#### Page No 139:

#### Question 11:

Choose the correct answer from the given four options:

A mason constructs a wall of dimensions 270cm × 300cm × 350cm with the bricks each of size 22.5cm × 11.25cm × 8.75cm and it is assumed that $\frac{1}{8}$ space is covered by the mortar. Then the number of bricks used to construct the wall is

(A) 11100

(B) 11200

(C) 11000

(D) 11300

#### Answer:

For wall,

Length, l = 270 cm

Breadth, b = 300 cm

Height, h = 350 cm

Volume of wall = 270 × (300) × (350) = 28350000 cm^{3} [Volume of cuboid = lbh]

As $\frac{1}{8}$ of this volume is covered by mortar, therefore $\frac{7}{8}$ of this volume is covered by bricks.

Volume of wall covered by bricks = $\frac{7}{8}$ (volume of wall)

= $\frac{7}{8}\times 28350000$

= 24806250 cm^{3}

For one brick,

Length, l = 22.5 cm

Breadth, b = 11.25 cm

Height, h = 8.75 cm

Volume of one brick = 22.5(11.25)(8.75) cm^{3} [Volume of cuboid = lbh ]

= 2214.844 cm^{3}

$\therefore \mathrm{Required}\mathrm{number}\mathrm{of}\mathrm{bricks}=\frac{\mathrm{Volume}\mathrm{of}\mathrm{wall}\mathrm{covered}\mathrm{by}\mathrm{bricks}}{\mathrm{Volume}\mathrm{of}\mathrm{one}\mathrm{brick}}$

$\Rightarrow \mathrm{Number}\mathrm{of}\mathrm{bricks}=\frac{24806250}{2214.844}=11200\left(\mathrm{approx}\right)$

Hence, the correct answer is option (B).

#### Page No 140:

#### Question 12:

Choose the correct answer from the given four options:

Twelve solid spheres of the same size are made by melting a solid metallic cylinder of base diameter 2 cm and height 16 cm. The diameter of each sphere is

(A) 4 cm

(B) 3 cm

(C) 2 cm

(D) 6 cm

#### Answer:

Volume of cylinder = π*r*^{2}*h*

For Given Solid cylinder,

Base diameter = 2 cm

Base radius,* r* = 1 cm [as radius = $\frac{1}{2}$ diameter]

Height, *h *= 16 cm

Volume of cylinder, V = π(1)^{2}(16) = 16π cm^{3} [Volume of sphere = $\frac{4}{3}\mathrm{\pi}{r}^{3}$]

Now, let the radius of sphere be *r*

Volume of 12 spheres of radius *r* = $12\times \frac{4}{3}\mathrm{\pi}{r}^{3}=16\mathrm{\pi}{r}^{3}$

Now,

Volume of 12 spheres = volume of cylinder [As volume remains same, when a metal body is melted and recast into another metal body]

⇒ 16π*r*^{3} = 16π

⇒* r*^{3} = 1

⇒ *r *= ∛1 = 1 cm

Diameter = 2*r *= 2 cm

Hence, the correct answer is Option(C).

#### Page No 140:

#### Question 13:

Choose the correct answer from the given four options:

The radii of the top and bottom of a bucket of slant height 45 cm are 28 cm and 7 cm, respectively. The curved surface area of the bucket is

(A) 4950 cm^{2}

(B) 4951 cm^{2}

(C) 4952 cm^{2}

(D) 4953 cm^{2}

#### Answer:

Clearly, the given bucket is in the form of frustum of a cone.

Curved surface area of frustum of a cone = π*l*(*r*_{1} + *r*_{2})

For given bucket,

Radius of top,* r*_{1} = 28 cm

Radius of bottom, *r*_{2} = 7 cm

Slant height, *l* = 45 cm

Therefore,

Curved surface area of bucket = π(45)(28 + 7)

= $\frac{22}{7}\times 45\times 35$

= 4950 cm^{2}

So, the curved surface area of bucket is 4950 cm^{2}.

Hence, the correct answer is Option(A).

#### Page No 140:

#### Question 14:

Choose the correct answer from the given four options:

A medicine-capsule is in the shape of a cylinder of diameter 0.5 cm with two hemispheres stuck to each of its ends. The length of entire capsule is 2 cm. The capacity of the capsule is

(A) 0.36 cm^{3}

(B) 0.35 cm^{3}

(C) 0.34 cm^{3}

(D) 0.33 cm^{3}

#### Answer:

Capacity of capsule = Volume of 2 hemispherical part + volume of cylindrical part

Now,

Diameter of capsule = 0.5 cm

Radius of capsule, *r* = 0.25 cm [as radius = 1/2 diameter]

Also,

Radius of hemispherical part = Radius of cylindrical part = *r*

Height of entire capsule = 2 cm

Let the height of cylindrical part be *h*.

Then

Height of entire capsule = radius of 2 hemispherical parts + height of cylindrical part

⇒ 2 = 2*r* + *h*

⇒ 2 = 2(0.25) + *h*

⇒ *h* = 1.5 cm

Volume of entire capsule = $2\times \frac{2}{3}\mathrm{\pi}\times {r}^{3}+\mathrm{\pi}\times {r}^{2}\times h$

= $\pi {r}^{2}\left(\frac{4r}{3}+h\right)$

= $\frac{22}{7}\times {\left(0.25\right)}^{2}\times \left(\frac{4}{3}\times \left(0.25\right)+1.5\right)\phantom{\rule{0ex}{0ex}}$

= 0.36 cm^{3}

Hence, the correct answer is Option(A).

#### Page No 140:

#### Question 15:

Choose the correct answer from the given four options:

If two solid hemispheres of same base radius *r *are joined together along their bases, then curved surface area of this new solid is

(A) 4π*r*^{2}

(B) 6π*r*^{2}

(C) 3π*r*^{2}

(D) 8π*r*^{2}

#### Answer:

When two hemispheres are joined together along their bases, a sphere of same base radius is formed.

And, Curved surface area of sphere is 4π*r*^{2}.

Hence, the correct answer is Option(A).

#### Page No 140:

#### Question 16:

Choose the correct answer from the given four options:

A right circular cylinder of radius *r *cm and height *h *cm (*h *> 2*r*) just encloses a sphere of diameter

(A) *r *cm

(B) 2*r *cm

(C) *h *cm

(D) 2*h *cm

#### Answer:

As the sphere just encloses in a cylinder,

The diameter of sphere will be equal to diameter of cylinder.

Diameter of base of cylinder = 2(radius of base) = 2*r*

So, Diameter of sphere = 2*r* cm

Hence, the correct answer is Option(B).

#### Page No 140:

#### Question 17:

Choose the correct answer from the given four options:

During conversion of a solid from one shape to another, the volume of the new shape will

(A) increase

(B) decrease

(C) remain unaltered

(D) be doubled

#### Answer:

When a solid is converted from one shape to another, the volume of new shape remains constant.

Hence, the correct answer is Option(C).

#### Page No 140:

#### Question 18:

Choose the correct answer from the given four options:

The diameters of the two circular ends of the bucket are 44 cm and 24 cm. The height of the bucket is 35 cm. The capacity of the bucket is

(A) 32.7 litres

(B) 33.7 litres

(C) 34.7 litres

(D) 31.7 litres

#### Answer:

Clearly, the given bucket is in the form of frustum of a cone.

For given bucket,

Diameter of top = 44 cm

Radius of top, r_{1} = 22 cm [as radius = $\frac{1}{2}$ diameter]

Diameter of bottom = 24 cm

Radius of bottom, r_{2} = 12 cm [as radius = $\frac{1}{2}$ diameter]

Height, h = 35 cm

Therefore,

Volume of Bucket = $\frac{1}{3}\mathrm{\pi h}[{\left({\mathrm{r}}_{1}\right)}^{2}+{\left({\mathrm{r}}_{2}\right)}^{2}+\left({\mathrm{r}}_{1}\right)\times \left({\mathrm{r}}_{2}\right)]$

= $\frac{1}{3}\times \mathrm{\pi}\times 35\times [{\left(22\right)}^{2}+{\left(12\right)}^{2}+\left(22\right)\times \left(12\right)]$

= $\frac{35\mathrm{\pi}}{3}\left[484+144+264\right]$

= $\frac{35\mathrm{\pi}\times 892}{3}$

= $\frac{35\times 892\times 22}{3\times 7}$

= 32706.6 cm^{3}

= 32.7 L

Hence, the correct answer is Option(A).

#### Page No 140:

#### Question 19:

Choose the correct answer from the given four options:

In a right circular cone, the cross-section made by a plane parallel to the base is a

(A) circle

(B) frustum of a cone

(C) sphere

(D) hemisphere

#### Answer:

We know that , if a cone is cut by a plane parallel to the base of the cone, the cross-section made is circle.

#### Page No 140:

#### Question 20:

Choose the correct answer from the given four options:

Volumes of two spheres are in the ratio 64 : 27. The ratio of their surface areas is

(A) 3 : 4

(B) 4 : 3

(C) 9 : 16

(D) 16 : 9

#### Answer:

Let the radius of spheres be *r*_{1} and *r*_{2} and we know that,

Volume of sphere of radius *r *= $\frac{4\mathrm{\pi}{r}^{3}}{3}$

Given, ratio of volumes = 64 : 27

$\Rightarrow \frac{{\displaystyle \frac{4\mathrm{\pi}{\left({\mathrm{r}}_{1}\right)}^{3}}{3}}}{{\displaystyle \frac{4\mathrm{\pi}{\left({\mathrm{r}}_{2}\right)}^{3}}{3}}}=\frac{64}{27}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{{\left({\mathrm{r}}_{1}\right)}^{3}}{{\left({\mathrm{r}}_{2}\right)}^{3}}=\frac{64}{27}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{\left({\mathrm{r}}_{1}\right)}{\left({\mathrm{r}}_{2}\right)}=\frac{4}{3}$

Now,

Surface area of sphere = 4π*r*^{2}

Ratio of surface area = $\frac{4\mathrm{\pi}{\left({\mathrm{r}}_{1}\right)}^{2}}{4\mathrm{\pi}{\left({\mathrm{r}}_{2}\right)}^{2}}\phantom{\rule{0ex}{0ex}}$

$=\frac{{\left({\mathrm{r}}_{1}\right)}^{2}}{{\left({\mathrm{r}}_{2}\right)}^{2}}\phantom{\rule{0ex}{0ex}}={\left(\frac{4}{3}\right)}^{2}\phantom{\rule{0ex}{0ex}}=16:9$

Hence, the correct answer is Option (D).

#### Page No 142:

#### Question 1:

Write ‘True’ or ‘False’ and justify your answer in this question:

Two identical solid hemispheres of equal base radius *r *cm are stuck together along their bases. The total surface area of the combination is 6π*r*^{2}.

#### Answer:

False

When two hemispheres are joined together along their bases, a sphere of same base radius is formed.

And CSA of sphere is 4π*r*^{2}.

#### Page No 142:

#### Question 2:

Write ‘True’ or ‘False’ and justify your answer in this question:

A solid cylinder of radius *r *and height *h *is placed over other cylinder of same height and radius. The total surface area of the shape so formed is 4π*rh *+ 4π*r*^{2}.

#### Answer:

False

As one cylinder is placed over another, so the base of first cylinder and top of other cylinder will not be covered in total surface area.

So,

Total surface area of shape formed = 2(Total surface of single cylinder) $-$ 2(Area of base of cylinder)

= 2(2π*rh* + 2π*r*^{2}) $-$ 2(π*r*^{2}) [As Total surface area of cylinder = 2π*rh* + 2π*r*^{2}*h*, where *r* = base radius and *h* = height]

= 4π*rh* + 2π*r*^{2}

#### Page No 142:

#### Question 3:

Write ‘True’ or ‘False’ and justify your answer in this question:

A solid cone of radius *r *and height *h *is placed over a solid cylinder having same base radius and height as that of a cone. The total surface area of the combined solid is $\pi r\left[\sqrt{{r}^{2}+{h}^{2}}+3r+2h\right].$

#### Answer:

False

When a solid cone is placed over a solid cylinder of same base radius, the base of cone and top of the cylinder will not be covered in total surface area.

As the height of cone and cylinder is same

Total surface area of shape formed = Total surface area of cone + Total Surface area of cylinder $-$ 2(Area of base)

Total surface area of cone = π*rl* + π*r*^{2 }

Total surface area of cylinder = 2π*rh* + 2π*r*^{2}

Here, when we placed a cone over a cylinder, then one base is common for both.

So, the total surface area of combined solid = π*rl* + π*r*^{2} + 2π*rh*

#### Page No 142:

#### Question 4:

Write ‘True’ or ‘False’ and justify your answer in this question:

A solid ball is exactly fitted inside the cubical box of side *a*. The volume of the ball is $\frac{4}{3}\pi {a}^{3}.$

#### Answer:

False

Let the radius of sphere be *r*

As the ball is exactly fitted into the cubical box.

Diameter of ball = Edge length of cube

$\Rightarrow $2*r* = *a* [As Diameter = 2(Radius)]

$\Rightarrow r=\frac{a}{2}$

As we know,

Volume of sphere = $\frac{4}{3}{\mathrm{\pi r}}^{3}$

$\Rightarrow $Volume of ball = $\frac{4}{3}\pi {\left(\frac{a}{2}\right)}^{3}$

= $\frac{1}{6}\times \mathrm{\pi}\times {\mathrm{a}}^{3}$

#### Page No 142:

#### Question 5:

Write ‘True’ or ‘False’ and justify your answer in this question:

The volume of the frustum of a cone is $\frac{1}{3}\pi h\left[{r}_{1}^{2}+{r}_{2}^{2}-{r}_{1}{r}_{2}\right],$ where *h *is vertical height of the frustum and *r*_{1}, *r*_{2} are the radii of the ends.

#### Answer:

False

As we know

Volume of frustum of a cone = $\frac{1}{3}\mathrm{\pi h}\left[{\left({\mathrm{r}}_{1}\right)}^{2}+{\left({\mathrm{r}}_{2}\right)}^{2}+{\mathrm{r}}_{1}{\mathrm{r}}_{2}\right]$

#### Page No 142:

#### Question 6:

Write ‘True’ or ‘False’ and justify your answer in this question:

The capacity of a cylindrical vessel with a hemispherical portion raised upward at the bottom as shown in the Fig. 12.7 is $\frac{\pi {r}^{2}}{3}\left[3h-2r\right].$

#### Answer:

True

The capacity of given shape = volume of cylinder $-$ volume of hemispherical portion

And the base radius of cylinder will be equal to radius of hemisphere.

Now, we know

Volume of a cylinder = π*r*^{2}*h*, where *r* = base radius and *h* = height of cylinder

And

Volume of hemisphere = $\frac{2}{3}\mathrm{\pi}{r}^{3}$

Capacity of given shape = π*r*^{2}*h $-$ *$\frac{2}{3}\mathrm{\pi}{r}^{3}$

= $\frac{{\mathrm{\pi r}}^{2}}{3}\left(3h-2r\right)$

So, given statement is true.

#### Page No 143:

#### Question 7:

Write ‘True’ or ‘False’ and justify your answer in this question: :

The curved surface area of a frustum of a cone is π*l *(*r*_{1} + *r*_{2}), where $l=\sqrt{{h}^{2}+{\left({r}_{1}+{r}_{2}\right)}^{2}}$ *r*_{1} and *r*_{2} are the radii of the two ends of the frustum and *h *is the vertical height.

#### Answer:

False

We know that,

Curved surface area of frustum = π*l*(*r*_{1} + *r*_{2})

Where, *r*_{1} and *r*_{2} are the radii of two ends (*r*_{1} > *r*_{2 }and *l* is slant height).

$l=\sqrt{\left({h}^{2}+{\left({r}_{1}-{r}_{2}\right)}^{2}\right)}$

[ In the statement, formula for calculating slant height is incorrect]

#### Page No 143:

#### Question 8:

Write ‘True’ or ‘False’ and justify your answer in this question: :

An open metallic bucket is in the shape of a frustum of a cone, mounted on a hollow cylindrical base made of the same metallic sheet. The surface area of the metallic sheet used is equal to the curved surface area of frustum of a cone + area of circular base + curved surface area of cylinder.

#### Answer:

Figure

As the resulting shape will be as in figure,

We will not count the area of base of frustum, top of frustum and top of cylinder , as no metal sheet is used there.

So required area will be = Curved surface area of frustum + area of base of cylinder + curved surface area of cylinder

#### Page No 146:

#### Question 1:

Three metallic solid cubes whose edges are 3 cm, 4 cm and 5 cm are melted and formed into a single cube. Find the edge of the cube so formed.

#### Answer:

We know that,

Volume of cube = a^{3}, where a = side of cube

Now,

Side of first cube, a_{1} = 3 cm

Side of second cube, a_{2} = 4 cm

Side of third cube, a_{3} = 5 cm

Now, Let the side of cube recast from melting these cubes is '*a*'.

As the volume remains same,

Volume of recast cube = (volume of 1^{st} + 2^{nd} + 3^{rd} cube)

$\Rightarrow {a}^{3}={\left({a}_{1}\right)}^{3}+{\left({a}_{2}\right)}^{3}+{\left({a}_{3}\right)}^{3}$

$\Rightarrow {a}^{3}={\left(3\right)}^{3}+{\left(4\right)}^{3}+{\left(5\right)}^{3}$

⇒ *a*^{3} = 27 + 64 + 125 = 216

⇒ *a* = 6 cm

So, side of cube so formed is 6 cm.

#### Page No 146:

#### Question 2:

How many shots each having diameter 3 cm can be made from a cuboidal lead solid of dimensions 9cm × 11cm × 12cm?

#### Answer:

Volume of cuboid = *lbh *

For cuboidal lead:

Length, *l* = 9 cm

Breadth, *b* = 11 cm

Height, *h* = 12 cm

Volume of lead = 9(11)(12) = 1188 cm^{3}

Volume of sphere = $\frac{4}{3}\mathrm{\pi}{r}^{3}$

For spherical shots,

Diameter = 3 cm

$\Rightarrow $Radius,* r* = 1.5 cm

Volume of one shot = $\frac{4}{3}\mathrm{\pi}{(1.5)}^{3}$

= $\frac{99}{7}{\mathrm{cm}}^{3}$

Now,

$\mathrm{No}.\mathrm{of}\mathrm{shots}\mathrm{that}\mathrm{can}\mathrm{be}\mathrm{made}=\frac{\mathrm{Volume}\mathrm{of}\mathrm{Lead}}{\mathrm{Volume}\mathrm{of}\mathrm{one}\mathrm{Shot}}=\frac{1188}{{\displaystyle \frac{99}{7}}}=84$

So, 84 bullets can be made from lead.

#### Page No 146:

#### Question 3:

A bucket is in the form of a frustum of a cone and holds 28.490 litres of water. The radii of the top and bottom are 28 cm and 21 cm, respectively. Find the height of the bucket.

#### Answer:

Clearly, Given bucket is in the form of frustum of a cone.

As we know

Volume of frustum of a cone = $\frac{1}{3}\mathrm{\pi h}\left[{\left({\mathrm{r}}_{1}\right)}^{2}+{\left({\mathrm{r}}_{2}\right)}^{2}+{\mathrm{r}}_{1}\times {\mathrm{r}}_{2}\right]$

For given bucket,

Volume of bucket = 28.490 L

Volume of bucket = 28490 cm^{3} [As, 1 L = 1000 cm^{3} ]

Radius of top, *r*_{1} = 28 cm

Radius of bottom, *r*_{2} = 21 cm

Let the height be* h*.

Using these values,

We have,

Volume of Bucket = $\frac{1}{3}\mathrm{\pi h}\left[{\left(28\right)}^{2}+{\left(21\right)}^{2}+28\times 21\right]$

$\Rightarrow 28490=\frac{1}{3}\times \frac{22}{7}\times h\left(784+441+588\right)$

$\Rightarrow h=\frac{28490\times 21}{22\times 1813}$

$\Rightarrow h=15\mathrm{cm}$

#### Page No 146:

#### Question 4:

A cone of radius 8 cm and height 12 cm is divided into two parts by a plane through the mid-point of its axis parallel to its base. Find the ratio of the volumes of two parts.

#### Answer:

Figure

Let ORN be the cone, when this is cone is divided into two parts by a plane through the mid-point of its axis parallel to its base, the upper and lower parts obtained are cone and a frustum respectively.

Given,

Height of cone = OM = 12 cm

As, the cone is divided from mid-point, let P be the mid-point of cone, then

OP = PM = 6 cm

Now, In △OPD and △OMN

∠POD = ∠POD [Common]

∠OPD = ∠OMN [Both 90°]

△OPD ~ △OMN [So, By Angle-Angle similarity criterion]

$\Rightarrow \frac{\mathrm{PD}}{\mathrm{MN}}=\frac{\mathrm{OP}}{\mathrm{OM}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{\mathrm{PD}}{8}=\frac{6}{12}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{\mathrm{PD}}{8}=\frac{1}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{PD}=4\mathrm{cm}$

[MN = 8 cm = radius of base of cone]

Now, For First part i.e. cone

Base Radius, *r* = PD = 4 cm

Height, *h* = OP = 6 cm

As we know, volume of cone for radius *r* and height is

$V=\frac{1}{3}{\mathrm{\pi r}}^{2}\mathrm{h}=\frac{1}{3}\mathrm{\pi}{\left(4\right)}^{2}\times 6=32\mathrm{\pi}$

Now, For second part, i.e. Frustum

Bottom radius, *r*_{1} = MN = 8 cm

Top Radius, *r*_{2} = PD = 4 cm

Height, *h* = PM = 6 cm

As we know,

$\mathrm{Volume}\mathrm{of}\mathrm{frustrum}\mathrm{of}\mathrm{cone}=\frac{1}{3}\mathrm{\pi h}[{\left({\mathrm{r}}_{1}\right)}^{2}+{\left({\mathrm{r}}_{2}\right)}^{2}+\left({\mathrm{r}}_{1}\times {\mathrm{r}}_{2}\right)]$

$\Rightarrow \mathrm{Volume}\mathrm{of}\mathrm{second}\mathrm{part}=\frac{1}{3}\mathrm{\pi}\left(6\right)\left[{\left(8\right)}^{2}+{\left(4\right)}^{2}+8\times 4\right]$

$\Rightarrow \mathrm{Volume}\mathrm{of}\mathrm{second}\mathrm{part}=2\mathrm{\pi}\left(112\right)$

So,

Volume of first part : Volume of second part = 32π : 224π = 1 : 7

#### Page No 146:

#### Question 5:

Two identical cubes each of volume 64 cm^{3} are joined together end to end. What is the surface area of the resulting cuboid?

#### Answer:

Figure

Let *a* be the side of one cube.

As two cubes are joined together, the surfaces that are joined together will not be included in the surface area of resulting cuboid.

So,

Surfaces area of resulting cuboid = 2(Total surface area of a cube) $-$ 2(area of single surface)

⇒ Surfaces area of resulting cuboid = 2(6*a*^{2}) $-$ 2(*a*^{2}) = 10*a*^{2 }[As total surface area of cube = 6*a*^{2} ]

Also,

Volume of cube = *a*^{3}

$\Rightarrow $64 = *a*^{3} [Given volume of cube is 64 cm^{3} ]

$\Rightarrow $*a* = 4 cm

Therefore,

Surface area of resulting cuboid = 10*a*^{2} = 10(4)^{2} = 160 cm^{2}

#### Page No 146:

#### Question 6:

From a solid cube of side 7 cm, a conical cavity of height 7 cm and radius 3 cm is hollowed out. Find the volume of the remaining solid.

#### Answer:

Figure

Since the conical cavity is hollowed out from the cube,

Volume of remaining solid = volume of cube $-$ volume of cone

Now,

For Cube

Side, *a* = 7 cm

Volume of cube = (7)^{3} = 343 cm^{3} [ Volume of cube = a^{3}]

For cone

Radius, *r* = 3 cm

Height, *h* = 7 cm

Volume of cone = $\frac{1}{3}\mathrm{\pi}{\left(3\right)}^{2}\left(7\right)=66{\mathrm{cm}}^{3}$ [Volume of cone = $\frac{1}{3}{\mathrm{\pi r}}^{2}\mathrm{h}$]

Volume of remaining solid = 343 $-$ 66 = 277 cm^{3}

#### Page No 146:

#### Question 7:

Two cones with same base radius 8 cm and height 15 cm are joined together along their bases. Find the surface area of the shape so formed.

#### Answer:

Figure

When two identical cones are joined along their base, the shape so formed will be as shown in figure

So,

Total surface area of shape formed = Curved area of first cone + Curved surface area of second cone

= 2(Surface area of cone) [since, both cones are identical]

Total Surface area of shape so formed = 2π*rl * [Surface area of cone = π*rl*]

Given,

Radius, *r* = 8 cm

Height, *h* = 15 cm

Slant height = $\sqrt{\left({r}^{2}\right)+\left({h}^{2}\right)}=\sqrt{{\left(8\right)}^{2}+{\left(15\right)}^{2}}=17\mathrm{cm}$

So,

Area = 2(3.14)(8)(17) = 855 cm^{2}.

#### Page No 146:

#### Question 8:

Two solid cones A and B are placed in a cylinderical tube as shown in the Fig.12.9. The ratio of their capacities are 2 : 1. Find the heights and capacities of cones. Also, find the volume of the remaining portion of the cylinder.

#### Answer:

Figure

Diameter of Cylinder = 6 cm

$\Rightarrow $Radius of cylinder = *r* = 3 cm

As both cones have equal radius

Radius of cone A = radius of cone B = *r* = 3 cm

Let the height of cone A be *h*_{1} and Cone B be *h*_{2}

Given,

Ratio of volume of cones is 2 : 1

i.e.$\frac{VolumeofconeA}{VolumeofconeB}=\frac{2}{1}$

$\Rightarrow \frac{{\displaystyle \frac{1}{3}}\pi {r}^{2}{h}_{1}}{{\displaystyle \frac{1}{3}}\pi {r}^{2}{h}_{2}}=\frac{2}{1}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{{h}_{1}}{{h}_{2}}=\frac{2}{1}\phantom{\rule{0ex}{0ex}}$

⇒ *h*_{1} = 2*h*_{2}

Now,

Total height of cylinder is 21 cm

$\Rightarrow $h_{1} + h_{2} = 21

$\Rightarrow $2h_{2} + h_{2} = 21

$\Rightarrow $3h_{2} = 21

$\Rightarrow $h_{2} = 7 cm

$\therefore $h_{1} = 2h_{2} = 2(7) = 14 cm

Volume of Cone A = $\frac{1}{3}{\mathrm{\pi r}}^{2}{\mathrm{h}}_{1}$

= $\frac{1}{3}\mathrm{\pi}{\left(3\right)}^{2}\times 14$

= 132 cm^{3}

Volume of Cone B = $\frac{1}{3}{\mathrm{\pi r}}^{2}{\mathrm{h}}_{2}$

= $\frac{1}{3}\mathrm{\pi}{\left(3\right)}^{2}\times 7$

= 66 cm^{3}

Volume of given cylinder = π(3)^{2}(21) [ Volume of cylinder = π*r*^{2}*h *]

= 594 cm^{3}

Volume of remaining solid = (Volume of cylinder) – (volume of cone A) – (volume of cone B)

= 594 $-$ 132 $-$ 66 = 396 cm^{3}

#### Page No 146:

#### Question 9:

An ice cream cone full of ice cream having radius 5 cm and height 10 cm as shown in the Fig.12.10. Calculate the volume of ice cream, provided that its $\frac{1}{6}$ part is left unfilled with ice cream.

#### Answer:

As ice cream is a combination of hemisphere and cone

Radius of cone = radius of hemisphere = *r* = 5 cm

Height of cone, *h* = 10 cm

Volume of ice cream = volume of hemispherical part + volume of conical part

$=\frac{2}{3}\pi {r}^{3}+\frac{1}{3}\pi {r}^{2}h$

$=\frac{1}{3}\pi {r}^{2}\left(2r+h\right)$

$=\frac{1}{3}\times \frac{22}{7}\times {\left(5\right)}^{2}\times (10+10)$

= 392.85 cm^{2}

Now, $\frac{1}{6}$part is left unfilled $\therefore $ $\frac{5}{6}$ part is filled with ice-cream.

Hence, the volume of ice-cream = $\frac{5}{6}\times 392.85$

= 327.4 cm^{3}.

#### Page No 147:

#### Question 10:

Marbles of diameter 1.4 cm are dropped into a cylindrical beaker of diameter 7 cm containing some water. Find the number of marbles that should be dropped into the beaker so that the water level rises by 5.6 cm.

#### Answer:

Let *x* no of marbles are dropped, so that water level rises by 5.6 cm.

The increase in volume of water in beaker = Volume of *x* marbles.

Now,

Required raise in height, *h* = 5.6 cm

Diameter of beaker = 7 cm

Radius of beaker, *r* = 3.5 cm

Required increase in volume = volume of cylinder of above dimensions = π*r*^{2}*h*

$=\mathrm{\pi}\times {\left(3.5\right)}^{2}\times 5.6$

Now, As diameter of marble is 1.4 cm

$\Rightarrow $Radius of marble, *r* = 0.7 cm

Volume of one marble = $\frac{4}{3}\mathrm{\pi}{\left(0.7\right)}^{3}$ cm^{3}

Volume of *x* marbles = $\frac{4}{3}\mathrm{\pi}{\left(0.7\right)}^{3}\times \left(\mathrm{x}\right)$ cm^{3}

So, we have,

$\frac{4x}{3}\mathrm{\pi}{\left(0.7\right)}^{3}=\mathrm{\pi}\times {\left(3.5\right)}^{2}\times 5.6$

$\Rightarrow x=\frac{\mathrm{\pi}\times {\left(3.5\right)}^{2}\times 5.6\times 3}{4\times \mathrm{\pi}\times {\left(0.7\right)}^{3}}$

Therefore, 150 marbles are required.

#### Page No 147:

#### Question 11:

How many spherical lead shots each of diameter 4.2 cm can be obtained from a solid rectangular lead piece with dimensions 66 cm, 42 cm and 21 cm.

#### Answer:

Volume of cuboid = *lbh* where, *l* = length, *b* = breadth and *h* = height

For cuboidal lead ;

Length, *l* = 66 cm

Breadth, *b* = 42 cm

Height, *h* = 21 cm

Volume of lead = 66(42)(21) = 58212 cm^{3}

Volume of sphere = $\frac{4}{3}\mathrm{\pi}{r}^{3}$ , where *r* is the radius of sphere

For spherical shots,

Diameter = 4.2 cm

Radius, *r* = 2.1 cm

Volume of one shot = $\frac{4}{3}\mathrm{\pi}{\left(2.1\right)}^{3}=38.808{\mathrm{cm}}^{3}$

No of shots can be made = $\frac{\mathrm{Volume}\mathrm{of}\mathrm{Lead}}{\mathrm{Volume}\mathrm{of}\mathrm{one}\mathrm{shot}}=\frac{58212}{38.808}=1500$

So, 1500 bullets can be made from lead.

#### Page No 147:

#### Question 12:

How many spherical lead shots of diameter 4 cm can be made out of a solid cube of lead whose edge measures 44 cm.

#### Answer:

Volume of cube = *a*^{3} where, *a* = side of cube

For cubical lead, side, *a* = 44 cm

Volume of lead = 44(44)(44) cm^{3}

Volume of sphere = $\frac{4}{3}\mathrm{\pi}{r}^{3}$

For spherical shots,

Diameter = 4 cm

Radius, *r* = 2 cm

Volume of one shot = $\frac{4\mathrm{\pi}}{3}\times {\left(2\right)}^{3}$

No. of shots can be made = $\frac{\mathrm{Volume}\mathrm{of}\mathrm{Lead}}{\mathrm{Volume}\mathrm{of}\mathrm{one}\mathrm{shot}}$

=$\frac{44\times 44\times 44}{{\displaystyle \frac{4}{3}}\times {\displaystyle \frac{22}{7}}\times 8}$

= $121\times 21$

= 2541

So, 2541 bullets can be made from lead.

#### Page No 147:

#### Question 13:

A wall 24 m long, 0.4 m thick and 6 m high is constructed with the bricks each of dimensions 25 cm × 16 cm × 10 cm. If the mortar occupies $\frac{1}{10}$th of the volume of the wall, then find the number of bricks used in constructing the wall.

#### Answer:

Volume of cuboid = *lbh*, where, *l* = length, *b* = breadth and *h* = height

For wall,

Length, *l* = 24 m

Breadth, *b* = 0.4 m

Height, *h* = 6 cm

Volume of wall = 24(0.4)(6) = 57.6 m^{3}

As $\frac{1}{10}$ of this volume is covered by mortar, therfore $\frac{9}{10}$ of this volume is covered by bricks.

Volume of wall covered by bricks = $\frac{9}{10}\left(57.6\right)$

= 51.84 cm^{3}

For one brick,

Length, *l* = 25 cm = 0.25 m [As 1 m = 100 cm]

Breadth, *b* = 16 cm = 0.16 m

Height, *h* = 10 cm = 0.10 m

Volume of one brick = *lbh* = 0.25(0.16)(0.10) m^{3}

No. of bricks = $\frac{\mathrm{Volume}\mathrm{of}\mathrm{wall}\mathrm{covered}\mathrm{by}\mathrm{bricks}}{\mathrm{Volume}\mathrm{of}\mathrm{one}\mathrm{brick}}$

No. of bricks = $\frac{51.84}{0.25\times 0.16\times 0.10}=12960$

Hence Total No of bricks used in constructing the wall is 12960.

#### Page No 147:

#### Question 14:

Find the number of metallic circular disc with 1.5 cm base diameter and of height 0.2 cm to be melted to form a right circular cylinder of height 10 cm and diameter 4.5 cm.

#### Answer:

For Cone,

Base Diameter = 4.5 cm

Base radius, *r* = 2.25 cm

Height of cone, *h* = 10 cm

We know that,

Volume of cone = $\frac{1}{3}\pi {r}^{2}h$

Volume of given cone = $\frac{1}{3}\mathrm{\pi}\times {\left(2.25\right)}^{2}\times 10$

Now, disk is in shape of a cylinder

For a single disk

Base diameter = 1.5 cm

Base radius, *r* = 0.75 cm

Volume of a disk = π (0.75)^{2}(0.2) = (0.75 × 0.75 × 0.2) π cm^{3}

Now,

No. of disks that can be made by melting cone = $\frac{\mathrm{Volume}\mathrm{of}\mathrm{cone}}{\mathrm{Volume}\mathrm{of}\mathrm{single}\mathrm{disk}}$

$\Rightarrow \mathrm{No}.\mathrm{of}\mathrm{disks}=\frac{{\displaystyle \frac{1}{3}}\mathrm{\pi}{\left(2.25\right)}^{2}10}{\mathrm{\pi}{\left(0.75\right)}^{2}\left(0.2\right)}$ = $3\times 50=150$

So, 150 disks can be made by melting the cone.

#### Page No 150:

#### Question 1:

A solid metallic hemisphere of radius 8 cm is melted and recasted into a right circular cone of base radius 6 cm. Determine the height of the cone.

#### Answer:

For hemisphere,

Radius, *r* = 8 cm

As we know,

Volume of hemisphere = $\frac{2}{3}\mathrm{\pi}{r}^{3}$

Volume of given hemisphere = $\frac{2}{3}\mathrm{\pi}{\left(8\right)}^{3}$

= $\frac{1024}{3}\mathrm{\pi}{\mathrm{cm}}^{3}$

For cone that is recast from hemisphere,

Base radius, *r* = 6 cm

We know that,

Volume of cone = $\frac{1}{3}\mathrm{\pi}{\left(r\right)}^{2}h$

Volume of cone = $\frac{1}{3}\mathrm{\pi}{\left(6\right)}^{2}h$

As the volume remains same, when a body is reformed to another body

Volume of cylinder = Volume of cone

12π*h* = 1024$\frac{\mathrm{\pi}}{3}$

*$\Rightarrow $h* = 28.44 cm

#### Page No 150:

#### Question 2:

A rectangular water tank of base 11 m × 6 m contains water upto a height of 5 m. If the water in the tank is transferred to a cylindrical tank of radius 3.5 m, find the height of the water level in the tank.

#### Answer:

Volume of water in tank = volume of cuboidal tank up to a height of 5 m

For cuboidal tank

Length, *l* = 11 m

Breadth, *b* = 6 m

Height, *h* = 5m

We know that,

Volume of tank = *lbh*

Where, *l*, *b* and *h* are the length, breadth and height of tank respectively

Volume of water = 11(6)(5) = 330 m^{3}

Also,

Let the cylindrical tank is filled up to a height of *h* m

Base radius of cylindrical tank, *r* = 3.5 m

As we know,

Volume of a cylinder = $\mathrm{\pi}{r}^{2}h$

Where *r* is base radius and *h* is the height of cylinder

Volume of water in cylindrical tank = $\mathrm{\pi}{\left(3.5\right)}^{2}h$

*$\Rightarrow 330=\mathrm{\pi}{\left(3.5\right)}^{2}h$*

*h* = 8.57 m

So, cylindrical tank is filled up to a height of 8.57 m.

#### Page No 150:

#### Question 3:

How many cubic centimetres of iron is required to construct an open box whose external dimensions are 36 cm, 25 cm and 16.5 cm provided the thickness of the iron is 1.5 cm. If one cubic cm of iron weighs 7.5 g, find the weight of the box.

#### Answer:

Let the length (*l*), breath(*b*) and height(*h*) be the external dimensions of an open box and thickness be x.

The volume of metal used in box = Volume of external box $-$ Volume of internal box

For external box,

Length, *l* = 36 cm

Breadth, *b* = 25 cm

Height, *h* = 16.5 cm

We know that,

Volume of cuboid = *lbh*

Volume of external box = 36(25)(16.5) = 14850 cm^{3}

Now,

As the box is open from top,

For internal box,

Length, l' = Length of external box $-$ 2(thickness of box) = 36$-$ 2(1.5) = 33 cm [i.e. the thickness of two sides is deduced]

Breadth, b' = Breadth of external box $-$ 2(thickness of box) = 25 $-$ 2(1.5) = 22 cm [i.e. the thickness of two sides is deduced]

Height, h' = Height of external box $-$ thickness of box = 16.5 $-$ 1.5 = 15 cm [i.e. the thickness of bottom is reduced]

Volume of internal box = 33(22)(15) = 10890

Volume of metal in box = 14850 $-$ 10890 = 3960 cm^{3}

Also,

3960 cm^{3} weighs 3960(7.5) = 29,700 g [1 cm^{3} weighs 7.5 g]

$\therefore $The weight of box is 29,700 g i.e. 29.7 kg.

#### Page No 150:

#### Question 4:

The barrel of a fountain pen, cylindrical in shape, is 7 cm long and 5 mm in diameter. A full barrel of ink in the pen is used up on writing 3300 words on an average. How many words can be written in a bottle of ink containing one fifth of a litre?

#### Answer:

Let us first calculate the volume of barrel of pen that is of cylindrical shape

For barrel,

Base diameter = 5 mm = 0.5 cm [As 1 cm = 10 mm]

Base radius, *r* = 0.25 cm

Height, *h* = 7 cm

As we know,

Volume of a cylinder = π*r*^{2}*h*

Where *r* is base radius and *h* is the height of cylinder

Volume of barrel = $\mathrm{\pi}{\left(0.25\right)}^{2}\times 7$

= 1.375 cm^{3}

So,

1.375 cm^{3} of ink can write 3300 words

No of words that can be written by 1 cm^{3} = $\frac{3300}{1.375}$ = 2400 words

[Also $\frac{1}{5}$ of a litre i.e. 0.2L = 200 cm^{3}] [1L = 1000 cm^{3} ]

No of words that can be written by 200 cm^{3} = 2400(200) = 480000 words

Hence, one-fifth of a liter ink can write 480000 words on an average.

#### Page No 150:

#### Question 5:

Water flows at the rate of 10m/minute through a cylindrical pipe 5 mm in diameter. How long would it take to fill a conical vessel whose diameter at the base is 40 cm and depth 24 cm?

#### Answer:

Let the time taken by pipe to fill vessel is *t* minutes

As water flows 10 m in 1 minute, it will flow 10*t* meters in *t* minutes.

Also,

Volume of conical vessel = Volume of water that passes through pipe in *t* minutes

Now, For conical pipe

Base Diameter = 40 cm

Base radius, *r* = 20 cm

Height, *h* = 24 cm

We know that,

Volume of conical vessel = $\frac{1}{3}\pi {r}^{2}h$

= $\frac{1}{3}\mathrm{\pi}{\left(20\right)}^{2}\times 24$ = 3200 $\mathrm{\pi}$ cm^{3}

For cylindrical pipe

Base diameter = 5 mm = 0.5 cm [As 1 cm = 10 mm]

Base radius, *r* = 0.25 cm

Height, *h* = 10*t* m = 1000*t* cm [As water covers 10t m distance in pipe]

As we know,

Volume of a cylinder = $\mathrm{\pi}{r}^{2}h$

Volume of water passed in pipe = $\mathrm{\pi}$(0.25)^{2}(1000*t*) = 62.5t$\mathrm{\pi}$ cm^{3}

So, we have

62.5*t*$\mathrm{\pi}$ = 3200

$\Rightarrow $62.5*t* = 3200

*$\therefore $t* = 51.2 minutes

t = 51 minutes 12 seconds [ as 0.2 minutes = 0.2(60) seconds = 12 seconds]

Hence, it will take 51 minutes 12 seconds to fill a conical vessel whose diameter at the base is 40 cm and depth 24 cm.

#### Page No 150:

#### Question 6:

A heap of rice is in the form of a cone of diameter 9 m and height 3.5 m. Find the volume of the rice. How much canvas cloth is required to just cover the heap?

#### Answer:

Given for conical heap,

Base Diameter = 9 cm

Base radius, *r* = 4.5 cm

Height, h = 3.5 cm

Slant height,

$l=\sqrt{{\left(r\right)}^{2}+{\left(h\right)}^{2}}$

= $\sqrt{{\left(4.5\right)}^{2}+{\left(3.5\right)}^{2}}$

= $\sqrt{\left(20.25\right)+\left(12.25\right)}$

= 5.7 cm

Volume of rice = Volume of conical heap

= $\frac{1}{3}\mathrm{\pi}{\left(r\right)}^{2}h$

= $\frac{1}{3}\mathrm{\pi}{\left(4.5\right)}^{2}\left(3.5\right)$ = 74.25 cm^{3}

Also,

Canvas requires to just cover heap = Curved surface area of conical heap

And we know,

Curved surface area of a cone = $\mathrm{\pi}$*rl*

Canvas required = $\mathrm{\pi}$(4.5)(5.7) = 80.61 cm^{2} [appx]

#### Page No 150:

#### Question 7:

A factory manufactures 120000 pencils daily. The pencils are cylindrical in shape each of length 25 cm and circumference of base as 1.5 cm. Determine the cost of colouring the curved surfaces of the pencils manufactured in one day at Rs 0.05 per dm^{2}.

#### Answer:

The pencil is in shape of cylinder.

Let the radius of base be *r* cm

Circumference of base = 1.5 cm

$\Rightarrow $2$\mathrm{\pi}$*r* = 1.5 cm [As Circumference of circle is 2$\mathrm{\pi}$*r*]

$\Rightarrow r=\frac{1.5}{2\mathrm{\pi}}$

Also, Height, *h* = 25 cm

As we know,

Curved surface area of cylinder = 2$\mathrm{\pi}$rh

Curved surface area of pencil = $2\mathrm{\pi}\times \left(\frac{1.5}{2\mathrm{\pi}}\right)\times \left(25\right)$

= 37.5 cm^{2}

Cost for coloring 1 dm^{2} = Rs. 0.05

Cost for coloring 0.375 dm^{2} (i.e. 1 pencil) = Rs. 0.01875 [37.5 cm^{2} = 0.375 dm^{2}]

Cost for coloring 120000 pencils = 120000 ×0.01875 = Rs. 2250

#### Page No 151:

#### Question 8:

Water is flowing at the rate of 15 km/h through a pipe of diameter 14 cm into a cuboidal pond which is 50 m long and 44 m wide. In what time will the level of water in pond rise by 21 cm?

#### Answer:

Let the time taken by pipe to fill pond is *t* hours

As water flows 15 km in 1 hour, it will flow 15*t* meters in *t* hours.

Also,

Volume of cuboidal pond up to height 21 cm = Volume of water that passes through pipe in *t* hours

Now, For cuboidal pond

Length, *l* = 50 m

Breadth, *b* = 44 m

Height, *h* = 21 cm = 0.21 m

We know that,

Volume of tank = *lbh*

Volume of water = 50(44)(0.21) = 462 m^{3}

For cylindrical pipe

Base diameter = 14 cm

Base radius, *r* = 7 cm = 0.07 m

Height, *h* = 15*t* km = 15000*t* m [1 km = 1000 m]

As we know,

Volume of a cylinder =$\mathrm{\pi}$*r*^{2}*h*

Volume of water passed in pipe = $\mathrm{\pi}$(0.07)^{2}(15000*t*)

= 231*t* cm^{3}

So, we have

231*t* = 462

$\Rightarrow $*t* = 2 hours

Time required to fill tank up to a height of 21 cm is 2 hours.

#### Page No 151:

#### Question 9:

A solid iron cuboidal block of dimensions 4.4 m × 2.6 m × 1m is recast into a hollow cylindrical pipe of internal radius 30 cm and thickness 5 cm. Find the length of the pipe.

#### Answer:

For cuboidal block

Length, *l* = 4 m

Breadth, *b* = 2.6 m

Height, *h* = 1 m

We know that,

Volume of tank = *lbh*

Volume of cuboid = 4.4(2.6)(1) = 11.44 m^{3}

Also,

As the volume remains same when a body is recast to another body.

We have

Volume of cylindrical pipe = 11.44 m^{3}

Internal radius, *r*_{2} = 30 cm = 0.3 m

Thickness = 5 cm

External radius, *r*_{1} = Internal radius + thickness = 30 + 5 = 35 cm = 0.35 m

Let the length of pipe be *h*

Also, we know

Volume of hollow cylinder = $\mathrm{\pi}h\left[{\left({r}_{1}\right)}^{2}-{\left({r}_{2}\right)}^{2}\right]$ (Where h is height and *r*_{1} and *r*_{2 }are external and internal diameters respectively)

So, we have

Volume of pipe = $\mathrm{\pi}$*h*[(0.35)^{2} $-$ (0.3)^{2}]

$\Rightarrow 11.44=\frac{22}{7}\times h\times 0.0325\phantom{\rule{0ex}{0ex}}\Rightarrow h=\frac{11.44\times 7}{22\times 0.0325}=112\mathrm{m}\phantom{\rule{0ex}{0ex}}$

So, the length of pipe is 112 m.

#### Page No 151:

#### Question 10:

500 persons are taking a dip into a cuboidal pond which is 80 m long and 50 m broad. What is the rise of water level in the pond, if the average displacement of the water by a person is 0.04m^{3}?

#### Answer:

Let the rise of water level in the pond be *h* meters, when 500 persons are taking a dip into a cuboidal pond.

Given that,

Average displacement by a person = 0.04 m^{3}

Average displacement by 500 persons = 500 ×0.04 = 20 m^{3}

So, the volume of water raised in pond is 20 m^{3}

Also,

Length of pond, *l* = 80 m

Breadth of pond , *b* = 50 m

Height is *h*

And we know,

Volume of water raised in pond = 80(50)(*h*)

20 m^{3} = 4000*h*

$\Rightarrow $*h* = 0.005 m = 0.5 cm [As 1 m = 100 cm]

So, Raise in water height is 0.5 cm.

#### Page No 151:

#### Question 11:

16 glass spheres each of radius 2 cm are packed into a cuboidal box of internal dimensions 16 cm × 8 cm × 8 cm and then the box is filled with water. Find the volume of water filled in the box.

#### Answer:

For one glass sphere

Radius , *r* = 2 cm

As we know

Volume of hemisphere = $\frac{4\mathrm{\pi}}{3}{r}^{3}$

= $\frac{4\mathrm{\pi}}{3}{\left(2\right)}^{3}$

= 33.52 cm^{3}.

For cuboidal box,

Length, *l* = 16 cm

Breadth, *b* = 8 cm

Height, *h* = 8 cm

As,

Volume of cuboid = *lbh* where *l*, *b* and *h* are length, breadth and height respectively.

Volume of cuboid box = 16(8)(8) = 1024 cm^{3}

Volume of water = Volume of Cuboid $-$ Volume of 16 spheres

Volume of water = 1024 $-$ 16(33.52) = 487.6 cm^{3}

So, Volume of water in bucket is 487.6 cm^{3}.

#### Page No 151:

#### Question 12:

A milk container of height 16 cm is made of metal sheet in the form of a frustum of a cone with radii of its lower and upper ends as 8 cm and 20 cm respectively.

Find the cost of milk at the rate of Rs. 22 per litre which the container can hold.

#### Answer:

Clearly, Given container is in the form of frustum of a cone.

As we know

Volume of frustum of a cone = $\frac{1}{3}\mathrm{\pi h}\left[{\left({\mathrm{r}}_{1}\right)}^{2}+{\left({\mathrm{r}}_{2}\right)}^{2}+{\mathrm{r}}_{1}\times {\mathrm{r}}_{2}\right]$

Where, *h* = height, *r*_{1} and* r*_{2} are radii of two ends (*r*_{1} > *r*_{2})

For given bucket,

Radius of lower end, *r*_{2} = 8 cm

Radius of upper end, *r*_{1} = 20 cm

Height of container, *h* = 16 cm

Using these values,

We have.

$\Rightarrow \mathrm{Volume}=\frac{1}{3}\mathrm{\pi}16\left[{\left(8\right)}^{2}+{\left(20\right)}^{2}+20\times 8\right]$

= 10459.42 cm^{3}

= 10.45942 L = 10.46 L [Appx]

Also,

Cost for 1 L milk = Rs 22

Cost for 10.46 L milk = 22(10.46) = Rs 230.12

#### Page No 151:

#### Question 13:

A cylindrical bucket of height 32 cm and base radius 18 cm is filled with sand. This 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:

For cylindrical bucket,

Radius, *r* = 18 cm

Height, *h* = 32 cm

As we know,

Volume of cylinder = $\mathrm{\pi}$*r*^{2}*h*

Volume of sand in bucket = $\mathrm{\pi}$(18)^{2}(32) cm^{3}

Also, For conical heap

Let the radius be* r* and height,* h* = 24 cm (given),

As we know,

Volume of Cone = $\frac{1}{3}\pi {r}^{2}h$

= $\frac{1}{3}{\mathrm{\pi r}}^{2}\times 24$

= $8\pi {r}^{2}$ cm^{3}

As the volume of sand is constant

Volume of sand in bucket = Volume of conical heap

$\Rightarrow $$\mathrm{\pi}$ (18)^{2 }(32) = 8$\mathrm{\pi}$*r*^{2}

$\Rightarrow $(18)(18)(4) = *r*^{2}

$\Rightarrow $*r* = 18(2) = 36 cm

Also, we know

*$\Rightarrow $l*^{2} = (24)^{2} + (36)^{2} = 576 + 1296 = 1876 [*l*^{2} = *h*^{2} + *r*^{2}]

*$\Rightarrow $ l* = 43.267 cm

So, radius and slant height of heap are 36 cm and 43.267 cm respectively.

#### Page No 151:

#### Question 14:

A rocket is in the form of a right circular cylinder closed at the lower end and surmounted by a cone with the same radius as that of the cylinder. The diameter and height of the cylinder are 6 cm and 12 cm, respectively. If the the slant height of the conical portion is 5 cm, find the total surface area and volume of the rocket [Use π = 3.14].

#### Answer:

The diagram is given as:

For upper conical part,

Radius of base, *r* = 3 cm

Slant height, *l* = 5 cm

As,

${h}^{2}={l}^{2}-{r}^{2}$

$\Rightarrow {h}^{2}=25-9=16$

$\Rightarrow $*h* = 4 cm

Also,

Volume of cone = $\frac{1}{3}\mathrm{\pi}{r}^{2}h$

= $\frac{1}{3}\mathrm{\pi}{\left(3\right)}^{2}\left(4\right)$

= $12\mathrm{\pi}$ cm^{3}

Curved surface area of cone = $\mathrm{\pi}$*rl *= π(3)(5) = 15$\mathrm{\pi}$ cm^{2}

For cylindrical part,

Radius of base = Radius of base of conical part = *r* = 3 cm

Height, *h* = 12 cm

Also,

Volume of cylinder = $\mathrm{\pi}$*r*^{2}*h* = $\mathrm{\pi}$(3)^{2}(12) = 108$\mathrm{\pi}$ cm^{3}

Curved surface area of cylinder = 2$\mathrm{\pi}$*rh* = 2$\mathrm{\pi}$(3)(12) = 72$\mathrm{\pi}$ cm^{2}

Volume of rocket = volume of conical part + volume of cylindrical part

Volume of rocket = 12$\mathrm{\pi}$ + 108$\mathrm{\pi}$ = 120$\mathrm{\pi}$

Curved surface area of cylinder = 2$\mathrm{\pi}$*r*h = 2$\mathrm{\pi}$(3)(12) = 72$\mathrm{\pi}$ cm^{2}

Volume of rocket = volume of conical part + volume of cylindrical part

Volume of rocket = 12$\mathrm{\pi}$ + 108$\mathrm{\pi}$ = 120$\mathrm{\pi}$

= 377.14 cm^{3}

Also,

Surface area of rocket = Curved surface area of conical part + Curved surface area of Cylindrical part + Surface area of base of rocket

Surface area of base of rocket = $\mathrm{\pi}$*r*^{2} = $\mathrm{\pi}$(3)^{2} = 9$\mathrm{\pi}$ cm^{2}

Therefore,

Surface area of rocket = 15$\mathrm{\pi}$ + 72$\mathrm{\pi}$ + 9$\mathrm{\pi}$ = 94$\mathrm{\pi}$ cm^{2}

= 295.43 cm^{2}

#### Page No 151:

#### Question 15:

A building is in the form of a cylinder surmounted by a hemispherical vaulted dome and contains $41\frac{19}{21}$ m^{3} of air. If the internal diameter of dome is equal to its total height above the floor, find the height of the building?

#### Answer:

Let the radius of surmounted hemispherical dome =* r*

Diameter of hemispherical dome = 2*r*

Given,

Total height of dome = 2*r*

Height of hemispherical part = Radius of hemispherical part =* r*

Height of cylindrical part =* r*

As we know,

Volume of cylinder = π*r*^{2}h

Where* r* is base radius and *h* is height of cylinder.

Volume of cylindrical part = π*r*^{2}*r* = π*r*^{3} cm^{3}

Also,

Volume of hemisphere = $\frac{2}{3}\mathrm{\pi}{r}^{3}$, where *r* is radius of hemisphere.

Volume of building = Volume of cylindrical part + volume of hemispherical part

= $\mathrm{\pi}{r}^{3}+\frac{2}{3}\mathrm{\pi}{r}^{3}$

= $\frac{5}{3}{\mathrm{\pi r}}^{3}{\mathrm{cm}}^{3}$

Volume of air in building = volume of building

$\Rightarrow $$41\frac{19}{21}$= $\frac{5}{3}{\mathrm{\pi r}}^{3}$

$\Rightarrow \frac{880}{21}=\frac{5}{3}\mathrm{\pi}{r}^{3}$

$\Rightarrow {r}^{3}=8\phantom{\rule{0ex}{0ex}}\Rightarrow r=2$

Hence, the height of building = 2*r =* 4m.

#### Page No 151:

#### Question 16:

A hemispherical bowl of internal radius 9 cm is full of liquid. The liquid is to be filled into cylindrical shaped bottles each of radius 1.5 cm and height 4 cm. How many bottles are needed to empty the bowl?

#### Answer:

Let *n* bottles are needed to empty the bowl,

Therefore,

Volume of bowl = Volume of *n* bottles

For hemispherical bowl,

Radius, *r* = 9 cm

As we know,

Volume of a hemispherical bowl = $\frac{2}{3}\mathrm{\pi}{r}^{3}$

= $\frac{2}{3}\mathrm{\pi}{\left(9\right)}^{3}$ = 486 $\mathrm{\pi}$ cm^{3}

For single cylindrical bottle,

Base radius, *r* = 1.5 cm

Height, *h* = 4 cm

As we know,

Volume of cylinder = $\mathrm{\pi}$*r*^{2}*h*

Volume of one bottle = $\mathrm{\pi}$(1.5)^{2}(4) = 9$\mathrm{\pi}$ cm^{3}

As,

Volume of bowl = volume of *n* bottles

486$\mathrm{\pi}$ = (9$\mathrm{\pi}$)*n*

$\Rightarrow n=\frac{486}{9}=54$

Hence, 54 bottles are needed to empty the bowl.

#### Page No 151:

#### Question 17:

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

#### Answer:

Whenever we placed a solid right circular cone in a right circular cylinder with full of water, then volume of a solid right circular cone is equal to the volume of water spilled from the cylinder and Total volume of water in a cylinder is equal to the volume of the cylinder.

Therefore, we have,

Volume of water left in the cylinder = (Volume of cylinder) – (Volume of cone)

For cylinder,

Base radius, *r* = radius of cone = 60 cm

Height, *h* = 180 cm

As we know,

Volume of cylinder = $\mathrm{\pi}$*r*^{2}*h*

Volume of cylinder =$\mathrm{\pi}$(60)^{2}(180)

= 2036571.43 cm^{3}

For cone

Base radius, *r* = 60 cm

Height, *h* = 120 cm

As we know,

Volume of cone = $\frac{1}{3}\mathrm{\pi}{r}^{2}h$

= $\frac{1}{3}\mathrm{\pi}{\left(60\right)}^{2}\left(120\right)$

= 452571.43 cm^{3}

So,

Volume of water left in cylinder = 2036571.43 $-$ 452571.43 = 1584000 cm^{3}

#### Page No 152:

#### Question 18:

Water flows through a cylindrical pipe, whose inner radius is 1 cm, at the rate of 80 cm/sec in an empty cylindrical tank, the radius of whose base is 40 cm. What is the rise of water level in tank in half an hour?

#### Answer:

Water flows in 1 sec = 80 cm

Water flows in $\frac{1}{2}$ hour = $80\times \frac{3600}{2}$ = 144000 cm [As 1 hour = 3600 seconds]

For cylindrical pipe,

Base radius,* r *= 1 cm

Height, *h* = water flowed in half hour = 144000 cm

As we know,

Volume of cylinder = $\mathrm{\pi}$*r*^{2}*h*

Volume of water flowed through pipe = $\mathrm{\pi}$(1)^{2}(144000) = 144000$\mathrm{\pi}$cm^{3}

For cylindrical tank,

Base radius, *r* = 40 cm

Let the height of water raised be *h* cm

As we know,

Volume of cylinder = $\mathrm{\pi}$*r*^{2}*h*

Volume of water in tank = $\mathrm{\pi}$(40)^{2}*h*

$\Rightarrow $144000$\mathrm{\pi}$ = 1600$\mathrm{\pi}$*h*

*h* = 90 cm

Hence, the level of water in cylindrical tank rises 90 cm in half an hour.

#### Page No 152:

#### Question 19:

The rain water from a roof of dimensions 22 m × 20 m drains into a cylindrical vessel having diameter of base 2 m and height 3.5 m. If the rain water collected from the roof just fill the cylindrical vessel, then find the rainfall in cm.

#### Answer:

Let the rainfall be *x* cm [i.e. 0.01*x* meters because 1 m = 100 cm ] .

For cylindrical vessel,

Diameter of base = 2 m

Base Radius = 1 m

Height, *h* = 3.5 m

As we know,

Volume of cylinder = π*r*^{2}*h*

So,

Volume of cuboidal vessel = π(1)^{2}(3.5)

= 11 m^{3}

Also,

For roof

Length, *l* = 22 m

Breadth, *b* = 20 m

Height, *h* = height of rainfall = 0.01*x* m

As we know,

Volume of cuboid = *lbh*

Volume of water on roof = 22(20)(0.01*x*) = 4.4*x* m^{3}

Given,

Volume of water on roof = volume of cuboidal vessel

4.4*x* = 11

*x* = 2.5 cm

Height of rainfall is 2.5 cm.

#### Page No 152:

#### Question 20:

A pen stand made of wood is in the shape of a cuboid with four conical depressions and a cubical depression to hold the pens and pins, respectively. The dimension of the cuboid are 10 cm, 5 cm and 4 cm. The radius of each of the conical depressions is 0.5 cm and the depth is 2.1 cm. The edge of the cubical depression is 3 cm. Find the volume of the wood in the entire stand.

#### Answer:

Given,

For cuboidal stand,

Length,* l* = 10 cm

Breadth, *b* = 5 cm

Height,* h* = 4 cm

We know that

Volume of a cuboid = *lbh*

So,

Volume of cuboidal stand = 10(5)(4) = 200 cm^{3}

For one conical depression,

Radius, *r* = 0.5 cm

Height, i.e. depth, *h* = 2.1 cm

We know that

Volume of cone = $\frac{1}{3}\pi {r}^{\mathit{2}}h$

Volume of conical depression = $\frac{1}{3}\mathrm{\pi}{\left(0.5\right)}^{2}\left(2.1\right)$

= 0.55 cm^{3}

For Cubical depression,

Side,* a *= 3 cm

We know that

Volume of cube = *a*^{3}, where *a* is the side of the cube.

Volume of cubical depression = (3)^{3} = 27 cm^{3}

Volume of wood in the entire stand = volume of cuboidal stand $-$ volume of 4 conical depression $-$ volume of one cubical depression.

Volume of wood = 200 $-$ 4(5.5) $-$ 27 = 170.8 cm^{3}

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