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#### 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

A pencil is a combination of: 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).

#### 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.

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).

#### 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

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).

#### 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

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).

#### 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

A gilli is a combination of: 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).

#### 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

The cork of a shuttle is hemispherical in shape 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).

#### 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

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).

#### Question 8:

Choose the correct answer from the given four options:
A hollow cube of internal edge 22 cm 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 accommodate is
(A) 142244
(B) 142344
(C) 142444
(D) 142544

Given that, the edge of cube = 22 cm.
Now,
Volume of given hollow cube = a3 = (22)3 = 10648 cm3

Filled space of cube = $\frac{7}{8}$ of total volume of cube
= $\frac{7}{8}$ × 10648
= 9317 cm3

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

Now, the volume of one marble = volume of sphere of radius r.
∴ Volume of one marble = $\frac{4}{3}\mathrm{\pi }{r}^{3}$
= $\frac{4}{3}×\frac{22}{7}×{\left(0.25\right)}^{3}$

$\therefore$ The number of marbles that the cube can accommodate is 142244 (approx).
Hence, the correct answer is option (A).

#### 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 8 cm. The height of the cone is
(A) 12 cm
(B) 14 cm
(C) 15 cm
(D) 18 cm

Volume of spherical shell = Volume of cone recast by melting                           .....(1) For the Spherical Shell,
Internal diameter, d1 = 4 cm
Internal radius, r1 = 2 cm                                                           [as radius = $\frac{1}{2}$diameter]
External diameter, d2 = 8 cm
External radius, r2 = 4 cm

Now,

⇒ Volume of cone = $\frac{224\mathrm{\pi }}{3}$ cm3                                          [Using (1)]

For cone,
Base diameter = 8 cm
Base radius, r = 4 cm                                                                  [as radius = $\frac{1}{2}$diameter]
Let the height of cone be cm.
Volume of given cone = $\frac{1}{3}\mathrm{\pi }{\left(4\right)}^{2}h$
$⇒\frac{224\mathrm{\pi }}{3}=\frac{1}{3}\mathrm{\pi }{\left(4\right)}^{2}h$
⇒ 16h = 224
h = 14 cm

So, the height of the cone is 14 cm.
Hence, the correct answer is option (B).

#### Question 10:

Choose the correct answer from the given four options:
A solid piece of iron in the form of a cuboid of dimensions 49 cm × 33 cm × 24 cm, is molded to form a solid sphere. The radius of the sphere is
(A) 21cm
(B) 23 cm
(C) 25 cm
(D) 19 cm

For the given cuboid,
Length, l = 49 cm,
Breadth, b = 33 cm and
Height, h = 24 cm

Volume of cuboid = lbh
= 49 × 33 × 24 cm3
Now, let the radius of sphere be r.

Here,
Volume of cuboid = volume of sphere
$⇒49×33×24=\frac{4}{3}\mathrm{\pi }{r}^{3}$
$⇒29106×\frac{22}{7}={r}^{3}$

Hence, the correct answer is option (A).

#### Question 11:

Choose the correct answer from the given four options:
A mason constructs a wall of dimensions 270 cm × 300 cm × 350 cm with the bricks each of size 22.5 cm × 11.25 cm × 8.75 cm 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

For the wall,
Length, l = 270 cm
Height, h = 350 cm

Now,
Volume of wall = lbh
= 270 × 300 × 350
= 28350000 cm3

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

For one brick,
Length, l = 22.5 cm
Height, h = 8.75 cm

∴ Volume of one brick = lbh
= 22.5 × 11.25 × 8.75 cm3
= 2214.844 cm3

Hence, the correct answer is option (B).

#### 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

Volume of cylinder = $\mathrm{\pi }$r2h

For the given solid cylinder,
Base diameter = 2 cm
Base radius, r = 1 cm                                                                     [as radius = $\frac{1}{2}$diameter]
Height, h = 16 cm

Also,
Volume of cylinder, V = $\mathrm{\pi }$(1)2(16) = 16$\mathrm{\pi }$ cm3

Now, let the radius of sphere be r.
As the volume of sphere  = $\frac{4}{3}\mathrm{\pi }{r}^{3}$,

Now, volume remains same, when a metal body is melted and recast into another metal body.
∴ Volume of 12 spheres = Volume of cylinder
⇒ 16$\mathrm{\pi }$r3 = 16$\mathrm{\pi }$
r3 = 1

⇒ Diameter = 2r = 2 cm
Hence, the correct answer is option (C).

#### 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 cm2
(B) 4951 cm2
(C) 4952 cm2
(D) 4953 cm2

A bucket is in the form of frustum of a cone. Curved surface area of frustum of a cone = πl(r1 + r2)
For the given bucket,
Radius of top, r1 = 7 cm
Radius of bottom, r2 = 28 cm
Slant height, l = 45 cm

Therefore,
Curved surface area of bucket = π(45)(28 + 7)
= $\frac{22}{7}×45×35$
= 4950 cm2
So, the curved surface area of bucket is 4950 cm2.
Hence, the correct answer is option (A).

#### 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 cm3
(B) 0.35 cm3
(C) 0.34 cm3
(D) 0.33 cm3

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 = $\frac{1}{2}$diameter]

Also,
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 = 2r + h
⇒ 2 = 2(0.25) + h
h = 1.5 cm

Hence, the correct answer is option (A).

#### 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) 4$\mathrm{\pi }$r2
(B) 6$\mathrm{\pi }$r2
(C) 3$\mathrm{\pi }$r2
(D) 8$\mathrm{\pi }$r2

When two hemispheres are joined together along their bases, a sphere of same base radius is formed.
And the curved surface area of sphere is 4$\mathrm{\pi }$r2.

Hence, the correct answer is option (A).

#### Question 16:

Choose the correct answer from the given four options:
A right circular cylinder of radius r cm and height h cm (h > 2r) just encloses a sphere of diameter
(A) r cm
(B) 2r cm
(C) h cm
(D) 2h cm

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) = 2r

So, Diameter of sphere = 2r cm

Hence, the correct answer is option (B).

#### 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) remains unaltered
(D) be doubled

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

Hence, the correct answer is option (C).

#### 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

A bucket is in the form of frustum of a cone.

For the given bucket,
Diameter of top = 44 cm
Radius of top, r1 = 22 cm                                             [as radius = $\frac{1}{2}$diameter]
Diameter of bottom = 24 cm
Radius of bottom, r2 = 12 cm                                      [as radius = $\frac{1}{2}$diameter]
Height, h = 35 cm

Therefore,

Hence, the correct answer is option (A).

#### 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

If a cone is cut by a plane parallel to the base of the cone, the cross-section made is circle.

#### 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

Let the radius of spheres be r1 and r2.
And the volume of sphere of radius r $\frac{4\mathrm{\pi }{r}^{3}}{3}$

Given, ratio of volumes =  64 : 27

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

Now,
Surface area of sphere = 4$\mathrm{\pi }$r2

Hence, the correct answer is option (D).

#### Question 1:

Two identical solid hemispheres of equal base radius r cm are stuck together along their bases. The total surface area of the combination is 6$\mathrm{\pi }$r2.

False

When two hemispheres are joined together along their bases, a sphere of same base radius is formed.
And CSA of sphere is 4$\mathrm{\pi }$r2.

#### Question 2:

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 4$\mathrm{\pi }$rh + 4$\mathrm{\pi }$r2.

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.

Now, the total surface area of cylinder = 2$\mathrm{\pi }$rh + 2$\mathrm{\pi }$r2h, where r = base radius and h = height.

So,
Total surface area of shape formed = 2(Total surface of single cylinder) − 2(Area of base of cylinder)
= 2(2$\mathrm{\pi }$rh + 2$\mathrm{\pi }$r2) − 2($\mathrm{\pi }$r2)
= 4$\mathrm{\pi }$rh + 2$\mathrm{\pi }$r2

#### Question 3:

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 $\mathrm{\pi }r\left[\sqrt{{r}^{2}+{h}^{2}}+3r+2h\right]$.

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 = $\mathrm{\pi }$rl + $\mathrm{\pi }$r
Total surface area of cylinder = 2$\mathrm{\pi }$rh + 2$\mathrm{\pi }$r2
∴ Total surface area of shape formed =  $\mathrm{\pi }$rl + $\mathrm{\pi }$r+ 2$\mathrm{\pi }$rh + 2$\mathrm{\pi }$r2 − 2$\mathrm{\pi }$r2
= $\mathrm{\pi }$rl + $\mathrm{\pi }$r+ 2$\mathrm{\pi }$rh

Here, when we placed a cone over a cylinder, then one base is common for both.
So, the total surface area of combined solid = $\mathrm{\pi }$rl + $\mathrm{\pi }$r2 +  2$\mathrm{\pi }$rh

#### Question 4:

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

False

Let the radius of sphere be r.
As the ball is exactly fitted into the cubical box.
∴ Diameter of ball = Length of the edge of the cube
⇒ 2r = a                                                             (∵ Diameter = 2 × Radius)
$⇒r=\frac{a}{2}$

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

#### Question 5:

The volume of the frustum of a cone is $\frac{1}{3}\mathrm{\pi }h\left[{r}_{1}^{2}+{r}_{2}^{2}-{r}_{1}{r}_{2}\right]$, where h is vertical height of the frustum and r1, r2 are the radii of the ends.

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

#### Question 6:

The capacity of a cylindrical vessel with a hemispherical portion raised upward at the bottom as shown in the Fig. 12.7 is $\frac{\mathrm{\pi }{r}^{2}}{3}\left[3h-2r\right]$. 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,
Volume of a cylinder = $\mathrm{\pi }$r2h, where r = base radius and h = height of cylinder
And
Volume of hemisphere = $\frac{2}{3}\mathrm{\pi }{r}^{3}$

Here,

So, given statement is true.

#### Question 7:

The curved surface area of a frustum of a cone is $\mathrm{\pi }$l (r1 + r2), where $l=\sqrt{{h}^{2}+{\left({r}_{1}+{r}_{2}\right)}^{2}}$r1 and r2 are the radii of the two ends of the frustum and h is the vertical height.

False

Curved surface area of frustum = $\mathrm{\pi }$l(r1 + r2)
where r1 and r2 are the radii of two ends (r1 > rand 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]

#### 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.

The resulting shape will be as in figure: The area of base of frustum, top of frustum and top of cylinder, has no metal sheet.
So, required area will be = curved surface area of frustum + area of base of cylinder + curved surface area of cylinder
Thus, the given statement is TRUE.

#### 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.

The volume of cube = a3, where a = side of cube

Now,
Side of first cube, a1 = 3 cm
Side of second cube, a2 = 4 cm
Side of third cube, a3 = 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 1st + 2nd + 3rd cube)
$⇒{a}^{3}={\left({a}_{1}\right)}^{3}+{\left({a}_{2}\right)}^{3}+{\left({a}_{3}\right)}^{3}$
$⇒{a}^{3}={\left(3\right)}^{3}+{\left(4\right)}^{3}+{\left(5\right)}^{3}$
a3 = 27 + 64 + 125
= 216
a = 6 cm
So, side of cube so formed is 6 cm.

#### Question 2:

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

Volume of cuboid = lbh

Length, l = 9 cm
Height, h = 12 cm

∴ Volume of lead = 9(11)(12)
= 1188 cm3
Now, volume of sphere with radius r$\frac{4}{3}\mathrm{\pi }{r}^{3}$

For spherical shots,
Diameter = 3 cm
⇒ Radius, r = 1.5 cm

Now,

#### 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.

A bucket is in the form of frustum of a cone. Now,
Volume of frustum of a cone = $\frac{1}{3}\mathrm{\pi }h\left[{\left({r}_{1}\right)}^{2}+{\left({r}_{2}\right)}^{2}+{r}_{1}×{r}_{2}\right]$

For the given bucket,
Volume of bucket = 28.490 L
= 28490 cm3                              (∵ 1 L = 1000 cm3)

Radius of top, r1 = 28 cm
Radius of bottom, r2 = 21 cm
Let the height be h.

Using these values,

Hence, the height of the bucket is 15 cm.

#### 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.

Let ORN be the cone, when it 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 = h = OM = 12 cm

Let P be the mid-point of cone, then OP = PM = 6 cm.

Now, in $△$OPD and $△$OMN,
∠MON = ∠POD                            (common)
∠OPD = ∠OMN = 90°
⇒ $△$OPD ~ $△$OMN                     (by Angle-Angle similarity)

Now, for the first part i.e., the cone
Base Radius, r1 = PD = 4 cm
Height, h1 = OP = 6 cm
$\begin{array}{rcl}\therefore V& =& \frac{1}{3}\mathrm{\pi }{r}^{2}h\\ & =& \frac{1}{3}\mathrm{\pi }{\left(4\right)}^{2}×6\\ & =& 32\mathrm{\pi }\end{array}$

Now, for the second part, i.e. the frustum
Bottom radius, r2 = MN = 8 cm
Top Radius, r1 = PD = 4 cm
Height, h = PM = 6 cm

So,
Volume of first part : Volume of second part  =  32$\mathrm{\pi }$ : 224$\mathrm{\pi }$  = 1 : 7

#### Question 5:

Two identical cubes each of volume 64 cm3 are joined together end to end. What is the surface area of the resulting cuboid?

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)
⇒ Surface area of resulting cuboid = 2(6a2) − 2(a2) = 10a2                  (∵ total surface area of cube = 6a2)

Also,
Volume of cube = a3
⇒ 64 = a3                                     (Given volume of cube is 64 cm3)
⇒ a = 4 cm
Therefore,
Surface area of resulting cuboid = 10a2 = 10(4)2 = 160 cm2

#### 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.

Since the conical cavity is hollowed out from the cube.
Volume of remaining solid = volume of cube − volume of cone

Now, for the cube,
Side, a = 7 cm
⇒ Volume of cube = (7)3                                 (∵ Volume of cube = a3)
= 343 cm3
For the cone,
Height, h = 7 cm
⇒ Volume of cone =                                       (∵ Volume of cone = $\frac{1}{3}\mathrm{\pi }{r}^{2}h$)
∴ Volume of remaining solid = 343 − 66 = 277 cm3

#### 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.

The total surface area of shape formed will be the sum total of the surface areas of the two cones. Now, the two cones are identical.

Thus,
Total surface area of shape formed = Curved surface area of first cone + Curved surface area of second cone
= 2(Surface area of the cone)

∴ Total Surface area of shape so formed = 2$\mathrm{\pi }$rl                             (∵ Surface area of cone = $\mathrm{\pi }$rl)

Here,
Height, h = 15 cm

So,
Area = 2$\left(\frac{22}{7}\right)$(8)(17) = 855 cm2.

#### 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. Diameter of Cylinder = 6 cm
⇒ 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 h1 and cone B be h2.
Given,
Ratio of volume of cones is 2 : 1
i.e.
$⇒\frac{\frac{1}{3}\mathrm{\pi }{r}^{2}{h}_{1}}{\frac{1}{3}\mathrm{\pi }{r}^{2}{h}_{2}}=\frac{2}{1}\phantom{\rule{0ex}{0ex}}⇒\frac{{h}_{1}}{{h}_{2}}=\frac{2}{1}\phantom{\rule{0ex}{0ex}}$
h1 = 2h2

Now, the total height of cylinder is 21 cm.
⇒ h1 + h2 = 21
⇒ 2h2 + h2 = 21
⇒ 3h2 = 21
⇒ h2 = 7 cm
∴ h1 = 2h2 = 2(7) = 14 cm

Volume of Cone A = $\frac{1}{3}\mathrm{\pi }{r}^{2}{h}_{1}$
= $\frac{1}{3}\mathrm{\pi }{\left(3\right)}^{2}×14$
= 132 cm3

Volume of Cone B = $\frac{1}{3}\mathrm{\pi }{r}^{2}{h}_{2}$
= $\frac{1}{3}\mathrm{\pi }{\left(3\right)}^{2}×7$
= 66 cm3

Volume of given cylinder = $\mathrm{\pi }$(3)2(21)               (∵ Volume of cylinder = $\mathrm{\pi }$r2h)
= 594 cm3

Volume of remaining solid = (Volume of cylinder) − (volume of cone A) − (volume of cone B)
= 594 − 132 − 66
= 396 cm3

#### 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. An 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
$=\frac{2}{3}\mathrm{\pi }{r}^{3}+\frac{1}{3}\mathrm{\pi }{r}^{2}h$
$=\frac{1}{3}\mathrm{\pi }{r}^{2}\left(2r+h\right)$
$=\frac{1}{3}×\frac{22}{7}×{\left(5\right)}^{2}×\left(10+5\right)$
= 392.85 cm2

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

Hence, the volume of ice-cream = $\frac{5}{6}×392.85$
= 327.4 cm3

#### 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.

Let x be the number of marbles that 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
= $\mathrm{\pi }$r2h
$=\mathrm{\pi }×{\left(3.5\right)}^{2}×5.6$

The diameter of marble is 1.4 cm.
⇒ Radius of marble, r = 0.7 cm

⇒ Volume of one marble = $\frac{4}{3}\mathrm{\pi }{\left(0.7\right)}^{3}$ cm3

⇒ Volume of x marbles = $\frac{4}{3}\mathrm{\pi }{\left(0.7\right)}^{3}×\left(x\right)$ cm3

So,
$\frac{4x}{3}\mathrm{\pi }{\left(0.7\right)}^{3}=\mathrm{\pi }×{\left(3.5\right)}^{2}×5.6$
$⇒x=\frac{\mathrm{\pi }×{\left(3.5\right)}^{2}×5.6×3}{4×\mathrm{\pi }×{\left(0.7\right)}^{3}}$
Therefore, 150 marbles are required.

#### 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?

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

Length, l = 66 cm
Height, h = 21 cm

⇒ Volume of lead = 66(42)(21) = 58212 cm3

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

For the spherical shots,
Diameter = 4.2 cm

⇒ Volume of one shot =

#### 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?

The volume of cube = a3, where a = side of cube.

For the cubical lead, side = a = 44 cm
∴ Volume of lead = 44 × 44 × 44 cm3

Also, the volume of sphere = $\frac{4}{3}\mathrm{\pi }{r}^{3}$
Now, for spherical shots,
Diameter = 4 cm
⇒ Radius, r = 2 cm
∴ Volume of one shot = $\frac{4\mathrm{\pi }}{3}×{\left(2\right)}^{3}$

Now,

#### 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.

Volume of cuboid = lbh, where
l = length,
h = height

For the wall,
Length, l = 24 m
Height, h = 6 m
∴ Volume of wall = 24(0.4)(6) = 57.6 m3

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

∴ Volume of wall covered by bricks = $\frac{9}{10}\left(57.6\right)$
= 51.84 cm3

For one brick,
Length, l = 25 cm = 0.25 m                       (∵ 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) m3

Hence, the total number of bricks used in constructing the wall is 12960.

#### 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.

For the Cone,
Base Diameter = 4.5 cm
Base radius, r = 2.25 cm
Height of cone, h = 10 cm

Now, volume of a cone = $\frac{1}{3}\pi {r}^{2}h$
Here,
Volume of given cone = $\frac{1}{3}\mathrm{\pi }×{\left(2.25\right)}^{2}×10$

Now, the disk is in the shape of a cylinder.

For a single disk,
Base diameter = 1.5 cm
Base radius, r = 0.75 cm
∴ Volume of a disk = $\mathrm{\pi }$(0.75)2(0.2) = (0.75 × 0.75 × 0.2)$\mathrm{\pi }$ cm3

Now,
No. of disks that can be made by melting cone =

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

#### Question 1:

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

For the given hemisphere,

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

Here,
Volume of given hemisphere = $\frac{2}{3}\mathrm{\pi }{\left(8\right)}^{3}$
=
Now, the cone formed has
Base radius, r = 6 cm
And
Volume of a cone = $\frac{1}{3}\mathrm{\pi }{\left(r\right)}^{2}h$

Here,
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
$\frac{1}{3}\mathrm{\pi }{\left(6\right)}^{2}h=\frac{1024\mathrm{\pi }}{3}\phantom{\rule{0ex}{0ex}}⇒h=\frac{1024×3}{3×6×6}$
$⇒$h = 28.44 cm

#### 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.

Volume of water in tank = volume of cuboidal tank up to a height of 5 m
For the cuboidal tank,
Length, l = 11 m
Height, h = 5 m

Now,
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 m3

Also, let the cylindrical tank is filled up to a height of h m.
Base radius of cylindrical tank, r = 3.5 m

Now,
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 = $\mathrm{\pi }{\left(3.5\right)}^{2}h$

∴ h = 8.57 m

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

#### 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.

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

Now,
Volume of metal used in box = Volume of external box − Volume of internal box

For the external box,
Length, l = 36 cm
Height, h = 16.5 cm

Also, volume of cuboid = lbh
∴ Volume of external box = 36(25)(16.5)
= 14850 cm3

The box is open from top.
So, for the 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 cm3

Also,
3960 cm3 weighs 3960 × 7.5 = 29,700 g                                [1 cm3 weighs 7.5 g]
Hence, the weight of box is 29,700 g i.e. 29.7 kg.

#### 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?

For the barrel,
Base diameter = 5 mm = 0.5 cm                                                   (∵ 1 cm = 10 mm)
⇒ Base radius, r = 0.25 cm
Height, h = 7 cm

Now,
Volume of a cylinder = πr2h
where, r is base radius and h is the height of cylinder.

Here,
Volume of barrel = $\mathrm{\pi }{\left(0.25\right)}^{2}×7$
= 1.375 cm3

So,
1.375 cm3 of ink can write 3300 words.
The number of words that can be written by 1 cm3$\frac{3300}{1.375}$ = 2400 words
[Also, $\frac{1}{5}$ of a litre i.e. 0.2 L  = 200 cm3]                                 (∵ 1 L = 1000 cm3)

∴ Number of words that can be written by 200 cm3 = 2400(200)
= 480000 words
Hence, one-fifth of a litre ink can write 480000 words on an average.

#### Question 5:

Water flows at the rate of 10 m/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?

Let the time taken by pipe to fill vessel is t minutes.
As water flows 10 m in 1 minute, it will flow 10t meters in t minutes.

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

Now, for the conical pipe,
Base Diameter = 40 cm
⇒ Base radius, r = 20 cm
Height, h = 24 cm

The volume of conical vessel = $\frac{1}{3}\mathrm{\pi }{r}^{2}h$
= $\frac{1}{3}\mathrm{\pi }{\left(20\right)}^{2}×24$
= 3200$\mathrm{\pi }$ cm3

For the cylindrical pipe,
Base diameter = 5 mm = 0.5 cm                             (∵ 1 cm = 10 mm)
⇒ Base radius, r = 0.25 cm
Height, h = 10t m = 1000t cm                                 (∵ water covers 10t m distance in pipe)

Now,
Volume of a cylinder = $\mathrm{\pi }{r}^{2}h$
∴ Volume of water passed in pipe = $\mathrm{\pi }$(0.25)2(1000t) = 62.5t$\mathrm{\pi }$ cm3

So,
62.5t$\mathrm{\pi }$ = 3200
⇒ 62.5t = 3200
⇒ t = 51.2 minutes
⇒ t = 51 minutes 12 seconds                         (∵ 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.

#### 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?

Given that for the conical heap,
Base Diameter = 9 cm
⇒ Base radius, r = 4.5 cm
Height, h = 3.5 cm

Now, the slant height is given as:

Therefore,
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 cm3

Also,
Canvas requires to just cover heap = Curved surface area of conical heap
And
Curved surface area of a cone = $\mathrm{\pi }$rl
Canvas required = $\mathrm{\pi }$(4.5)(5.7)
= 80.61 cm2 (approx.)

#### 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 dm2.

The pencil is in shape of cylinder.
Let the radius of base be r cm.

Now, the circumference of base = 1.5 cm
$⇒$2$\mathrm{\pi }$r = 1.5 cm                                  (∵ Circumference of circle is 2$\mathrm{\pi }$r)
$⇒r=\frac{1.5}{2\mathrm{\pi }}$

Also, the height, h = 25 cm.

Now,
Curved surface area of cylinder = 2$\mathrm{\pi }$rh
Curved surface area of pencil = $2\mathrm{\pi }×\left(\frac{1.5}{2\mathrm{\pi }}\right)×\left(25\right)$
= 37.5 cm2

Cost for coloring 1 dm2 = ₹0.05
∴ Cost for coloring 0.375 dm2 (i.e. 1 pencil) = ₹0.01875                      (∵ 37.5 cm2 = 0.375 dm2)
Cost for coloring 120000 pencils = 120000 × 0.01875
= ₹2250

#### 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?

Let the time taken by pipe to fill pond is t hours.
As water flows 15 km in 1 hour, it will flow 15t 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 the cuboidal pond,
Length, l = 50 m
Height, h = 21 cm = 0.21 m

Also, the volume of a cuboidal tank = lbh
∴ Volume of water = 50(44)(0.21)
= 462 m3

For the cylindrical pipe,
Base diameter = 14 cm
Base radius, r = 7 cm = 0.07 m
Height, h = 15t km = 15000t m                                      [1 km = 1000 m]

Also, the volume of a cylinder =$\mathrm{\pi }$r2h.
∴ Volume of water passed in pipe = $\mathrm{\pi }$(0.07)2(15000t)
= 231t cm3

So,
231t = 462
⇒ t = 2 hours
Time required to fill tank up to a height of 21 cm is 2 hours.

#### 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.

For the cuboidal block,
Length, l = 4 m
Height, h = 1 m

Now,
Volume of a tank = lbh
∴ Volume of cuboid = 4.4(2.6)(1)
= 11.44 m3

As the volume remains same when a body is recast to another body,
∴ Volume of cylindrical pipe = 11.44 m3

Now, the internal radius, r2 = 30 cm = 0.3 m
Thickness = 5 cm = 0.05 m
External radius, r1 = Internal radius + thickness = 30 + 5 = 35 cm = 0.35 m

Let the length of pipe be h.

Also,
Volume of hollow cylinder =
where, h is height and r1 and r2 are external and internal diameters respectively

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

So, the length of pipe is 112 m.

#### 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 0.04 m3?

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 m3
Average displacement by 500 persons = 500 × 0.04 = 20 m3
So, the volume of water raised in pond is 20 m3.

Now,
Length of pond, l = 80 m
Breadth of pond , b = 50 m
Height is h

And
Volume of water raised in pond = 80(50)(h)
= 4000h
⇒ 20 m3 = 4000h
⇒ h = 0.005 m
= 0.5 cm                    (∵ 1 m = 100 cm)
So, the rise in water height is 0.5 cm.

#### 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.

For one glass sphere,
Radius , r = 2 cm
And
Volume of hemisphere = $\frac{4\mathrm{\pi }}{3}{r}^{3}$
= $\frac{4\mathrm{\pi }}{3}{\left(2\right)}^{3}$
= 33.52 cm3

For the cuboidal box,
Length, l = 16 cm
Height, h = 8 cm

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

Here,
Volume of cuboid box = 16(8)(8)
= 1024 cm3
Volume of water = Volume of Cuboid − Volume of 16 spheres
∴ Volume of water = 1024 − 16(33.52)
= 487.6 cm3
So, the volume of water in bucket is 487.6 cm3.

#### 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.

The given container is in the form of frustum of a cone.
Now,
Volume of frustum of a cone = $\frac{1}{3}\mathrm{\pi }h\left[{\left({r}_{1}\right)}^{2}+{\left({r}_{2}\right)}^{2}+{r}_{1}×{r}_{2}\right]$
where, h = height, r1 and r2 are radii of two ends (r1 > r2

For the given bucket,
Radius of lower end, r2 = 8 cm
Radius of upper end, r1 = 20 cm
Height of container, h = 16 cm

Putting these values,
$\begin{array}{rcl}& ⇒& \mathrm{Volume}=\frac{1}{3}\mathrm{\pi }16\left[{\left(8\right)}^{2}+{\left(20\right)}^{2}+20×8\right]\\ & =& \frac{1}{3}\mathrm{\pi }16\left[64+400+160\right]\\ & =& \frac{1}{3}×\frac{22}{7}×16×624\end{array}$

Also,
Cost for 1 L milk = ₹22
Cost for 10.46 L milk = 22(10.46) = ₹230.12

#### 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.

For the cylindrical bucket,
Height, h = 32 cm

Now,
Volume of cylinder = $\mathrm{\pi }$r2h
∴ Volume of sand in bucket = $\mathrm{\pi }$(18)2(32) cm3

Also, For conical heap
Let the radius be r and height, h = 24 cm (given).

Now,
Volume of Cone = $\frac{1}{3}\mathrm{\pi }{r}^{2}h$
= $\frac{1}{3}\mathrm{\pi }{r}^{2}×24$
=  $8\mathrm{\pi }{r}^{2}$ cm3

As the volume of sand is constant.
Volume of sand in bucket = Volume of conical heap
⇒ $\mathrm{\pi }$(18)(32) = 8$\mathrm{\pi }$r2
⇒ (18)(18)(4) = r2
⇒ r = 18(2) = 36 cm

Also,
⇒ l2 = (24)2 + (36)2                         (∵ l2 = h2 + r2)
= 576 + 1296
= 1872
l = 43.267 cm
Hence, the radius and slant height of heap are 36 cm and 43.267 cm respectively.

#### 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 $\mathrm{\pi }$ = 3.14].

The diagram is given as: For the upper conical part,
Radius of base, r = 3 cm
Slant height, l = 5 cm

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

h = 4 cm
Also,
Volume of cone = $\frac{1}{3}\mathrm{\pi }{r}^{2}h$
= $\frac{1}{3}\mathrm{\pi }{\left(3\right)}^{2}\left(4\right)$
= $12\mathrm{\pi }$ cm3

Curved surface area of cone = $\mathrm{\pi }$rl = $\mathrm{\pi }$(3)(5)
= 15$\mathrm{\pi }$ cm2

For the cylindrical part,
Radius of base = Radius of base of conical part = r = 3 cm
Height, h = 12 cm

Also,
Volume of cylinder = $\mathrm{\pi }$r2h = $\mathrm{\pi }$(3)2(12)
= 108$\mathrm{\pi }$ cm3
Curved surface area of cylinder = 2$\mathrm{\pi }$rh = 2$\mathrm{\pi }$(3)(12) = 72$\mathrm{\pi }$ cm2
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 cm3

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 = $\mathrm{\pi }$r2 = $\mathrm{\pi }$(3)2 = 9$\mathrm{\pi }$ cm2
Therefore,
Surface area of rocket = 15$\mathrm{\pi }$ + 72$\mathrm{\pi }$ + 9$\mathrm{\pi }$ = 96$\mathrm{\pi }$ cm2
= 301.71 cm2

#### Question 15:

A building is in the form of a cylinder surmounted by a hemispherical vaulted dome and contains $41\frac{19}{21}$ m3 of air. If the internal diameter of dome is equal to its total height above the floor, find the height of the building.

Let the radius of surmounted hemispherical dome = r
⇒ Diameter of hemispherical dome = 2r
Given that,
Total height of dome = 2r
Height of hemispherical part = Radius of hemispherical part = r
Height of cylindrical part = r Also,
Volume of cylinder = $\mathrm{\pi }$r2h
where, r is base radius and h is height of cylinder.

Here,
Volume of cylindrical part = $\mathrm{\pi }$r2r = $\mathrm{\pi }$r3 cm3

Now,
Volume of  hemisphere = $\frac{2}{3}\mathrm{\pi }{r}^{3}$, where r is radius of hemisphere.
And
Volume of building = Volume of cylindrical part + volume of hemispherical part
= $\mathrm{\pi }{r}^{3}+\frac{2}{3}\mathrm{\pi }{r}^{3}$
=

Again,
Volume of air in building = volume of building
$⇒$$41\frac{19}{21}$ =
$⇒\frac{880}{21}=\frac{5}{3}\mathrm{\pi }{r}^{3}$
$⇒{r}^{3}=8\phantom{\rule{0ex}{0ex}}⇒r=2$

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

#### 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?

Let n bottles are needed to empty the bowl.
Therefore,
Volume of bowl = Volume of  n bottles

For the hemispherical bowl,
Now,
Volume of a hemispherical bowl = $\frac{2}{3}\mathrm{\pi }{r}^{3}$
= $\frac{2}{3}\mathrm{\pi }{\left(9\right)}^{3}$
= 486$\mathrm{\pi }$ cm3

For a single cylindrical bottle,
Base radius, r = 1.5 cm
Height, h = 4 cm

Now,
Volume of cylinder = $\mathrm{\pi }$r2h

Here,
Volume of one bottle = $\mathrm{\pi }$(1.5)2(4)
= 9$\mathrm{\pi }$ cm3
As,
Volume of bowl = volume of n bottles
∴ 486$\mathrm{\pi }$ = (9$\mathrm{\pi }$)n
$⇒n=\frac{486}{9}=54$
Hence, 54 bottles are needed to empty the bowl.

#### 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.

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,
Volume of water left in the cylinder = (Volume of cylinder) − (Volume of cone)

For cylinder,
Height, h = 180 cm

Now,
Volume of cylinder = $\mathrm{\pi }$r2h
Volume of cylinder = $\mathrm{\pi }$(60)2(180)
= 2036571.43 cm3
For the cone
Base radius, r = 60 cm
Height, h = 120 cm

Now,
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 cm3
So,
Volume of water left in cylinder = 2036571.43 − 452571.43 = 1584000 cm3

#### 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?

Water flows in 1 sec = 80 cm
Water flows in $\frac{1}{2}$ hour = $80×\frac{3600}{2}$ = 144000 cm                            (∵ 1 hour = 3600 seconds)

For the cylindrical pipe,
Base radius, r = 1 cm
Height, h = water flowed in half hour = 144000 cm

Now,
Volume of cylinder = $\mathrm{\pi }$r2h
Volume of water flowed through pipe = $\mathrm{\pi }$(1)2(144000)
= 144000$\mathrm{\pi }$cm3

For the cylindrical tank,
Base radius, r = 40 cm

Let the height of water raised be h cm.

Now,
Volume of cylinder = $\mathrm{\pi }$r2h
Volume of water in tank = $\mathrm{\pi }$(40)2h
⇒ 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.

#### 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.

Let the rainfall be x cm.
Thus,
0.01x meters because 1 m = 100 cm.

For the cylindrical vessel,
Diameter of base = 2 m
Height, h = 3.5 m

Now,
Volume of cylinder = $\mathrm{\pi }$r2h
So,
Volume of cylindrical vessel = $\mathrm{\pi }$(1)2(3.5)
= 11 m3

For the roof,
Length, l = 22 m
Height, h = height of rainfall = 0.01x m

Now,
Volume of cuboid = lbh
∴ Volume of water on roof = 22(20)(0.01x)
= 4.4x m3

Given that,
Volume of water on roof = volume of cuboidal vessel
⇒ 4.4x = 11
⇒ x = 2.5 cm
Hence, the height of rainfall is 2.5 cm.

#### 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.

Given that,
For cuboidal stand,
Length, l = 10 cm
Height, h = 4 cm

Now,
Volume of a cuboid = lbh

So,
Volume of cuboidal stand = 10(5)(4)
= 200 cm3  For one conical depression,
Height, i.e. depth, h = 2.1 cm
Also,
Volume of cone = $\frac{1}{3}\mathrm{\pi }{r}^{\mathit{2}}h$
∴ Volume of conical depression = $\frac{1}{3}\mathrm{\pi }{\left(0.5\right)}^{2}\left(2.1\right)$
= 0.55 cm3

For the cubical depression,
side = a = 3 cm
Now,
Volume of cube = a3, where a is the side of the cube.
∴ Volume of cubical depression = (3)3
= 27 cm3
Finally,
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
= 151 cm3

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