CBSE 10th Maths Chapter 13 - Surface Areas and Volumes-Exercise-13.3

It is given that radius of the sphere (R) = 4.2 cm

Also, the radius of the cylinder (r) = 6 cm

Now, let the height of the cylinder = h

It is given that the sphere is melted into a cylinder.

So, the volume of the sphere = Volume of the cylinder

∴ ($\dfrac{4}{3}$)×π×R$^{3}$ = π×$r^{2}$ ×h.

h = 2.74 cm

For Sphere 1:

Radius $(r_1)$ = 6 cm

∴ Volume $(V_1)$ = $(\dfrac{4}{3})×π×(r_1)^{3}$

For Sphere 2:

Radius $(r_2) $= 8 cm

∴ Volume $(V_2)$ = $(\dfrac{4}{3})×π×(r_2)^{3}$

For Sphere 3:

Radius $(r_3)$ = 10 cm

∴ Volume $(V_3)$ = $(\dfrac{4}{3})× π× (r_3)^{3}$

Also, let the radius of the resulting sphere be “r”

Now,

The volume of the resulting sphere = $V_1+V_2+V_3$

$(\dfrac{4}{3})×π×r^{3} $= $(\dfrac{4}{3})×π×(r_1)^{3}+(\dfrac{4}{3})×π×(r_2)^{3} +(\dfrac{4}{3})×π× (r_3)^{3}$

$r^{3}$ = 63+83+103

$r^{3}$ = 1728

r = 12 cm

It is given that the shape of the well is the shape of a cylinder with a diameter of 7 m

So, radius = $\dfrac{7}{2}$ m

Also, Depth (h) = 20 m

The volume of the earth dug out will be equal to the volume of the cylinder

Let the height of the platform = H

The volume of soil from the well (cylinder) = Volume of soil used to make such a platform

π×$r^{2}$×h = Area of platform × Height of the platform

We know that the dimension of the platform is = 22×14

So, the Area of the platform = 22×14 $m^{2}$

∴ π×$r^{2}$×h = 22×14×H

⇒ H = 2.5 m

The shape of the well will be cylindrical, as given below.

Given, depth $(h_{1})$ of well = 14 m

Diameter of the circular end of the well =3 m

So, Radius $(r_1)$ = $\dfrac{3}{2}$ m

Width of the embankment = 4 m

From the figure, it can be said that the embankment will be a cylinder having an outer radius $(r_2)$ as 4+($\dfrac{3}{2})$ = $\dfrac{11}{2}$ m and an inner radius $(r_1)$ as $\dfrac{3}{2}m$

Now, let the height of the embankment be $h_2$

∴ The volume of soil dug from the well = Volume of earth used to form the embankment

π×$(r_1)^{2} ×h_1$ = π×$((r_2)^{2}-(r_1)^{2}) \times h_2$

Solving this, we get,

The height of the embankment $(h_2)$ is 1.125 m.

The number of cones will be = $\dfrac{Volume \:\:of\:\: cylinder}{Volume \:\:of \:\:ice\:\: cream \:\:cone}$

For the cylinder part,

Radius = $\dfrac{12}{2}$ = 6 cm

Height = 15 cm

∴ Volume of cylinder = $π×r^{2}×h$ = 540π

For the ice cone part,

Radius of conical part = $\dfrac{6}{2}$ = 3 cm

Height = 12 cm

Radius of hemispherical part = $\dfrac{6}{2}$= 3 cm

Now,

The volume of the ice cream cone = Volume of the conical part + Volume of the hemispherical part

= $(\dfrac{1}{3})×π \times r^{2} \times h+(\dfrac{2}{3}) \times π \times r^{3}$

= 36π +18π

= 54π

∴ Number of cones = ($\dfrac{540π}{54π}$)

= 10

It is known that the coins are cylindrical in shape.

So, height ($h_1$) of the cylinder = 2 mm = 0.2 cm

Radius (r) of circular end of coins = $\dfrac{1.75}{2}$ = 0.875 cm

Now, the number of coins to be melted to form the required cuboids be “n”

So, Volume of n coins = Volume of cuboids

n × π × $r^{2} \times h_1$ = l × b × h

n×π×$(0.875)^{2}$×0.2 = 5.5×10×3.5

Or, n = 400

The diagram will be as-

Given,

Height $(h_1)$ of cylindrical part of the bucket = 32 cm

Radius $(r_1)$ of circular end of the bucket = 18 cm

Height of the conical heap $(h_2)$ = 24 cm

Now, let “$r_2$” be the radius of the circular end of the conical heap.

We know that volume of the sand in the cylindrical bucket will be equal to the volume of sand in the conical heap.

∴ The volume of sand in the cylindrical bucket = Volume of sand in the conical heap

π×$(r_1)^{2}×h_1$ = $(\dfrac{1}{3})×π×(r_2)^{2}×h_2$

π×$18^{2}$×32 = ($\dfrac{1}{3}$)×π ×$(r_2)^{2}$×24

Or, $r_2$= 36 cm

And,

Slant height (l) =$ \sqrt{(362+242)}$ = 12$\sqrt{13 }$cm.

It is given that the canal is the shape of a cuboid with dimensions as:

Breadth (b) = 6 m and Height (h) = 1.5 m

It is also given that

The speed of canal = 10 km/hr

Length of canal covered in 1 hour = 10 km

Length of canal covered in 60 minutes = 10 km

Length of canal covered in 1 min = ($\dfrac{1}{60}$)x10 km

Length of canal covered in 30 min (l) = ($\dfrac{30}{60})$x10 = 5km = 5000 m

We know that the canal is cuboidal in shape. So,

The volume of the canal = lxbxh

= 5000x6x1.5 $m^{3}$

= 45000 $m^{3}$

Now,

The volume of water in the canal = Volume of area irrigated

= Area irrigated x Height

So, Area irrigated = 56.25 hectares

∴ The volume of the canal = lxbxh

45000 = Area irrigated x 8 cm

45000 = Area irrigated x ($\dfrac{8}{100}$)m

Or, Area irrigated = 562500 $m^{2}$ = 56.25 hectares.

Consider the following diagram-

Radius $(r_1)$ of circular end of pipe = $\dfrac{20}{200}$=0.1m

Area of cross-section=π x$(r_1)^{2}$

=πx$(0.1)^{2}$

=0.01π$m^{2}$

Speed of water=3km/h

=$\dfrac{3000}{60}$

=50 metre/min

The volume of water that flows in 1 minutes from pipe = 50×0.01π =0.5π$m^{3}$

The volume of water that flows in t minutes from pipe = t×0.5π m$m^{3}$

Radius $(r_2)$ of circular end of cylindrical tank =$\dfrac{10}{2}$ = 5 m

Depth $(h_2)$ of cylindrical tank = 2 m

Let the tank be filled completely in t minutes.

The volume of water filled in the tank in t minutes is equal to the volume of water flowed in t minutes from the pipe.

The volume of water that flows in t minutes from pipe = Volume of water in tank

t×0.5π = π×$r_2^{2} \times h_2$

Or, t = 100 minutes