**Maths Class 10
Exercise 13.3**

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

**Solution: **Radius of the sphere (*r*) = 4.2 cm

Volume
of the sphere = 4/3 × Ï€*r*^{3}

= 4/3 ×
Ï€(4.2)^{3} cm^{3}

Radius of the cylinder (R) = 6 cm

Let
the height of the cylinder be *H* cm.

Then,
the volume of the cylinder = Ï€*R*^{2}*H*

= Ï€(6)^{2}*H* cm^{3}

According
to the question,

Volume
of the sphere = Volume of the cylinder

4/3
× Ï€(4.2)^{3} = Ï€(6)^{2}*H*

H
= 4(4.2)^{3}/3(6)^{2}

H
= 2.74 cm

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

**Solution: **Let the radii of the smaller spheres
be *r*_{1}, *r*_{2} and *r*_{3} and let the radius of the
resulting sphere be *r* cm.

According
to the question,

Volume
of the resulting sphere = Volume of the three smaller spheres

4/3
× Ï€*r*^{3 }=^{ }4/3 × Ï€*r*_{1}^{3 }+ 4/3 × Ï€*r*_{2}^{3} + 4/3 × Ï€*r*_{3}^{3}

4/3
× Ï€*r*^{3 }=^{ }4/3 × Ï€(6)^{3
}+ 4/3 × Ï€(8)^{3} + 4/3 × Ï€(10)^{3}

*r*^{3 }=^{ }(6)^{3 }+ (8)^{3}
+ (10)^{3}

*r*^{3} = 216 + 512 + 1000

*r*^{3} = 1728

*r*^{3} = (12)^{3}

*r* = 12 cm

Hence, the radius of the resulting
sphere is 12 cm.

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

**Solution: **Diameter of the well = 7 m

Radius
of the well (*r*) = 7/2 m

Depth
of the well (*h*) = 20 m

Length
of the platform (*l*) = 22 m
and breadth of the platform (*b*) =
14 m

Let
the height of the platform be *H* m.

According
to the question,

Volume
of the earth dug from well = Volume of the earth spread out on the platform

Ï€*r*^{2}*h* = *l* × *b* × *H*

22/7
× 7/2 × 7/2 × 20 = 22 × 14 × *H*

*H* = 2.5 m

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

**Solution: **Diameter of the well = 3 m

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

Depth
of the earth dug from the well (*h*) =
14 m

Width
of the embankment = 4 m

Radius
of the well with embankment (*r*_{1}) = 3/2 + 4 = 11/2 m

Let
the height of the embankment be *h*_{1} m.

According
to the question,

Volume
of the embankment = Volume of the earth dug from the well

Ï€[(*r*_{1})^{2} – (*r*)^{2}]*h*_{1} = Ï€*r*^{2}*h*

[(11/2)^{2} – (3/2)^{2}]*h*_{1} = (3/2)^{2} × 14

[121/4 – 9/4]*h*_{1} = 9/4 × 14

112/4 × *h*_{1} = 9/4 × 14

*h*_{1} = 1.125 m

Hence, the height of the embankment is
1.125 m.

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

**Solution: **Diameter of the right circular cylinder = 12
cm

Radius
of the right circular cylinder (*r*) = 12/2 = 6 cm and its
height (*h*) = 15 cm

Diameter of the cone = 6 cm

Radius
of the cone (*r*_{1}) =
6/2 = 3 cm and height (*h*_{1}) =
12 cm

Radius
of the hemispherical shape = 3 cm

Volume
of ice cream in a cone = 1/3 × Ï€*r*_{1}^{2}*h*_{1
}+_{ }2/3 × Ï€*r*_{1}^{3}

Let
the number of cones that can be filled with ice cream
be *n*.

According
to question,

Volume
of *n* cones = Volume of the right
circular cylinder

*n* × [1/3 × Ï€*r*_{1}^{2}*h*_{1 }+_{ }2/3 × Ï€*r*_{1}^{3}] = Ï€*r*^{2}*h*

*n* × Ï€[1/3 × (3)^{2 }× 12 +_{
}2/3 × (3)^{3}] =
Ï€
× (6)^{2} × 15

*n* × [4 × (3)^{2} +_{
}2 × (3)^{2}] =
(6)^{2} × 15

*n* × 54
= 36 × 15

*n* = 540/54

*n* =
10

Hence,
10 cones can be filled with ice cream.

**6. How many silver coins, 1.75 cm in diameter
and of thickness 2 mm, must be melted to form a cuboid of dimensions 5.5 cm ×
10 cm × 3.5 cm?**

**Solution: **Diameter of the silver coin = 1.75 cm

Radius of the
silver coin (*r*) = 1.75/2 = 7/8 cm
and its thickness (*h*) = 2 mm = 1/5 cm

Dimensions
of the cuboid = 5.5 cm × 10 cm × 3.5 cm

Let
the number of coins to be melted is *n*.

According
to question,

Volume
of *n* silver coins = Volume of
the cuboid

*n* ×
Ï€*r*^{2}*h* = 5.5 ×
10 × 3.5

*n* ×
Ï€(7/8)^{2}*
*× 1/5 = 5.5 ×
10 × 3.5

*n* ×
22/7 × 49/64* *× 1/5 = 5.5 ×
10 × 3.5

*n* = 400

Hence,
400 silver coins must be melted.

**7. A cylindrical bucket, 32 cm high and with
radius of base 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.**

**Solution:
**Radius of the
base of the cylindrical bucket (*r*) =
18 cm and its height (*h*) = 32 cm.

Volume
of the cylindrical bucket = Ï€*r*^{2}*h*

=
Ï€(18)^{2}*
*× 32* *

=
10368Ï€ cm^{3}

Height of the conical heap (*h*_{1}) = 24 cm

Let
the radius of the conical heap be *r*_{1} cm.

Volume
of the conical heap = 1/3 × Ï€*r*_{1}^{2}*h*_{1}

=
1/3 × Ï€ × *r*_{1}^{2 }× 24

= 8Ï€*r*_{1}^{2} cm^{3}

According
to question, Volume of bucket = Volume of conical heap

10368Ï€ = 8Ï€*r*_{1}^{2}

*r*_{1}^{2} = 10368/8

*r*_{1}^{2 }= 1296

*r*_{1} = 36 cm

Now,
slant height of the heap (*l*) =

**8. Water in a canal 6 m wide and 1.5 m deep is
flowing with a speed of 10 km/h. How much area will it irrigate in 30 minutes,
if 8 cm of standing water is needed?**

**Solution: **Width of the canal = 6 m

Depth of the canal = 1.5 m

Speed
of the flow of water = 10 km/h

=
10 × 1000 m/h

=
10000 m/h

=
10000/60 m/min

=
500/3 m/min

Distance
of water flown in 30 minutes = 500/3 × 30 m

Volume
of water that flows in 30 minutes = 6 × 1.5 × 5000 = 45000 m^{3}

Volume of water
that flows in 30 minutes = Area it will irrigate × height of water

45000^{ }=
A × 8/100

A
= 4500000/8

A
= 562500 m^{3}

A
= 562500/10000 hectares

A
= 56.25 hectares

Hence,
it will irrigate 562500 m^{3} or 56.25 hectares.

**9. A farmer connects a pipe of internal
diameter 20 cm from a canal into a cylindrical tank in her field, which is 10 m
in diameter and 2 m deep. If water flows through the pipe at the rate of 3
km/h, in how much time will the tank be filled?**

**Solution: **Diameter of the cylindrical tank = 10
m

Radius
of the cylindrical tank (*r*) = 10/2 =
5 m and depth of the cylindrical tank (*h*) = 2 m

Volume
of the cylindrical tank = Ï€*r*^{2}*h*

=
Ï€(5)^{2 }× 2

= 50Ï€ m^{3}

Rate of flow of
water (*h*_{1}) = 3 km/h

= 3000 m/h

= 3000/60 m/min

= 50 m/min

Internal
diameter of the pipe = 20 cm

Radius
of the pipe (*r*_{1}) = 10 cm = 0.1 m

Volume
of water that flows in the pipe per minute = Ï€*r*_{1}^{2}*h*_{1}

=
Ï€(0.1*)*^{2 }× 50

= Ï€/2 m^{3}

Required
time = 50Ï€ ÷ Ï€/2

=
100 minutes