Step-by-step explanation:
Diameter of spherical ball of lead = 3 cm
∴
R
a
d
i
u
s
(
r
)
=
3
2
c
m
∴
V
o
l
u
m
e
=
4
3
π
r
3
=
4
3
π
×
3
2
×
3
2
×
3
2
c
m
3
=
9
π
2
c
m
3
Diameter of smaller ball
=
3
2
c
m
∴
Radius
(
r
1
)
=
3
4
and volume
=
4
3
π
×
3
4
×
3
4
×
3
4
c
m
3
=
9
16
π
c
m
3
Diameter of second smaller ball = 2 cm
and radius
(
r
2
)
=
2
2
=
1
c
m
∴
V
o
l
u
m
e
=
4
3
π
×
1
×
1
×
1
=
4
3
π
c
m
3
∴
Volume of third smaller ball
9
2
π
−
(
9
16
π
+
4
3
π
)
=
(
9
2
−
9
16
−
4
3
)
π
c
m
3
=
216
−
27
−
64
48
=
125
48
π
c
m
3
∴
Radius of the third ball
=
(
V
o
l
u
m
e
×
3
4
π
)
1
3
=
(
125
π
×
3
48
×
4
π
)
1
3
=
(
125
64
)
1
3
=
[
(
5
4
)
3
]
1
3
=
5
4
=
1.25
c
m
∴
Diameter of third ball
=
2
×
r
a
d
i
u
s
=
2
×
1.25
=
2.5
c
m
