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SSC CGL Tier 2 Exam 25 Oct 2015 Paper 1
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© examsnet.com
Question : 86
Total: 100
The radii of two solid iron spheres are 1 cm and 6 cm respectively. A hollow sphere is made by melting the two spheres. If the external radius of the hollow sphere is 9 cm, then its thickness (in cm) is
2
1.5
0.5
1
Validate
Solution:
(4) Volume of sphere
=
4
3
π
r
3
∴ Total volume of both spheres
=
4
3
π
(
r
1
3
+
r
2
3
)
=
4
3
π
(
1
3
+
6
3
)
=
4
3
π
(
1
+
216
)
=
(
4
π
3
×
217
)
cu
.
cm
If the internal radius of hollow sphere = r cm, then
∴ Volume of the iron of this sphere
=
4
3
π
(
9
3
−
r
3
)
cu
.
cm
According to the question,
4
3
π
(
9
3
−
r
3
)
=
4
π
3
×
217
⇒
729
−
r
3
=
217
⇒
r
3
=
729
−
217
=
512
⇒
r
3
=
(
8
)
3
⇒
r
=
8
c
m
∴ Required thickness = 9 – r = 9 – 8 = 1 cm.
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