SSC CGL Tier 2 Exam 21 Sep 2014 Paper

(2) Let H and R be the height and radius of bigger cone respectively and h and r that of smaller cone. From triangles AOB and AMN.
∠A is common and MN || OB.
∴ Triangles AOB and AMN are similar,
∴

AO

AM

=

BO

MN

⇒

30

h

=

R

r

...(i)
Volume of smaller cone =

1

3

πr2h
Volume of bigger cone =

1

3

πR2H
∴ According to the question,

1

3

πr2h=(

1

3

πR2H)×

1

27

⇒r2h=

R2H

27

⇒27r2h=R2H
⇒

27h

H

=

R2

r2

⇒

27h

H

=(

30

h

)2[From(i)]
⇒

27h

H

=

900

h2

⇒27h3=900H=900×30
⇒h3=

900×30

27

=1000
⇒h=3√1000=10cm
∴ Required height = 30 – 10 = 20 cm