SSC CGL Tier II 25 Oct 2015 Paper 1

Question : 78 of 100

Marks: +1, -0

A plane divides a right circular cone into two parts of equal volume. If the plane is parallel to the base, then the ratio, in which the height of the cone is divided, is

$1:√^3 2$
$1:√2$
$1:√^3 2$
$1:√^3 2 -1$

Validate

Solution:

(4)

OA' = h units
AA' = H units
AB = R units
A'B' = r units.
A'B' || AB
∠ OA'B' =∠OAB
∠OB'A' = ∠OBA
∴ ΔOAB ~ ΔOA'B'
∴

{\OA'}/{\OA}={\A'\B'}/{\AB}

⇒{h}/{ {H}+h}= {r}/{ {R}}

According to the question,

\ {1}/{3} \π r^{2} h= {1}/{3} \π R^{2}(H+h)- {1}/{3} \π r^{2} h

⇒{2}/{3} \π r^{2} h= {1}/{3} \π R^{2}(H+h)

⇒ 2 {r^{2}}/{ {R}^{2}}= {H+h}/{h}

⇒ 2 .{h^{2}}/{(H+h)^{2}}= {H+h}/{h}

⇒{(H+h)^{3}}/{h^{3}}=2

⇒{H+h}/{h}=√^3{2}

⇒{H}/{h}+1=√^3{2}

⇒{H}/{h}= {√^3{2}-1}/{1}

∴

{h}/{H}=1 : √^3{2}-1

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