NTSE Stage 2 SAT 2018 Exam Question Paper

Question : 51 of 100

Marks: +1, -0

There is a right circular cone of height h and vertical angle 60º. A sphere when placed inside the cone, it touches the curved surface and the base of the cone. The volume of sphere is

$\;\;{4}/{3}πh^{3}$
$\;\;{4}/{9}πh^{3}$
$\;\;{4}/{27}πh^{3}$
$\;\;{4}/{81}πh^{3}$

Validate

Solution:

Since vertical angle of the cone = 60º

h=\;\;{√{3}}/{2}l→l=\;\;{2h}/{√{3}}

∵∠{OPS}=30^{∘}

∴ In ΔOPS,PS

=l/2=h/√3

......(1)
In ΔPOS,

tan30°={OS}/{PS}

⇒\;\;{1}/{√{3}}=\;\;{{OS}}/{{PS}}⇒{OS}=\;\;{{PS}}/{√{3}}=\;\;{h}/{√{3}√{3}}=\;\;{h}/{3}

Therefore, the radius of sphere is

r=OS=h/3

Hence, the required volume of sphere

=4/3πr^3

=\;\;{4}/{3}π(\;\;{h}/{3})^{3}=\;\;{4πh^{3}}/{81}

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