NTSE Stage 2 SAT 2019 Exam Question Paper

Question : 48 of 100

Marks: +1, -0

Let the volume of a solid sphere be

288πcm^3

.A horizontal plane cuts the sphere at a distance of 3 cm from the centre so that the ratio of the curved surface areas of the two parts of the sphere is 3 : 1. The total surface area of the bigger part of the sphere (in

cm^2

) is

36π
108π
135π
144π

Validate

Solution:

\;\;{4}/{3}πr^{3}=288π

(∵ r = radius of sphere)

⇒r=6

.....(1)
OA is the radius of sphere

∆AOC,OA^{2}=AC^{2}+{OC}^{2}⇒AC^{2}=6^{2}-3^{2}

⇒AC=3√3

cm (radius of circle formed)
Curved surface area of bigger part = 3x
Curved surface area of smaller part = x

⇒3x+x=4πr^{2}⇒x=πr^{2}

⇒x=36π

[From(1)]
Total surface area of bigger part

=3x+π(AC)^{2}=108π+π(3√{3})^{2}

=(108+27)π=135π{cm}^{2}

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