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UPSC CDS 2 2021 Math Paper

Directions (Q.Nos. 46-48) Consider the following for the next three questions that follow.

A solid consisting of a right circular cone of radius

x

and height

2x

standing on a hemisphere of radius

x

(take

\pi = \frac{22}{7}

)

Question : 47 of 100

Marks: +1, -0

What is the approximate total surface area of the solid

Solution: 👈: Video Solution

We have,

Cone :

r_{1}=x

and

h=2 x

Hemisphere

: r_{2}=x

Total surface area of solid

=

curved surface area of cone

+

curved surface area of hemisphere

=\pi r_{1} l+2\pi r_{2}^{2}

=\pi \cdot x \cdot A B+2\pi x^{2}

In right angled triangle

A B C

A B^{2} \;\; = O A^{2}+O B^{2}

\Rightarrow A B^{2} \;\; = (2x)^{2}+x^{2}

\;\; =4x^{2}+x^{2}=5x^{2}

\Rightarrow A B \;\; = \sqrt{5} x

Put

A B = \sqrt{5} x

into Eq (i), we get

Total surface area of solid

\;\; = \pi \cdot x \cdot \sqrt{5} x + 2\pi x^{2}

\;\; = 3.14 \times 2.23 x^{2} + 6.28 x^{2}

\;\; = 7.0022 x^{2} + 6.28 x^{2}

\;\; = 13.2822 x^{2} \approx 13.3 x^{2}

Hence, the approximate total surface area of the solid is

13.3 x^{2}

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