SSC CGL Tier II 12 Apr 2015 Paper

Question : 4 of 100

Marks: +1, -0

From a solid right circular cylinder of length 4 cm and diameter 6 cm, a conical cavity of the same height and base is hollowed out. The whole surface of the remaining solid (in square cm.) is

$48 \π$
$15 \π$
$63 \π$
$24 \π$

Validate

Solution:

(1) Whole surface of the remaining solid

=2 \π r h+\π r^{2}+\π r l

where

l

= slant height of cone

I=√{r^{2}+h^{2}}=√{3^{2}+4^{2}}

=√{9+16}=√{25}=5 {cm}

∴ Required area

= (2 × π × 4 × 3 + π × 3 × 3 + π ×3 × 5)

square cm

= (24π + 9π + 15π)

square cm.

= 48π

square cm

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