SSC CGL Tier II 25 Oct 2015 Paper 1

Question : 70 of 100

Marks: +1, -0

A cylinder with base radius 8 cm and height 2 cm is melted to form a cone of height 6 cm. The radius of the cone will be

Solution:

(2) Volume of cylinder =

πr^2h

= (π × 8 × 8 × 2) cu. cm. = 128 π cu. cm.
If the radius of the base of cone be R cm. then

1/3πR^2H=128π

\⇒{R}^{2} × 6=128× 3

⇒{R}^{2}= {128× 3}/{6}=64

⇒ R=√{64}=8 {cm}

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