SSC CGL Tier II 21 Sep 2014 Paper

Question : 11 of 100

Marks: +1, -0

A ball of lead 4 cm in diameter is covered with gold. If the volume of the gold and lead are equal, then the thickness of gold [given

√^3 2 = 1.259

] = is approximately

Solution:

(4) Volume of lead

= {4}/{3} \π r^{3}= {4}/{3} \π × 2^{3}

If the thickness of gold be x cm, then
Volume of gold

= {4}/{3} \π((2+x)^{3}-2^{3} ) {\cu.\cm}

∴{4}/{3} \π((2+x)^{3}-2^{3} )

= {4}/{3} \π×2^{3}

⇒(2+x)^{3}-2^{3}=2^{3}

⇒(2+x)^{3}=8+8=16

⇒(2+x)^{3}=2^{3} . 2

⇒ 2+x=2 × √^3 {2}

⇒ 2+x=2 × 1.259=2.518

∴ x=2.518-2=0.518 {cm}

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