504 cones each of diameter 3.5 …

{tex}{/tex}According to the question,we have the following information.
Volume of cone {tex}= \frac { 1 } { 3 } \pi r ^ { 2 } h{/tex}
Volume of metal in 504 cones
{tex}= 504 \times \frac { 1 } { 3 } \times \frac { 22 } { 7 } \times \frac { 3.5 } { 2 } \times \frac { 3.5 } { 2 } \times 3{/tex}
Volume of Sphere {tex}= \frac { 4 } { 3 } \pi r ^ { 3 }{/tex}

{tex}= \frac { 4 } { 3 } \times \frac { 22 } { 7 } \times r ^ { 3 }{/tex}

 Now,Volume of sphere =Volume of  metal in 504 cones

{tex}\frac { 4 } { 3 } \times \frac { 22 } { 7 } \times r ^ { 3 } = 504 \times \frac { 1 } { 3 } \times \frac { 22 } { 7 } \times \frac { 35 } { 20 } \times \frac { 35 } { 20 } \times 3{/tex}

or, {tex}r ^ { 3 } = \left( \frac { 21 } { 2 } \right) ^ { 3 }{/tex}

or, r = 10.5 cm

{tex}\therefore{/tex} Diameter = 2×10.5=21 cm

Surface area {tex}= 4 \pi r ^ { 2 }{/tex}

{tex}= 4 \times \frac { 22 } { 7 } \times 10.5 \times 10.5{/tex}

{tex}= 1386 \mathrm { cm } ^ { 2 }{/tex}