How many silver coins, 1.75 cm in diameter and of thickness 2 mm

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 Multiple Choice QuestionsShort Answer Type

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431.

How many silver coins, 1.75 cm in diameter and of thickness 2 mm, must be melted to form a cuboid of dimensions 5.5 cm × 10 cm × 3.5 cm?


Let r cm be the radius and h cm be the thickness of the silver coin (cylindrical in shape). Then,


straight r space equals space fraction numerator 1.75 over denominator 2 end fraction space cm space space and space straight h space equals space fraction numerator 2 space mm over denominator 10 end fraction space equals space 0.2 space cm
Now,      Volume = πr squared straight h

    equals space straight pi open parentheses fraction numerator 1.75 over denominator 2 end fraction close parentheses open parentheses fraction numerator 1.75 over denominator 2 end fraction close parentheses space straight x space 0.2 space equals open parentheses fraction numerator straight pi space straight x space 0.6125 over denominator 4 end fraction close parentheses space cm cubed
equals space 0.153125 space straight pi
equals space 0.153125 space straight x space 22 over 7 space equals space fraction numerator 3.36875 over denominator 7 end fraction equals 0.478125 space cm cubed
Let l, b and h are respectively the length, breadth and height of the cuboid. Then
l = 5.5 cm, b = 10 cm and h = 3.5 cm
Now, Volume = l x b x h
= (5.5 x 10 x 3.5) cm3 = 192.5 cm3
Hence, Required number of silver coins

equals fraction numerator 192.5 over denominator 0.48125 end fraction equals 400
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 Multiple Choice QuestionsLong Answer Type

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 Multiple Choice QuestionsShort Answer Type

440.

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