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A coin made of alloy of aluminum and silver measures 2 x 15 mm (it is 2 mm thick and its diameter is 15 mm). If the weight of the coin is 30 grams and the volume of aluminum in the alloy equals that of silver, what will be the weight of a coin measuring 1 x 30 mm made of pure aluminum if silver is twice as heavy as aluminum?

A coin made of alloy of aluminum and silver measures 2 x 15 mm (it is 2 mm thick and its diameter is 15 mm). If the weight of the coin is 30 grams and the volume of aluminum in the alloy equals that of silver, what will be the weight of a coin measuring 1 x 30 mm made of pure aluminum if silver is twice as heavy as aluminum?

36 grams 40 grams 42 grams 48 grams 50 grams

I would go with (B) = 40 grams. And here is why -

Let weight of aluminum per \(mm^3\) = w Then weight of silver per \(mm^3\) = 2w (since we are told silver is twice as heavy as aluminium)

Now Volume of first coin = \(PI*(7.5)^2*2\) (since 15 is diameter => radius = 7.5) But we are told this coin contains equal volumes of aluminium and silver. HEnce \(1/2*PI*(7.5)^2*2*W\) + \(1/2*PI*(7.5)^2*2*2W\) = 30 gms From this we can calculate that W = 40/225*PI

Now volumne of second coin = \(PI*15^2*1\) And we are told this coin is entirely made of aluminium. Hence its weight = \(PI*15^2*1*W\) = 40 grams