A solid rectangular piece of iron measures 6 m by 6 cm by 2 cm. Find the weight of this piece, if 1 $c m^{3}$ of iron weighs 8 gm.

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Solution

Dimensions of a piece of rectangluar iron = $6 m \times 6 m \times 2 c m$
$∴$ Volume = $600 \times 6 \times 2 c m^{3} = 7200 c m^{3}$
Weight of 1 $c m^{3}$ = 8 gm
$∴$ Total weight of the piece = $7200 \times 8$ gm
= $57600 g m = \frac{57600}{1000} k g = 57.6$ kg

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A solid rectangular piece of iron measures 6 m by 6 cm by 2 cm. Find the weight of this piece, if 1 cm3 of iron weighs 8 gm.

Dimensions of a piece of rectangluar iron = 6 m×6 m×2 cm ∴ Volume = 600×6×2 cm3=7200 cm3 Weight of 1 cm3 = 8 gm ∴ Total weight of the piece = 7200×8 gm = 57600 gm=576001000 kg=57.6 kg

A solid rectangular piece of iron measures 6 m by 6 cm by 2 cm. Find the weight of this piece, if 1 $c m^{3}$ of iron weighs 8 gm.

Dimensions of a piece of rectangluar iron = $6 m \times 6 m \times 2 c m$
$∴$ Volume = $600 \times 6 \times 2 c m^{3} = 7200 c m^{3}$
Weight of 1 $c m^{3}$ = 8 gm
$∴$ Total weight of the piece = $7200 \times 8$ gm
= $57600 g m = \frac{57600}{1000} k g = 57.6$ kg