Question

A piece of iron of density 7.8×10³kg/m³ and volume 100cm³ is totally immersed in water. Calculate the weight of iron piece in air.

Answer 1

Answer:

Density of iron, d= 7.8 × 10³ kg/m³

Volume of iron piece, V = 100 cm³

$100{cm}^{2} = 100 \times {10}^{ - 6} {m}^{3} = {10}^{ - 4} {m}^{3}$

Mass of iron piece, M = V × d

$= {10}^{ - 4} {m}^{ - 3} \times 7.8 \times {10}^{ 3} \: kg {m}^{ - 3}$

$= 0.78kg$

Weight of iron piece, W = Mg = (0.78kg) (10m/s²)

= 7.8 N

Hope it helps you ✌

Answer 2

$\huge \underline \bold{Given:}$

Density of iron piece:

$= 7.8 \times {10}^{3} kg \: {m}^{ - 3}$

Volume of iron piece:

$= 100 {cm}^{3} \\ = 100 \times 10 {}^{ - 6} {m}^{3} \\ = {10}^{ - 4} {m}^{3}$

$\huge \underline \bold{To \: find:}$

Weight of iron piece in air

$\huge \underline \bold{Solution}$

$= volume \times density \: of \: iron \times g \\ = {10}^{ - 4} \times (7.8 \times {10}^{3} ) \times 10 \\ = 7.8N$

Hence, the volume of iron piece in air is 7.8N.

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