A piece of iron weighs $44.5 g f$ in air and $39.5 g f$ in water. The R.D. of iron is $X$ . Find $X$

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Solution

The correct option is C $9$
weight in air $= 44.5 g f = 0.445 g N$
weight in water $= 39.5 g f = 0.395 g N$
volume of the iron $= V_{i}$
density of the iron $= d$
$V_{i} \times d_{g} = 0.445 g$
$\Rightarrow V_{i} \times d = 0.445$
density of water $= 1000 k g / m^{3}$
$V_{i} \times 1000 \times g = 0.445 g - 0.395 g$
$\Rightarrow 1000 V = 0.05$
$\Rightarrow V = 0.00005$
$d = 0.445 / 0.00005$
DENSITY OF IRON: $8900 k g / m^{3} = 8.9 g / c m^{3}$

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A piece of iron weighs 44.5gf in air and 39.5gf in water. The R.D. of iron is X. Find X

The correct option is C 9weight in air =44.5gf=0.445gNweight in water =39.5gf=0.395gNvolume of the iron =Vi density of the iron =dVi×dg=0.445g⇒Vi×d=0.445density of water =1000kg/m3 Vi×1000×g=0.445g−0.395g⇒1000V=0.05⇒V=0.00005d=0.445/0.00005 DENSITY OF IRON: 8900kg/m3=8.9g/cm3

A piece of iron weighs $44.5 g f$ in air and $39.5 g f$ in water. The R.D. of iron is $X$ . Find $X$