A piece of iron weighs 44-5 gf in the air- If the density of iron is 8-9- 103- kg-m-3- find the weight of iron piece when immersed in water-

Density of water (d) = 10^3 kg m^ 3 Density of iron (d’) = 8.9× 10^3 kg m^ 3 Weight in water = Weight in air × [1 d/d’] = 44.5 kgf × [1 10^3 kg m^ 3/ 8. ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Variation of Pressure in Fluids

A piece of ir...

Question

A piece of iron weighs 44.5 gf in the air. If the density of iron is $8.9 \times 10^{3} k g m^{- 3}$ , find the weight of iron piece when immersed in water.

Open in App

Solution

Density of water (d) $= 10^{3} k g m^{- 3}$

Density of iron (d’) $= 8.9 \times 10^{3} k g m^{- 3}$

Weight in water = Weight in air $\times [1 - \frac{d}{d^{'}}]$

$= 44.5 k g f \times [1 - \frac{10^{3} k g m^{- 3}}{8.9 \times 10^{3} k g m^{- 3}}]$

$= 39.5 k g f$

The weight of iron piece in water is 39.5 kgf

Suggest Corrections

Similar questions

Q. A piece of weighs 44.5gf in air. If the density of iron is 8.9*10^3kg/m^3,find weight of iron piece when immersed in water.

Q. A piece of iron of density

7800 k g / m^{3}

and volume

100 c m^{3}

is completely immersed in water. Calculate the weight of the iron piece in air.

[Take,

g = 10 m / s^{2}

]

A piece of iron has density 7.8/

c m^{3}

& volume

10^{- 4} m^{3}

. It is totally immersed in the water calculate weight of iron in water

A piece of iron of density $7.8 \times 10^{3} k g m^{- 3}$ and volume 100 $c m^{3}$ is totally immersed in water ( $ρ$ = 1000 kg $m^{- 3})$ . Calculate its apparent weight in water.

Q. 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

Join BYJU'S Learning Program

A piece of iron weighs 44.5 gf in the air. If the density of iron is 8.9×103 kg m−3, find the weight of iron piece when immersed in water.

Density of water (d) =103 kg m−3 Density of iron (d’) =8.9×103 kg m−3 Weight in water = Weight in air ×[1−dd′] =44.5 kgf×[1−103 kg m−38.9×103 kg m−3] =39.5 kgf The weight of iron piece in water is 39.5 kgf

A piece of iron weighs 44.5 gf in the air. If the density of iron is $8.9 \times 10^{3} k g m^{- 3}$ , find the weight of iron piece when immersed in water.