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

7.8 N

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8 N

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6.8 N

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6 N

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Solution

The correct option is C 6.8 N
Given that,

Volume of body $V = 10^{- 4} m^{3}$

Density of iron, $ρ_{i} = 7800 k g m^{- 3}$

Density of water $ρ_{w} = 1000 k g m^{- 3}$

Weight in air, $W_{a} = ρ_{i} V g = 7800 \times 10^{- 4} \times 10 = 7.8 N$

We know that,

Weight in water = weight in air – buoyancy force

$W_{w} = W_{a} - ρ_{w} V g$

$W_{w} = 7.8 - 1000 \times 10^{- 4} \times 10$

$W_{w} = 6.8 N$

Hence, the weight of body in water is $6.8 N$

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A piece of iron has density 7.8/cm3 & volume 10−4m3. It is totally immersed in the water calculate weight of iron in water

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