A piece of iron of density $7800 k g / m^{3}$ and volume $100 c m^{3}$ is completely immersed in water. Calculate apparent weight of iron piece in water.
[Take, $g = 10 m / s^{2}$ .]

6.8 N

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1 N

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7.8 N

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8.8 N

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Solution

The correct option is A $6.8 N$
Apparent weight of iron = Weight of iron in air - weight of iron in water

As we know , Density = $\frac{m a s s}{v o l u m e}$

So mass of iron piece in air = (volume) (density) = $(10^{- 4}) (7800) k g = 0 .78 k g$

Weight of iron piece in air $= m g = 7.8 N$

weight of iron piece in water = upthrust = $V ρ g$ = $10^{- 4} (1000) 10 N = 1 N$

so apparent weight of the iron piece $= 7.8 N - 1 N = 6.8 N$

so option (A) is correct

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A piece of iron of density 7800kg/m3 and volume 100cm3 is completely immersed in water. Calculate apparent weight of iron piece in water.[Take, g=10m/s2.]

A piece of iron of density $7800 k g / m^{3}$ and volume $100 c m^{3}$ is completely immersed in water. Calculate apparent weight of iron piece in water.
[Take, $g = 10 m / s^{2}$ .]