One gamble measured in air has a weight of 100 N. When immersed in the water, its weight is 75 N. How much is the dice side? The density of the water is $1000 \frac{k g}{m^{3}}$ $1000 (kg) / m^3$ .

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$1000 \frac{k g}{m^{3}}$ $1000 (kg) / m^3$

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Answer 1 · 2018-03-18T14:31:43

We can say that the weight of the dice decreased because of the buoyancy force of water on it.

So,we know that, buoyancy force of water acting on a substance = It's weight in air - weight in water

So,here the value is $100 - 75 = 25 N$ $100-75=25 N$

So,this much force had acted on the whole volume $V$ $V$ of the dice,as it was fully immersed.

So,we can write, $V \cdot ρ \cdot g = 25$ $V*rho*g=25$ (where, $ρ$ $rho$ is the density of water)

Given, $ρ = 1000 K g m^{- 3}$ $rho=1000 Kg m^-3$

So, $V = \frac{25}{1000 \cdot 9.8} = 0.00254 m^{3} = 2540 c m^{3}$ $V=25/(1000*9.8)=0.00254 m^3=2540 cm^3$

For a dice,if its one side length is $a$ $a$ its volume is $a^{3}$ $a^3$

So, $a^{3} = 2540$ $a^3=2540$

or, $a = 13.63 c m$ $a=13.63 cm$

so,its side will be $a^{2} = {13.63}^{2} = 185.76 c m^{2}$ $a^2=13.63^2=185.76 cm^2$