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 $1000 (kg) / m^3$ .

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$1000 (kg) / m^3$

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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 $100-75=25 N$

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

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

Given, $rho=1000 Kg m^-3$

So, $V=25/(1000*9.8)=0.00254 m^3=2540 cm^3$

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

So, $a^3=2540$

or, $a=13.63 cm$

so,its side will be $a^2=13.63^2=185.76 cm^2$