Question

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` .

`1000 (kg) / m^3`

Answer 1

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`