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#

1 Answer
Mar 18, 2018

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#