a cubical block of ice floating in water has to support a metal piece weighing 0.5 kg?

Question

a cubical block of ice floating in water has to support a metal piece weighing 0.5 kg .what can be the minimum edge of the block so that it does not sink in water? sp.gravity of ice is 0.9

Answer 1

17.13 cm (approx)

Answer 2

Basically, you need the ice+metal component to be less dense than water. The way it's worded leaves me to believe that the "minumum" edge it wants is one where the density of the ice+metal is equal to the density of water. This sets up our equation.

We also need the density of water here which is not given, but is common enough that you should probably know it: 1g per cubic centimeter. (This mean's ice is .9g per cubic centimeter)

To make the units match, we'll change the units of the metal from 0.5 kg to 500g.

(500 + 0.9*X^3)/(X^3) = 1 Where X is the cube's edge in cm.

500 + 0.9*X^3 = X^3
500 = 0.1*X^3
5000 = X^3
X = 17.1 cm

The minimum edge of the ice block is 17.1 cm.

Answer 3

because something floating displaces a mass of water equivalent to the mass of the article and ice is water there must be no substitute in water point simply by fact the ice melts. in case you ice weighs 50 grams it displaces 50 grams or 50 cc of water. simply by fact the 50 gram piece of ice melts, it particularly is going to become 50 grams or 50 cc of water. The water point with the floating ice and after the ice has melted is precisely an identical, 50 cc bigger than with out the ice.

Answer 4

the surface area of the ice must be more

Answer 5

Let side be a

So, axax0.1ax1000 = 0.5

a^3 = 0.5/100 = 0.005

So, a = 0.17 m