Question

A cube of size 10 cm is floating in equilibrium in a tank of water. When a mass of 10 gm is placed on the cube , the depth of cube inside water increases by(g=10 m/s 2 , density of water= 1000 kg/m 3 ) a) 0.1 mm b) 1 m c) 1 mm d) 0.1 m please explain answer in detail.

Answer 1

Let x be the initial depth up to which the cube is sunk in water. Let a be the density of the cube 

 

 x into 10 into 10 into 1 into g   = 10x10x10xaxg ---- 1 (mass=volume  x density) 

==> x = 10a 

Let x' be the new depth 

x' into 100 into g = 1000 x a x g + 10g ---- 2  

Subtracting (1) from (2) we will get 

100x'  = 100x + 10  

x' - x = 1/10 cm  = 1mm