Question

A cube of size 10cm is floating in equilibrium in a tank of water. When a mass of 10gm is placed on the cube, the depth of the cube inside water increases by ( g = 10m/s^2 and density of water = 1000kg/m^3).
(A) 0.1mm
(B) 1m
(C) 1mm
(D) 0.1m


PLEASE EXPLAIN THE ANSWER IN DETAIL WITH APPROPRIATE STATEMENTS AND STEPS.

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

Hope this helps you

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