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 increase by $(g = 10 m / s^{- 2}$ , density of water = $1000 k g / m^{2})$

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 increase by

(g = 10 m / s^{- 2}

, density of water =

1000 k g / m^{2})

Toppr Admin · Accepted Answer

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 increase by (g=10m/s−2, density of water = 1000kg/m2)0.1 mm4 m1 mm0.1 m

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 increase by $(g = 10 m / s^{- 2}$ , density of water = $1000 k g / m^{2})$
0.1 mm
4 m
1 mm
0.1 m