A cube (density 0.5gm/cc) of side 10cm is floating in water kept in a cylinder beaker of base area #1500cm^2#. When a mass m is kept on the cube, level of water rises in the beaker by 2mm. Find mass m?

To find the mass ( m ), we can use the principle of buoyancy. When an object floats in a fluid, the weight of the fluid displaced by the object is equal to the weight of the object itself.

First, let's calculate the volume of the cube. Since it's a cube, each side has a length of 10 cm. Therefore, the volume (( V )) of the cube is ( 10 \times 10 \times 10 = 1000 ) cm³.

Now, let's calculate the weight of the cube. The weight (( W )) is equal to the volume multiplied by the density (( \rho )) and the acceleration due to gravity (( g )). Given that the density of the cube is ( 0.5 ) gm/cc and the acceleration due to gravity is ( 9.8 ) m/s² (or ( 980 ) cm/s²), we have:

[ W = V \times \rho \times g = 1000 \times 0.5 \times 980 = 490000 ) dyne

Now, let's calculate the increase in volume of water when the mass ( m ) is added on top of the cube. Since the water level rises by 2 mm, or ( 0.2 ) cm, and the base area of the beaker is ( 1500 ) cm², the volume of water displaced (( V_w )) is:

[ V_w = \text{base area} \times \text{height} = 1500 \times 0.2 = 300 ) cm³

According to the principle of buoyancy, the weight of the water displaced by the cube and the added mass ( m ) equals the weight of the cube and the added mass. Therefore:

[ W + m = \text{weight of water displaced} ]

[ 490000 + m = \text{density of water} \times \text{volume of water displaced} \times g ]

Given that the density of water is ( 1 ) gm/cc, we can solve for ( m ):

[ m = \text{density of water} \times \text{volume of water displaced} \times g - W ]

[ m = 1 \times 300 \times 980 - 490000 ]

[ m = 294000 - 490000 ]

[ m = -196000 \text{ dyne} ]

Therefore, the mass ( m ) is ( -196000 ) dyne.