this is so urgent plz answer
A CUBE OF wood floating in water supports 200g mass resting at centre of its toop face . when mass is removed, cube rises 2cm . find out volume of cube
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i think the answer is
10 cm
10 cm
VedantVR1:
how
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The answer should be 1000 cm³.
Let the side of the cube be ‘a’ cm and its mass be ‘m’.
So, density of the cube is, ρ = m/a3
Let ‘h’ be the immersed height of the cube.
So, volume immersed = ha2
And, mass of water displaced = (ρw)ha2
For floatation, weight of water displaced = weight of cube
=> (ρw)ha2g = mg
=> m = (ρw)ha2
Again, when 200 g is added to the cube the cube is immersed further by 2 cm.
So, the new volume of water displaced = (h+2)a2
The new mass of water displaced = (ρw)(h+2)a2
Thus,
(ρw)(h+2)a2g = (m+200)g
=> (ρw)(h+2)a2 = (ρw)ha2+200
For water. ρw = 1 g/cm3
So,
(h+2)a2 = ha2 + 200
=> ha2 + 2a2 = ha2 + 200
=> a = 10 cm
This is the length of a side of the cube
Let the side of the cube be ‘a’ cm and its mass be ‘m’.
So, density of the cube is, ρ = m/a3
Let ‘h’ be the immersed height of the cube.
So, volume immersed = ha2
And, mass of water displaced = (ρw)ha2
For floatation, weight of water displaced = weight of cube
=> (ρw)ha2g = mg
=> m = (ρw)ha2
Again, when 200 g is added to the cube the cube is immersed further by 2 cm.
So, the new volume of water displaced = (h+2)a2
The new mass of water displaced = (ρw)(h+2)a2
Thus,
(ρw)(h+2)a2g = (m+200)g
=> (ρw)(h+2)a2 = (ρw)ha2+200
For water. ρw = 1 g/cm3
So,
(h+2)a2 = ha2 + 200
=> ha2 + 2a2 = ha2 + 200
=> a = 10 cm
This is the length of a side of the cube
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