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A solid, homogeneous sphere with a mass of m0, a radius of r0 and a density of ρ0 is placed in a container of water. In...

A solid, homogeneous sphere with a mass of m0, a radius of r0 and a density of ρ0 is placed in a container of water. Initially the sphere floats and the water level is marked on the side of the container. What happens to the water level, when the original sphere is replaced with a new sphere which has different physical parameters? Notation: r means the water level rises in the container, f means falls, s means stays the same. possible answers: R=rise, F=Fall, S=same, F or S, R or S, R or F or S

The new sphere has a radius of r > r0 and a mass of m < m0.

The new sphere has a radius of r > r0 and a density of ρ < ρ0.

The new sphere has a mass of m < m0 and a density of ρ > ρ0

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As long as the ball floats. The buoyant force will be equal to the weight of the ball Weight of theball-F -pVg Whese, p Density of Fluid con stant) o-Vol of fuid displaced Since density of fluid (p)and acceleration dueto gravity(g)isconstant Weight of theball oc V Also, as V increases, level rises and if V decreases, level falls When > ndm <mo Clearly, the weight of the ball is less. The volume displaced by the ball will be low, and therefore, the level of water will Fall. Although, density of the ball changes, but the amount of buoyant force is directly proportional to mass only Answer F

When r>10 and ρ<p, As the radius of ball increase, and its density decreases, implies that the mass is either constant or can also be decreasing Therefore, mSm, Since Volume displaced is directly proportional to mass, volume displaced can stay the same or may decrease. Therefore, the level of water may stay Same or Fall Answer S or F When m <moana ρ> In this case, the mass is decreasing, the volume displaced will be less. Therefore, the level will Fall Answer IF

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