A solid, homogeneous sphere with a mass of m_0, a radius of r_0 and a density of

Question

A solid, homogeneous sphere with a mass of m_0, a radius of r_0 and a density of rho_0 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. Combination answers like 'r or f or s' are possible answers in some of the cases. The new sphere has a radius of r > r_0 and a density of rho r_0 and a mass of m

Explanation / Answer

buoyant force Fb = d*V*g     where d is density of the fluid

in our case d would be density of water, V = volume of displaced water, g = gravitional pull

so clearly Fb is directly proportional to V

case 1) f

since density of the new object is less, so it will float on the water(means it will replace less volume of water than the first one)

water level will fall down

case 2) f, s

r > r0 and m < m0

density of the new sphere will be = (mass)/(volume) = m/((4/3)*pi*r3)

we can see that density of this new sphere can be less than the first sphere or can remain same

case 3) r

density of the second sphere is more than the first sphere, so it will replace more water now than the first sphere

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.