A spherical ball is submerged in water in a cylindrical container. The radius of the cylindrical container is 5 cm, and the height is 30 cm. What is the volume of water displaced by the ball if its radius is 4 cm?

Correct — D. A fully submerged ball pushes aside its own volume of water, so find the sphere's volume.

Formula: V = 4⁄3 π r³
Cube the radius: 4³ = 64
Multiply: 4⁄3 × 64 = 256⁄3, so V = 256⁄3 π cm³

Water displaced = 256⁄3 π cm³ ≈ 268 cm³ → option (d)

Why the others are wrong

• (a)
  640⁄3 π cm³ — 640⁄3 π is 2.5 times the ball's own volume. A submerged ball displaces exactly its volume, so this would need r³ = 160, not 4³ = 64.
• (b)
  320⁄3 π cm³ — 320⁄3 π is more water than the ball occupies (1.25 times 256⁄3 π). In 4⁄3 π r³ it needs r³ = 80, but a 4 cm radius gives r³ = 64.
• (c)
  160⁄3 π cm³ — 160⁄3 π is less than the ball's 256⁄3 π, which would mean part of the ball stayed above the water. The stem says submerged, so the whole volume counts.

Displacement: a solid pushed completely under a liquid moves aside a volume of liquid equal to its own volume. So the water displaced is simply the sphere's volume, 4⁄3 π r³.

A body that is only partly under the surface displaces only its submerged part. That is why the word submerged in the stem matters.

Key facts

• Volume of a sphere = 4⁄3 π r³.
• A fully submerged solid displaces a volume of liquid equal to its own volume.
• Rise in level in a cylindrical container = volume displaced ÷ its base area πR².

Common traps

• Putting the container's 5 cm radius into the sphere formula, which gives 500⁄3 π.
• Using the surface area 4πr² = 64π instead of the volume 4⁄3 π r³.

No directly related past PYQ was found.