Volume and Surface Areas Quiz Set 014

The volume of the pipe is π × (6² - 4²) × 49, which is $22/7$ × 20 × 49. Weight = vol × density = $22/7$ × 20 × 49 × 2 = 6160 gm.

The radius of the base of the cone r = 9 cm, and height h = 40 cm. The volume is $1/3$π$(r^2 × h)$ = $1/3$π$(9^2 × 40)$ = 1080π.

Let the radius of the sphere be R. The volumes are equal. So $4/3$ π R³ = $1/3$ π 4² × 2. Cancelling π/3 from both sides, R³ = ${4 × 4 × 2}/4$ = 8 = 2³. So R = 2 cm.

Let the dimensions of the cuboid be L, 2L and L. Its total surface area is 2(L × 2L + 2L × L + L × L) = 10L². When it is placed on the table, the bottom face is hidden, so net visible area is 10L² - 2L² = 8L². When the cubical block of side L is placed on this block its(i.e., cube's) own L² and L² of the cuboidal block is hidden. So it hides 2L². Thus, the net visible is 8L² + 6L² - 2L² = 12L² = 12 × 7² = 588 sq. units.

The height of the can will be filled to 3 cm. The volume of collected water is same as the volume of cylinder with radius 8 cm and height 3 cm., which equals π8² × 3 = 192π cu. cm.