Volume and Surface Areas Quiz Set 006

The well-known relation for a cube is $S/V = 6/a$, which can be re-arranged to get $a = 6 × V/S$ = ${186/41}$, which is same as: $4{22/41}$.

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 × 5² = 300 sq. units.

The wire is a cylinder with r = 1/10 cm, and length L. The volume of tin is equal to the volume of the wire. So π ${1/10} × {1/10} × L$ = 33, which gives L = ${10 × 10 × 33 × 7}/22$ = 1050 cm.

The volume of the pipe is π × (15² - 13²) × 56, which is $22/7$ × 56 × 56. Weight = vol × density = $22/7$ × 56 × 56 × 9 = 88704 gm.

The weight W is equal to the weight of water displaced by it. So W = ${3 × 8 × (3/100)}$ × 1000 = 720 kg.