Question

A hemispherical bowl is packed in a cuboidal box. The bowl just fits in the box. Inner radius of the bowl is 10 cm. Outer radius of the bowl is 10.5 cm.

Based on the above, answer the following questions:

1. Find the dimensions of the cuboidal box.  (1)
2. Find the total outer surface area of the box.  (1)
3. 1. Find the difference between the capacity of the bowl and the volume of the box. (Use π = 3.14).   (2)
      OR
   2. The inner surface of the bowl and the thickness is to be painted. Find the area to be painted.  (2)

Answer 1

i. Given, inner radius of bowl (r) = 10 cm

Outer radius (R) = 10.5 cm

Dimension of cuboidal box:

length (l) = 2 × outer radius

= 2R

= 2 × 10.5

= 21 cm

Width (b) = 2 × 10.5

= 21 cm

Height (h) = R = 10.5 cm

ii. Total outer surface area of the box = 2(lb + bh + hl)

= 2(21 × 21 + 21 × 10.5 + 10.5 × 21)

= 2(441 + 220.5 + 220.5)

= 2(882)

= 1764 cm²

iii. a. Volume of box = lbh

= 21 × 21 × 10.5

= 4630.5 cm²

Volume of hemisphere (bowl) = `2/3 πr^3`

= `2/3 xx 3.14 xx 10 xx 10 xx 10`

= `6280/3`

= 2093.33 cm³

Difference in volume = V_box – V_bowl

= 4630.50 – 2093.33

= 2537.17 cm³

OR

b. Inner surface area of hemisphere (bowl) = 2πr²

= 2 × 3.14 × 10 × 10

= 628 cm²

Inner surface area of ring = π(R² – r²)

= π[(10.5)² – 10²]

= π[110.25 – 100]

= 3.14 × 10.25

= 32.185 cm²

The total surface area to be painted = 628 + 32.185

= 660.185 cm²