Question

Mishika and Sahaj created a bird-bath from a cylindrical log of wood by scooping out a hemispherical depression. Cylinder length is 2 m (0.6 m in earth) and diameter is 1.4 m.
(i) Radius of depression?
(ii) Volume of water in hemisphere in terms of $\pi$?
(iii) (a) Total surface area of log above ground? OR
(iii) (b) Volume of log above ground?

Hint:

When a shape is "scooped out," the volume decreases, but the surface area actually increases because the interior of the hole is now exposed.

Answer 1

Given:
Length of cylinder = 2 m
Part inside earth = 0.6 m
Height above ground = 2 − 0.6 = 1.4 m
Diameter of cylinder = 1.4 m
Radius of cylinder = 1.4 / 2 = 0.7 m

1) Radius of Hemispherical Depression:
- The depression is hemispherical and fits exactly on the top.
- Therefore, radius = radius of cylinder
- Radius = 0.7 m

2) Volume of Water in Hemisphere:
- Formula: Volume of hemisphere = (2/3)πr³
- = (2/3)π(0.7)³
- = (2/3)π(0.343)
- = (2 × 0.343 / 3)π
- = (0.686 / 3)π
- = 0.2287π m³

3) (a) Total Surface Area of Log Above Ground:
- Only curved surface above ground is considered.
- Formula: Curved Surface Area = 2πrh
- r = 0.7 m, h = 1.4 m
- = 2π × 0.7 × 1.4
- = 1.96π m²

OR

3) (b) Volume of Log Above Ground:
- Formula: Volume of cylinder = πr²h
- = π(0.7)²(1.4)
- = π(0.49)(1.4)
- = 0.686π m³

Conclusion:
Radius of depression = 0.7 m.
Volume of water in hemisphere = 0.2287π m³.
Surface area above ground = 1.96π m² OR Volume above ground = 0.686π m³.