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

Step 1: Understanding the Concept:
The bird-bath is a combination of a cylinder and a hemisphere. We must account for the portion of the cylinder above the ground.
Step 2: Key Formula or Approach:
1. Radius $r = d/2$.
2. Volume of hemisphere = $\frac{2}{3}\pi r^3$.
3. TSA of bird-bath = CSA of cylinder + Area of base + CSA of hemisphere.
Step 3: Detailed Explanation:
1. (i) Radius: $d = 1.4$ m $\implies r = 0.7$ m.
2. (ii) Volume of water:
- $V = \frac{2}{3}\pi(0.7)^3 = \frac{2}{3}\pi(0.343) = 0.228\pi$ m$^3$ (approx $\frac{343}{1500}\pi$).
3. (iii) (a) TSA above ground:
- Height above ground $h = 2 - 0.6 = 1.4$ m.
- TSA = CSA(cyl) + CSA(hemi) = $2\pi rh + 2\pi r^2$ (Top is open).
- TSA = $2\pi(0.7)(1.4) + 2\pi(0.7)^2 = 1.96\pi + 0.98\pi = 2.94\pi$ m$^2$.
4. (iii) (b) OR Volume of wood:
- $V = V_{cyl} - V_{hemi} = \pi r^2h - \frac{2}{3}\pi r^3$.
- $V = \pi(0.7)^2(1.4) - 0.228\pi = 0.686\pi - 0.228\pi = 0.458\pi$ m$^3$.
Step 4: Final Answer:
(i) 0.7 m. (ii) $0.228\pi$ m$^3$. (iii)(a) $2.94\pi$ m$^2$ or (iii)(b) $0.458\pi$ m$^3$.