Respuesta :
Since in the above case, the beaker has two sections each with different radius and height, we will divide this problem into two parts.
We will calculate the volume of both the beakers separately and then add them up together to get the volume of the beaker.
Given, π = 3.14
Beaker 1:
Radius (r₁) = 2 cm
Height (h₁) = 3 cm
Volume (V₁) = π r₁² h₁ = 3.14 x 2² x 3 = 37.68 cm³
Beaker 2:
Radius (r₂) = 6 cm
Height (h₂) = 4 cm
Volume (V₂) = π r₂² h₂ = 3.14 x 6² x 4 = 452.16 cm³
Volume of beaker = V₁ + V₂ = 37.68 + 452.16 = 489.84 cm³
We will calculate the volume of both the beakers separately and then add them up together to get the volume of the beaker.
Given, π = 3.14
Beaker 1:
Radius (r₁) = 2 cm
Height (h₁) = 3 cm
Volume (V₁) = π r₁² h₁ = 3.14 x 2² x 3 = 37.68 cm³
Beaker 2:
Radius (r₂) = 6 cm
Height (h₂) = 4 cm
Volume (V₂) = π r₂² h₂ = 3.14 x 6² x 4 = 452.16 cm³
Volume of beaker = V₁ + V₂ = 37.68 + 452.16 = 489.84 cm³
so we just computate the volumes of 2 cylinders and add them
Vcylinder=hpir^2
top one is r=2 and h=3, so V=3pi2^2 or 3pi4 or 12pi cm^3
bottom one is r=6 and h=2 so V=4pi6^2 or 4pi36 or 144pi cm^3
total
12pi+144pi=156pi=total volume
aprox pi=3.14
total volume=489.84
C is answer