Let d 1 = d_1= d 1 = the diameter of the base of the first container, h 1 = h_1= h 1 = the heigth of the first container and V 1 = V_1= V 1 = the volume of the first container.
Let d 2 = d_2= d 2 = the diameter of the base of the second container, h 2 = h_2= h 2 = the heigth of the second container and V 2 = V_2= V 2 = the volume of the second container.
Then
h 1 = 2 d 1 , V 1 = π ( d 1 2 ) 2 h 1 = 1 2 π d 1 3 = 1 16 π h 1 3 h_1=2d_1, V_1=\pi (\dfrac{d_1}{2})^2h_1=\dfrac{1}{2}\pi d_1^3=\dfrac{1}{16}\pi h_1^3 h 1 = 2 d 1 , V 1 = π ( 2 d 1 ) 2 h 1 = 2 1 π d 1 3 = 16 1 π h 1 3
h 2 = 3 d 2 , V 1 = π ( d 2 2 ) 2 h 2 = 3 4 π d 2 3 = 1 36 π h 2 3 h_2=3d_2, V_1=\pi (\dfrac{d_2}{2})^2h_2=\dfrac{3}{4}\pi d_2^3=\dfrac{1}{36}\pi h_2^3 h 2 = 3 d 2 , V 1 = π ( 2 d 2 ) 2 h 2 = 4 3 π d 2 3 = 36 1 π h 2 3
Given
V 1 = V 2 = 12 L = 12000 c m 3 V_1=V_2=12L=12000cm^3 V 1 = V 2 = 12 L = 12000 c m 3
3 4 π d 2 3 = 12000 c m 3 \dfrac{3}{4}\pi d_2^3=12000cm^3 4 3 π d 2 3 = 12000 c m 3
d 2 = 4 ( 12000 ) 3 π 3 c m d_2=\sqrt[3]{\dfrac{4(12000)}{3\pi}}cm d 2 = 3 3 π 4 ( 12000 ) c m
d 2 = 20 2 π 3 c m ≈ 17.205 c m d_2=20\sqrt[3]{\dfrac{2}{\pi}}cm\approx 17.205\ cm d 2 = 20 3 π 2 c m ≈ 17.205 c m The diameter of the second container is 20 2 π 3 c m ≈ 17.205 c m . 20\sqrt[3]{\dfrac{2}{\pi}}cm\approx 17.205\ cm. 20 3 π 2 c m ≈ 17.205 c m .