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sonali singh

· started a discussion

· 1 Months ago

problem (Question: 90241.0 )

please explain the formula

Question:
Some water is contained in a cylindrical vessel of diameter 12 cm. A heavy sphere of diameter 6 cm is placed in the vessel. By how much does the water level rise if no water overflows?
Options:
A) 1 cm 
B) 2 cm 
C) 3 cm 
D) 4 cm
E) 5 cm
Solution:
\(h =\cfrac{4r^3_1}{3r^2_1}=\cfrac{4}{3}\times\cfrac{3^3}{6^2}= 1cm\)

Alternate sol:


Volume of cylinder = Volume of Sphere


pi R^2 h = 4/3 pi r^3


h = (4 pi r^3)/(3 pi R^2) = (4 r^3)/(3 R^2) = (4 * 3^3)/(3 * 6^2) = (4*27)/(36*3) = 108/108 = 1 cm


kanhaiya jha

· commented

· 1 Months ago

Volume of cylinder = Volume of Sphere

pi R^2 h = 4/3 pi r^3

h = (4 pi r^3)/(3 pi R^2) = (4 r^3)/(3 R^2) = (4 * 3^3)/(3 * 6^2) = (4*27)/(36*3) = 108/108 = 1 cm
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