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Question
A cylindrical tube of radius 12cm contains water to a depth of 20cm. A spherical ball of radius 9cm is dropped into the tube and thus level of water is raised by hcm. What is the value of h.
Solution
Given that radius of cylindrical tube (r1) = 12cm
Let height of cylindrical tube (h)
Volume of a cylinder = `pir_1^2h`
v1 = π(12)2 x h ............(1)
Given spherical ball radius (r2) = 9cm
Volume of sphere = `4/3pir_2^3`
`v_2=4/3xxpixx9^2` ..............(2)
Equating (1) and (2)
v1 = v2
`pi(12)^2xxh=4/3xxpixx9^3`
`h=(4/3xxpixx9^3)/(pi(12)^2)`
h = 6.75cm
Level of water raised in tube (h) = 6.75cm
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