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Question
If water is poured into an inverted hollow cone whose semi-vertical angle is 30°, so that its depth (measured along the axis) increases at the rate of`( 1"cm")/sec`. Find the rate at which the volume of water increasing when the depth is 2 cm.
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Solution
Let r be radius, h be the height, θ be the semi-vertical angle and V be the volume of the water at any time t.
Given : `"dh"/dt = (1"cm")/sec`, θ = 30°
Now, V = `(1)/(3)pir^2h`
But, tan 30° = `r/h`
∴ `(1)/sqrt(3) = r/h`
∴ r = `h/sqrt(3)`
∴ V = `(1)/(3)pi(h/sqrt(3))^2h = pi/(9)h^3`
Differentiating w.r.t. t, we get,
`"dV"/dt = pi/(9) xx 3h^2 "dh"/dt = pi/(3)h^2"dh"/dt`
When h = 2cm, then
`"dV"/dt = pi/(3) xx (2)^2 xx 1 = (4pi)/(3)`
Hence, the volume of water is increasing at the rate of `((4pi)/3)"cm"^3/sec`
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