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प्रश्न
A balloon, which always remains spherical, has a variable diameter `3/2 (2x + 1)` Find the rate of change of its volume with respect to x.
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उत्तर
Diameter of the balloon, `d = 3/2 (2x + 1)`
∴ The radius of the ballon, `r = d/2`
`= 1/2 {3/2 (2x + 1)} = 3/4 (2x + 1)`
So, the volume V of the balloon,
`V = 4/3 pi "(radius)"^3 = 4/3 pi {3/4 (2x + 1)}^3`
`= (9pi)/16 (2x + 1)^3` ....(i)
Differentiating (i) w.r.t. x, we get
`(dV)/dx = (9pi)/16 xx 3 (2x + 1)^2 xx2`
`= (27 pi)/8 (2x+ 1)^2`
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