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A balloon, which always remains spherical on inflation, is being inflated by pumping in 900 cubic centimetres of gas per second.

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प्रश्न

A balloon, which always remains spherical on inflation, is being inflated by pumping in 900 cubic centimetres of gas per second. Find the rate at which the radius of the balloon increases when the radius is 15 cm.

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उत्तर

Let at ant instant of time t, the redius of the balloon be r and its volume be V, then

`V = 4/3 pir^3`                      ....(i)

Differentiating (i) w.r.t.t, we get,

`(dV)/dt = (4/3 pi) (3r^2  (dr)/dt)`

= 900 cm3/sec = `(4/3 pi) {3 (15 cm)^2  (dr)/dt}`   ....`(∵ (dV)/ (dr) = 900  cm^3 //sec.)`

`= (dr)/dt = 900/ (4pi xx (15)^2)` cm/sec

`1/pi` cm/sec

∴ Rate of increase of the radius of the balloon  = `1/pi` cm/sec.

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अध्याय 6: Application of Derivatives - Exercise 6.1 [पृष्ठ १९८]

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एनसीईआरटी Mathematics Part 1 and 2 [English] Class 12
अध्याय 6 Application of Derivatives
Exercise 6.1 | Q 8 | पृष्ठ १९८

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