The volume of spherical balloon being inflated changes at a constant rate. If initially its radius is 3 units and after 3 seconds it is 6 units. Find the radius of balloon after t seconds.

Options

A. `(63t + 27)^(1//3)`
B. `(27t+63)^(1//3)`
C. `(27t+9)^(1//3)`
D. `(9 t +63)^(1//3)`

Answer

a

Explanation

Let the rate of change of the volume of the ballon be k. where k is a constant .
` d/(dt) `(volume )= constant
` rArr d/(dt) (4/3 pir^(3))= k [ :' "volume of shape " = 4/3 pir^(3)]`
` rArr (4/3pi) (3r(dr)/(dt))=k `
On separating the variables , we get
` 4 pir^(2) dr = kdt " "` ... (i)
On integrating both sides , we get
` 4pir^(2) dr = kdt" "` ...(i)
` 4pi int r^(2) dr = k int dt `
` rArr 4pi^(2) (r^(3))/3 = kt +C rArr 4pir^(3) = 3 (kt+C)" "` ... (ii) ltbRgt Now, at t= 0 , r = 3 intially
` :. 4pi(3)^(3) = 3(K xx 0+C)`
` rArr 108 pi = 3 C rArr C = 36 pi`
Also , when t = 3 , then r = 6
From Eq. (ii)
` 4pi(6)^(3) = 3(kxx 3 +C) rArr 864pi = 3 (3k +36pi)`
` rArr 3k = 288pi - 36pi = 252pi rArr k = 84pi`
On substituting the values of k and C in Eq. (ii) , we get
` 4pir^(3) = 3 (84 pit +36pi)`
` rArr 4pir^(3) = 4pi (63t+27)`
` rArr r^(3) = 63t +27`
` rArr r = (63t+27)^(1//3)` units
which is the required radius of the ballon at time t.