QJEEM22Mac86f: JEE Main 2022 Mathematics - Class 12 JEEM22Mac86f Solution: The surface area of a balloon of spherical shape being…

We know,

Surface area of balloon (s) = 4 $\pi$ r²

$\therefore$ ${{ds} \over {dt}} = {d \over {dt}}(4\pi {r^2})$

$\Rightarrow {{ds} \over {dt}} = 4\pi (2r) \times {{dr} \over {dt}}$

$\Rightarrow {{ds} \over {dt}} = 8\pi r \times {{dr} \over {dt}}$

Given that, surface area of balloon is increasing in constant rate.

$\therefore$ ${{ds} \over {dt}}$ = constant = k (Assume)

$\Rightarrow k = 8\pi r\,.\,{{dr} \over {dt}}$

$\Rightarrow \int {k\,dt = \int {8\pi r\,dr} }$

$\Rightarrow kt = 8\pi \times {{{r^2}} \over 2} + c$

$\Rightarrow kt = 4\pi {r^2} + c$ ..... (1)

Given at t = 0, radius r = 3

So, $0 = 4\pi ({3^2}) + c$

$\Rightarrow c = - 36\pi$

$\therefore$ Equation (1) becomes

$kt = 4\pi {r^2} - 36\pi$

Also given, at t = 5, radius r = 7

$\therefore$ $k(5) = 4\pi {(7)^2} - 36$

$\Rightarrow k = 32\pi$

$\therefore$ Equation (1) is

$32\pi t = 4\pi {r^2} - 36\pi$

Now at $t = 9$

$\Rightarrow 32\pi (9) = 4\pi {r^2} - 36\pi$

$\Rightarrow 8 \times 9 = {r^2} - 9$

$\Rightarrow {r^2} = 81 \Rightarrow r = 9$