Ant on a balloon

Let at any time $t$ , $\omega$ denote the angular velocity of the ant, $\theta$ denote the angle covered by the ant and $r$ denote the radius of the balloon. The velocity (given) is denoted by $v$ , initial radius (given) by $r_0$ and rate of increase of radius (given) by $\alpha$ .

$\therefore \frac{dr}{dt} = \alpha$

$\implies \int\limits_{r_0}^{r} dr = \int\limits_{0}^{t} \alpha dt$

$\implies r = \alpha t + r_0$

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$\because \omega = \frac{v}{r}$

$\implies\frac{d\theta}{dt} = \frac{v}{\alpha t + r_0}$

$\implies d\theta = \frac{v}{\alpha t + r_0}dt$

$\implies\int\limits_{0}^{2\pi} d\theta = \int\limits_{0}^{T}\frac{v}{\alpha t + r_0}dt$

$\implies2\pi = \frac{v}{a} \ln(\frac{\alpha T}{r_0} + 1)$

$\implies \boxed{T = \frac{r_0}{\alpha} \Big(e^{\frac{2\alpha\pi}{v}} - 1\Big)}$

Substituting the values of known quantities gives the answer to be equal to $e^{2\pi} - 1 \approx 534.5s$