The surface area of a balloon of spherical shape being inflated, increases at a constant rate. If initially, the radius of balloon is $$3$$ units and after $$5$$ seconds, it becomes $$7$$ units, then its radius after $$9$$ seconds is

The surface area of a sphere is $$S = 4\pi r^2$$. The surface area increases at a constant rate, so $$\frac{dS}{dt} = k$$ (constant).

Express S as a linear function of time.

Since $$\frac{dS}{dt} = k$$, we have $$S(t) = S(0) + kt$$.

Use the given conditions.

At $$t = 0$$: $$r = 3$$, so $$S(0) = 4\pi(3^2) = 36\pi$$.

At $$t = 5$$: $$r = 7$$, so $$S(5) = 4\pi(7^2) = 196\pi$$.

$$k = \frac{196\pi - 36\pi}{5} = \frac{160\pi}{5} = 32\pi$$

Find the radius at t = 9.

$$S(9) = 36\pi + 9 \times 32\pi = 36\pi + 288\pi = 324\pi$$

$$4\pi r^2 = 324\pi$$

$$r^2 = 81$$

$$r = 9$$

Answer: Option A (r = 9)