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 :

<p>We know,</p> <p>Surface area of balloon (s) = 4$\pi$r²</p> <p>$\therefore$ ${{ds} \over {dt}} = {d \over {dt}}(4\pi {r^2})$</p> <p>$\Rightarrow {{ds} \over {dt}} = 4\pi (2r) \times {{dr} \over {dt}}$</p> <p>$\Rightarrow {{ds} \over {dt}} = 8\pi r \times {{dr} \over {dt}}$</p> <p>Given that, surface area of balloon is increasing in constant rate.</p> <p>$\therefore$ ${{ds} \over {dt}}$ = constant = k (Assume)</p> <p>$\Rightarrow k = 8\pi r\,.\,{{dr} \over {dt}}$</p> <p>$\Rightarrow \int {k\,dt = \int {8\pi r\,dr} }$</p> <p>$\Rightarrow kt = 8\pi \times {{{r^2}} \over 2} + c$</p> <p>$\Rightarrow kt = 4\pi {r^2} + c$ ..... (1)</p> <p>Given at t = 0, radius r = 3</p> <p>So, $0 = 4\pi ({3^2}) + c$</p> <p>$\Rightarrow c = - 36\pi$</p> <p>$\therefore$ Equation (1) becomes</p> <p>$kt = 4\pi {r^2} - 36\pi$</p> <p>Also given, at t = 5, radius r = 7</p> <p>$\therefore$ $k(5) = 4\pi {(7)^2} - 36$</p> <p>$\Rightarrow k = 32\pi$</p> <p>$\therefore$ Equation (1) is</p> <p>$32\pi t = 4\pi {r^2} - 36\pi$</p> <p>Now at $t = 9$</p> <p>$\Rightarrow 32\pi (9) = 4\pi {r^2} - 36\pi$</p> <p>$\Rightarrow 8 \times 9 = {r^2} - 9$</p> <p>$\Rightarrow {r^2} = 81 \Rightarrow r = 9$</p>