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  • 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 radiu

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:

Option: 1

9


Option: 2

10


Option: 3

11


Option: 4

12


Answers (1)

best_answer

Let r : radius
\mathrm{s= 4\pi r^{2}}

\mathrm{\frac{ds}{dt}= 8\pi\, r\left ( \frac{dr}{dt} \right )= \lambda \left ( constant \right )}
\mathrm{4\pi r^{2}= \lambda \, t+c}

\mathrm{at\: t= 0,\; r= 3\quad \Rightarrow \quad c= 36\, \pi }
\mathrm{at\: t= 5,\; r= 7\quad \Rightarrow \quad \lambda = 32\, \pi }

\mathrm{\therefore \lambda ^{2}= 8\, t+9}

\mathrm{\therefore at\: t= 9}
        \mathrm{r^{2}= 81}
        \mathrm{r=9\: unit}
The correct answer is option (A)

Posted by

Gautam harsolia

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