JEE Main 2022 · 24 Jun · Shift 1 · Mathematics Q3

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MathematicsSingle correct+4 / -1Question 3 of 90

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 :

Answer and worked solution

Official answer: A

Constant rate. $\dfrac{d}{dt}(4\pi r^2)=k$ , so $r^2$ is linear in $t$ : $r^2=9+ct$ .

Fit. At $t=5$ , $49=9+5c$ , so $c=8$ .

At $t=9$ . $r^2=9+72=81$ , $r=9$ .

(A)

Topics:Rate of change

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JEE Main 2022 · 24 Jun · Shift 1, Mathematics question 3