JEE Main 24 June 2022 Shift 1 Solved Paper

We know,
Surface area of balloon (s)=4πr2
∴‌

ds

dt

=‌

d

dt

(4πr2)
⇒‌

ds

dt

=4π(2r)×‌

dr

dt

⇒‌

ds

dt

=8πr×‌

dr

dt

Given that, surface area of balloon is increasing in constant rate.
∴‌

ds

dt

= constant =k (Assume)
⇒k=8πr⋅‌

dr

dt

⇒∫kdt=∫8πrdr
⇒kt=8π×‌

r2

2

+c
⇒kt=4πr2+c.....(1)
Given at t=0, radius r=3
So, 0=4π(32)+c
⇒c=−36π
∴ Equation (1) becomes
kt=4πr2−36π
Also given, at t=5, radius r=7
∴k(5)=4π(7)2−36
⇒k=32π
∴ Equation (1) is
32πt=4πr2−36π
Now at t=9
⇒32π(9)=4πr2−36π
⇒8×9=r2−9
⇒r2=81⇒r=9