A) 5 min
B) 10 min
C) 20 min
D) 80 min
Correct Answer: B
$ \Rightarrow $ $ t _1\propto \frac{1}{\frac{\pi }{4}d_1^{2}} $ and $ t _2\propto \frac{1}{\frac{\pi }{4}d_2^{2}} $
$ \Rightarrow $ $ \frac{t _2}{t _1}={{( \frac{d _1}{d _2} )}^{2}} $
$ \therefore $ $ t _2=40{{( \frac{d}{2d} )}^{2}}=10\min $