Manipal Entrance Test 2020 Physics Paper

Number of half-lives in 40 min
n=‌

40

T1/2

=‌

40

10

=4
Amount of the substance remaining after n half- lives
N=N0(‌

1

2

)n=N0(‌

1

2

)4=‌

N0

16

where, N0 is original amount of the substance.
Amount of the substance decayed in 40min.
=(‌

15

16

×100%) of N0
=15×6.25% of N0
=93.75% of N0