MHTCET 13 May 2023 Shift 2 Solved Paper

Mass of substance at time ‌‌ ‌t‌ is ‌m
Rate of change of mass =‌

dm

dt

∝m
‌

dm

dt

‌=−km;k⟶ constant
∫‌

dm

m

‌=−∫k‌dt
log‌m‌=−kt+c...(i)
t=0,m‌=m0
‌ At ‌‌‌log‌m0‌=c
log‌m‌=−kt+log(m0)....(ii)
At time t=h,m=‌

1

2

m0
From Eq. (ii), log(‌

m0

2

)=−kh+log(m0)
‌⇒‌‌log(m0)−log(‌

m0

2

)=kh
‌⇒‌‌log‌2=kh⇒k=‌

1

h

‌log‌2
Put value of k in Eq. (ii), we get
log‌m=(−‌

1

h

‌log‌2)t+log‌m0...(iii)
⇒log(m)−log‌m0‌=−‌

t

h

‌log‌2
⇒‌‌log‌

m

m0

‌=log(2−‌

t

h

)
‌

m

m0

‌=2−‌

t

h

⇒m=2−t∕h⋅m0
Initial decay rate
‌

dm

dt

‌=−km=−‌

1

h

‌log‌2×m
‌

dm

dt

‌=−‌

m

h

‌log‌2
So, initial mass =m0 then
‌

dm

dt

=−‌

m0

h

‌log‌2