UPSEE 2018 Solved Paper

∵ Activity of a radioactive sample,
A=A0(

1

2

)

1

t1∕2

where, A0= initial activity.
In first case,

A0

5

=A0(

1

2

)

10

t1∕2

1

5

=(

1

2

)

10

t1∕2

.........(i)
In second case,
A=A0(

1

2

)

20

t1∕2

.........(ii)
From Eqs. (i) and (ii), we get

1∕5

A

=

( 1 2 ) 10 t1∕2

A0( 1 2 ) 20 t1∕2

⇒

1

5A

=

1

A0( 1 2 ) 10 t1∕2

⇒

1

5A

=

1

A0 5

[from Eq. (i) ]
∴A=

A0

5×5

=

A0

25