Let the quantity of mass at any time t be A.

According to the question,

⇒ where k is a constant

⇒

⇒

Integrating both sides, we have

⇒ ∫ = – k∫dt

⇒ log|A| = – kt + c……(1)

Given, the Initial quantity of masss be A₀ when the t = 0 sec

Putting the value in equation (1)

∴ log|A| = – kt + c

⇒ log| A₀| = 0 + c

⇒ c = log| A₀| ……(2)

Putting the value of c in equation (1) we have,

log|A| = – kt + log| A₀|

⇒ log|A| – log| A₀| = – k t []

⇒ log ( = – kt ……(3)

Let the mass becomes half at time t1, A =

From equation(3),we have

∴ – kt = log (

⇒ – k×t1 = log (

⇒ – k×t1 =

⇒ – k×t1 = – log 2

⇒ t1 =

∴ Required time = where k is the constant of proportionality.