Question

A wet substance in the open air loses its moisture at a rate proportional to the moisture content. If a sheet, hung in the open air, loses half its moisture during the first hour, then $90%$ of the moisture will be lost in .............. hours.

• A. $2 \log_2 10$
• B. $\frac{4 \log 10}{\log 2}$
• C. $\log_2 10$
• D. $\frac{3 \log 10}{\log 2}$

Hint:

Radioactive decay formula $N = N_0(1/2)^{t/T}$ works perfectly for "half-life" problems like this.

Answer 1

The Correct Option is C

Step 1: Form Differential Equation
Let $M$ be moisture content. $\frac{dM}{dt} = -kM \implies M(t) = M_0 e^{-kt}$.
Step 2: Find $k$
At $t=1$, $M = M_0/2 \implies \frac{1}{2} = e^{-k} \implies k = \log_e 2$.
So, $M(t) = M_0 e^{-t \ln 2} = M_0 (2^{-t})$.
Step 3: Solve for 90% loss
90% lost means 10% remains: $0.1 M_0 = M_0 (2^{-t})$.
$10^{-1} = 2^{-t} \implies 10 = 2^t \implies t = \log_2 10$.
Final Answer: (C)