Q.9.The decay rate of radioactive material at any time t is proportional to its mass at that time. The mass is 27 grams when t = 0. After three hours, it was found that 8 grams are left. Then the substance left after one more hour is?
\text{A.} \frac{27}{8} \text{ grams} \text{B.} \frac{81}{4} \text{ grams} \text{C.} \frac{16}{3} \text{ grams} \text{D.} \frac{16}{9} \text{ grams} \text{Answer: C} \text{Explanation:} \text{Let } x \text{ be the mass of the material at time } t.\therefore \frac{\mathrm{d}x}{\mathrm{d}t}=-\mathrm{k}x, (negative sign indicates decay.)
\therefore \int\frac{\mathrm{d}x}{x}=-\mathrm{k}\int\mathrm{d}t \therefore \log|x|=-kt + C \text{When } t=0, x=27 \therefore C=\log 27 \therefore \log |x| = -kt + \log 27When t=3, x=8
k= log\left ( \frac{3}{2} \right )When t=4, we get
log |x| = -4 log\left ( \frac{3}{2} \right ) + log 27
log |x| = log\; \left ( \frac{16}{3} \right )
x=\frac{16}{3}\; grams