A solution containing active cobalt $^{60}_{27}\text{Co}$ having an activity of $0.8\ \mu\text{Ci}$ and decay constant $\lambda$ is injected in an animal's body. When $1\ \text{cm}^3$ blood is drawn from the animal's body after $10\ \text{h}$ of injection, the activity was found to be $300\ \text{dpm}$ (decays per minute), then the total volume of blood in the animal's body is close to $\left[1\ \text{Ci} = 3.7\times10^{10}\ \text{dps (decays per second) and at } t = 10\ \text{h the value of } e^{-\lambda t} = 0.84\right]$
Answer: (C) $5\ \text{L}$
Initial activity:
$$A_0 = 0.8\times10^{-6}\times3.7\times10^{10} = 2.96\times10^4\ \text{dps} = 1.776\times10^6\ \text{dpm}$$
After $10\ \text{h}$:
$$A = 0.84\,A_0 = 1.49\times10^6\ \text{dpm}$$
Assuming the cobalt spreads uniformly through the blood, $300\ \text{dpm}$ per $\text{cm}^3$ gives
$$V = \frac{1.49\times10^6}{300} \approx 4.97\times10^3\ \text{cm}^3 \approx 5\ \text{L}$$
Solution by Sreeraj P, M.Sc Physics