$(a)\;6 l\qquad(b)\;8 l\qquad(c)\;2 l\qquad(d)\;1 l$

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Answer : (a) 6 l

Explanation :

The activity varries according to equation $\;A (t) = A_{0} \;e^{-\lambda t}$

or $\;A(t)=A_{0} A_{0} \;(\large\frac{1}{2})^{\large\frac{t}{T_{\large\frac{1}{2}}}}$

$A (t=5 hr)=296 \;disintegrations/min$

$\large\frac{296}{60}=A_{0}\; (\large\frac{1}{2})^{\large\frac{5}{15}}$

$A_{0}=\large\frac{296}{60} \times2^{\large\frac{1}{3}}=6.22\;disintegration/sec $

= the initial activity of 1cc of blood .

If total volume of blood is $V_{cc}.$

$6.22 V=3.7\times10^{6} \times10^{-6}=6 l\;.$

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