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JEE Main & Advanced JEE Main Paper (Held On 15 April 2018) Slot-I

  • question_answer
    A solution containing active cobalt \[_{27}^{60}Co\]having activity of \[0.8\mu Ci\] and decay constant \[\lambda \] is injected in an animal's body. If \[1c{{m}^{3}}\] of blood is drawn from the animal's body after 10 hrs of injection, the activity found was 300 decays per minute. What is the volume of blood that is flowing in the body? (\[1Ci=3.7\times {{10}^{10}}\]decay per second and at \[t=10hrs\]\[{{e}^{-\lambda t}}=0.84\])                        [JEE Online 15-04-2018]

    A) litres                

    B) litres                

    C) litres                

    D) litres

    Correct Answer: D

    Solution :

    The activity equation can be written as \[-\frac{dN}{dt}=\lambda {{N}_{o}}{{e}^{-\lambda t}}\] given that \[\lambda {{N}_{o}}=0.8\mu {{C}_{i}}\] Putting the values, \[\lambda {{N}_{o}}=2.96\times {{10}^{4}}\] Let the volume of the blood flowing be V, the activity would reduce by a factor of\[\frac{{{10}^{-3}}}{V}\] Hence \[\frac{\lambda {{N}_{o}}{{10}^{-3}}}{V}{{e}^{-\lambda t}}=300/60\] (Both R.H.S. and L.H.S. are decay/s) Putting the values of \[{{e}^{-\lambda t}}\] and \[\lambda {{N}_{o}}\] we get \[V=5litre\]


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