A solution containing active cobalt _ 27 ^ 60 Co having activity of 0.8 μ Ci and decay constant λ is injected in an animal's body. If 1 ~cm^3 of blood is drawn from the animal's body after 10 hrs of injection, the

Correct answer: D. 5 litres Let initial activity =No=0.8 μ ci 0.8 × 3.7 × 10^4 dps Activity in 1 ~cm^3 of blood at t=10 hr , n=(300)/(60) dps=5 dps N= Activity of whole blood at time t=10 hr . Total volume of the blood in the person, V =(N)/(n) =(N_0 e-λ t)/(n)=(0.8 × 3.7 × 10^4 × 0.7927)/(5) 5 litre

JEE Main 201815 AprilMorning ShiftOnline

Exam: JEE Main
Year: 2018
Date: 15 April
Shift: Morning Shift
Mode: Online
Subject: Physics
Chapter: Nuclear Physics

JEE Main 2018 · 15 April · Morning Shift · Online · Physics · Nuclear Physics

A solution containing active cobalt _ 27 ^ 60 Co having activity of...

A solution containing active cobalt

{ }_{27}^{60} \mathrm{Co}

having activity of

0.8 \mu \mathrm{Ci}

and decay constant

\lambda

is injected in an animal's body. If

1 \mathrm{~cm}^3

of blood is drawn from the animal's body after

10 \mathrm{hrs}

of injection, the activity found was 300 decays per minute. What is the volume of blood that is flowing in the body?

\left(1 \mathrm{Ci}=3.7 \times 10^{10}\right.decay per second and

\left.\mathrm{t}=10 \mathrm{hrs}^{-\lambda \mathrm{t}}=0.84\right)

Options

A6 litres
B7 litres
C4 litres
D5 litres

Correct answer

D. 5 litres

Step-by-step solution

Let initial activity

=\mathrm{No}=0.8 \mu \mathrm{ci}

0.8 \times 3.7 \times 10^4 \mathrm{dps}

Activity in

1 \mathrm{~cm}^3

of blood at

\mathrm{t}=10 \mathrm{hr}

\mathrm{n}=\frac{300}{60} \mathrm{dps}=5 \mathrm{dps}

\mathrm{N}=

Activity of whole blood at time

\mathrm{t}=10 \mathrm{hr}

. Total volume of the blood in the person, V

=\frac{N}{n}

=\frac{N_0 e-\lambda t}{n}=\frac{0.8 \times 3.7 \times 10^4 \times 0.7927}{5} \cong 5

litre

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