If $$\frac {dp}{dt}=\frac{1}{2p}$$ and $$p=1$$ when $$t=0$$, then $$p=3$$, when t is​

Correct option is B

​$$\textbf{Solution:}\\ \quad \frac{dp}{dt} = \frac{1}{2p}, \quad p(0) = 1 \\[8pt]\textbf{Step 1: Separate the variables} \\2p \, dp = dt \\[8pt]\textbf{Step 2: Integrate both sides} \\\int 2p \, dp = \int dt \Rightarrow p^2 = t + C \\[8pt]\textbf{Step 3: Use initial condition } p = 1 \text{ when } t = 0 \\1^2 = 0 + C \Rightarrow C = 1 \\[8pt]\text{So the solution is: } p^2 = t + 1 \Rightarrow t = p^2 - 1 \\[8pt]\textbf{Step 4: Find } t \text{ when } p = 3 \\t = 3^2 - 1 = 9 - 1 = 8 \\[10pt]\boxed{\text{Final Answer: (B) } t = 8}$$

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