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A mixture consists of two radioactive materials $A_1$ and $A_2$ with half lives of $20 \mathrm{~s}$ and $10 \mathrm{~s}$ respectively. Initially the mixture has $40 \mathrm{~g}$ of $A_1$ and $160 \mathrm{~g}$ of $A_2$. The amount of the two in the mixture will become equal after
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The correct answer is:
$40 \mathrm{~s}$
For $40 \mathrm{~g}$ amount
$40 \mathrm{~g} \underset{\text { half -life }}{\stackrel{20 \mathrm{~s}}{\longrightarrow}} 20 \mathrm{~g} \stackrel{20 \mathrm{~s}}{\longrightarrow} 10 \mathrm{~g}$
For $160 \mathrm{~g}$ amount
$\begin{array}{l}
160 \mathrm{~g} \stackrel{10 \mathrm{~s}}{\longrightarrow} 80 \mathrm{~g} \stackrel{10 \mathrm{~s}}{\longrightarrow} 40 \mathrm{~g} \\
40 \mathrm{~g} \stackrel{10 \mathrm{~s}}{\longrightarrow} 20 \mathrm{~g} \stackrel{10 \mathrm{~s}}{\longrightarrow} 10 \mathrm{~g}
\end{array}$
So, after $40 \mathrm{~s} A_1$ and $A_2$ remains same.
$40 \mathrm{~g} \underset{\text { half -life }}{\stackrel{20 \mathrm{~s}}{\longrightarrow}} 20 \mathrm{~g} \stackrel{20 \mathrm{~s}}{\longrightarrow} 10 \mathrm{~g}$
For $160 \mathrm{~g}$ amount
$\begin{array}{l}
160 \mathrm{~g} \stackrel{10 \mathrm{~s}}{\longrightarrow} 80 \mathrm{~g} \stackrel{10 \mathrm{~s}}{\longrightarrow} 40 \mathrm{~g} \\
40 \mathrm{~g} \stackrel{10 \mathrm{~s}}{\longrightarrow} 20 \mathrm{~g} \stackrel{10 \mathrm{~s}}{\longrightarrow} 10 \mathrm{~g}
\end{array}$
So, after $40 \mathrm{~s} A_1$ and $A_2$ remains same.
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