The graph in the figure shows how the count-r | vedclass

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(D) The graph is a straight line plotted between $\ln A$ and $t$. The equation of a straight line is $y = mx + C$.Here,$y = \ln A$ and $x = t$.The int

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$A = 12 e^{-0.1t}$

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(D) The graph is a straight line plotted between $\ln A$ and $t$. The equation of a straight line is $y = mx + C$.Here,$y = \ln A$ and $x = t$.The intercept $C$ on the $\ln A$ axis is $2.6$.The slope $m$ is calculated as $m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{0 - 2.6}{26 - 0} = -\frac{2.6}{26} = -0.1$.Thus,the equation of the line is $\ln A = -0.1t + 2.6$.Taking the exponential of both sides,we get $A = e^{-0.1t + 2.6} = e^{2.6} \cdot e^{-0.1t}$.Given $\ln 12 = 2.6$,it follows that $e^{2.6} = 12$.Substituting this value,we get $A = 12 e^{-0.1t}$.

The graph in the figure shows how the count-rate $A$ of a radioactive source as measured by a Geiger counter varies with time $t$. The relationship between $A$ and $t$ is (Assume $\ln 12 = 2.6$):

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