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(A) The half-life of the substance is given as $t_{1/2} = T$.If the original mass is $N_0$,then after disintegrating $\frac{7}{8}$th of its mass,the r

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$3T$

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(A) The half-life of the substance is given as $t_{1/2} = T$.If the original mass is $N_0$,then after disintegrating $\frac{7}{8}$th of its mass,the remaining mass $N$ is:$N = N_0 - \frac{7}{8}N_0 = \frac{1}{8}N_0$.We know that the remaining mass after $n$ half-lives is given by $N = N_0 \left(\frac{1}{2}ight)^n$.Substituting the values: $\frac{1}{8}N_0 = N_0 \left(\frac{1}{2}ight)^n$.$\left(\frac{1}{2}ight)^3 = \left(\frac{1}{2}ight)^n$,which implies $n = 3$.The total time taken is $t = n 	imes T = 3T$.

The half-life of a radioactive substance is $T$. The time taken for disintegrating $\frac{7}{8}$th part of its original mass will be

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