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(D) Given that the substance decays by $10 \%$ in $16 	ext{ hours}$, the amount remaining after $16 	ext{ hours}$ is $90 \%$ or $0.9$ of the initial

Vedclass · Accepted Answer

$72.9$

Vedclass · Answer

(D) Given that the substance decays by $10 \%$ in $16 	ext{ hours}$, the amount remaining after $16 	ext{ hours}$ is $90 \%$ or $0.9$ of the initial amount.Total time given is $2 	ext{ days} = 2 	imes 24 	ext{ hours} = 48 	ext{ hours}$.The number of $16 	ext{ hour}$ intervals in $48 	ext{ hours}$ is $n = \frac{48}{16} = 3$.The fraction of the substance remaining after $n$ intervals is given by $N = N_0 	imes (0.9)^n$.Substituting the values, $N = N_0 	imes (0.9)^3$.$N = N_0 	imes 0.729$.Therefore, the percentage remaining is $0.729 	imes 100 = 72.9 \%$.Thus, the correct option is $D$.

If a radioactive substance decays $10 \%$ in every $16 \text{ hours}$, then the percentage of the radioactive substance that remains after $2 \text{ days}$ is

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