c
No. of active nuclei present in original sample \(=4 \times 10^{16}\)

Half-life of element \(=10\) days.

No. of half-life in 30 days \(=\frac{30}{10}=3\)

No. of nuclei present after 30 days \(=\frac{4 \times 10^{16}}{(2)^{3}} \Rightarrow 0.5 \times 10^{16}\)

No. of decayed nuclei = (No. of active nuclei present originally) - (No. of Active nuclei present after \(30\) days)

\(=4 \times 10^{16}-0.5 \times 10^{16}\)

\(3.5 \times 10^{16}\)