MHT CET · Maths · Differential Equations
A metal has half life period of 10 days, \(A\) sample originally has a mass of \(1000 \mathrm{mg}\),
then the mass remaining after 50 days is
- A \(\frac{225}{8} m g\)
- B \(\frac{125}{8} \mathrm{mg}\)
- C \(\frac{125}{4} \mathrm{mg}\)
- D \(\frac{225}{4} m g\)
Answer & Solution
Correct Answer
(C) \(\frac{125}{4} \mathrm{mg}\)
Step-by-step Solution
Detailed explanation
A mass has half life period of 10 days. It means every ten days, mass remaining is half of the mass before 10 days.
Initial mass \(=1000 \mathrm{mg}\).
\(\therefore\) Mass after, 10 days \(=\frac{1}{2} \times 1000=500 \mathrm{mg}\)
Mass after 20 days \(=\frac{1}{2} \times 500=250 \mathrm{mg}\)
Mass after 30 days \(=\frac{1}{2} \times 250=125 \mathrm{mg}\)
Mass after 40 days \(=\frac{1}{2} \times 125=\frac{125}{2} \mathrm{mg}\)
Mass after 50 days \(=\frac{1}{2} \times \frac{125}{2}=\frac{125}{4} \mathrm{mg}\)
Initial mass \(=1000 \mathrm{mg}\).
\(\therefore\) Mass after, 10 days \(=\frac{1}{2} \times 1000=500 \mathrm{mg}\)
Mass after 20 days \(=\frac{1}{2} \times 500=250 \mathrm{mg}\)
Mass after 30 days \(=\frac{1}{2} \times 250=125 \mathrm{mg}\)
Mass after 40 days \(=\frac{1}{2} \times 125=\frac{125}{2} \mathrm{mg}\)
Mass after 50 days \(=\frac{1}{2} \times \frac{125}{2}=\frac{125}{4} \mathrm{mg}\)
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