Question:
An ice ball melts. The rate at which it melts is proportional to the amount of the ice at that time. If half quantity of ice melts in \(20\) minutes, then the time when \((1/8)^{th}\) part of original amount of the ice is left is:
An ice ball melts. The rate at which it melts is proportional to the amount of the ice at that time. If half quantity of ice melts in \(20\) minutes, then the time when \((1/8)^{th}\) part of original amount of the ice is left is:
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Set up dN/dt = -kN, find k from the half-life of 20 minutes, then find the time when N is one eighth of the original.
Updated On: Nov 12, 2026
- \(60\ \text{minutes}\)
- \(30\ \text{minutes}\)
- \(20\ \text{minutes}\)
- \(40\ \text{minutes}\)
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The Correct Option is A
Solution and Explanation
Step 1: Form the equation
Rate of melting is proportional to the amount: \(\frac{dN}{dt}=-kN\), so \(N=N_{0}e^{-kt}\).
Step 2: Use the half-quantity condition
At \(t=20\), \(N=\frac{N_{0}}{2}\), so \(e^{-20k}=\frac{1}{2}\).
Step 3: Find the time for one eighth
\(e^{-kt}=\frac{1}{8}=\left(\frac{1}{2}\right)^{3}=e^{-60k}\).
Step 4: Result
\(t=60\) minutes.
Final Answer:
One eighth of the ice is left after 60 minutes. \[ \boxed{\text{(A) }60\ \text{minutes}} \]
Rate of melting is proportional to the amount: \(\frac{dN}{dt}=-kN\), so \(N=N_{0}e^{-kt}\).
Step 2: Use the half-quantity condition
At \(t=20\), \(N=\frac{N_{0}}{2}\), so \(e^{-20k}=\frac{1}{2}\).
Step 3: Find the time for one eighth
\(e^{-kt}=\frac{1}{8}=\left(\frac{1}{2}\right)^{3}=e^{-60k}\).
Step 4: Result
\(t=60\) minutes.
Final Answer:
One eighth of the ice is left after 60 minutes. \[ \boxed{\text{(A) }60\ \text{minutes}} \]
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