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A mixture consists of two radioactive materials A₁ and A₂ with half lives of 20s and 10s respectively. Initially the mixture has 40 g of A₁ and 160 g of A₂. The amount of the two in the mixture will become equal after:
Options
(a) 60 s
(b) 80 s
(c) 20 s
(d) 40 s
Correct Answer:
40 s
Explanation:
Let, the amount of the two in the mixture will become equal after t years.
The amount of A₁, which remains after t years N₁ = N₀₁ / (2)ᵗ/²⁰
The amount of A₂, which remains after t years N₂ = N₀₂ / (2)ᵗ/¹⁰
According to the problem N₁ = N₂
40 / (2)ᵗ/²⁰ = 160 / (2)ᵗ/¹⁰ ⇒ 2ᵗ/²⁰ = 2(ᵗ/¹⁰⁻²)
t/20 = t/10 – 2 ⇒ t/20 – t/10 = 2
t/20 = 2 ⇒ t = 40 s
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- Two bodies of masses m₁ and m₂ and having velocities v₁ and v₂ respectively
- If longitudinal strain for a wire is 0.03 and its poisson’s ratio is 0.5
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Topics: Radioactivity
(83)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The dimensional formula for Young’s modulus is
- In a uniform circular motion,
- Two bodies of masses m₁ and m₂ and having velocities v₁ and v₂ respectively
- If longitudinal strain for a wire is 0.03 and its poisson’s ratio is 0.5
- A electric dipole is placed at an angle of 30⁰ to a non-uniform electric field.
Topics: Radioactivity (83)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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