⇦ | ⇨ |

**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

### Related Questions:

- A body of mass 0.25 kg is projected with muzzle velocity 100 m/s from a tank
- A particle moves from position r₁ = 3i + 2j -6k to position r₂ = 14i + 13 j+ 9 k
- The respective speeds of the molecules are 1,2,3,4 and 5 km/s.
- The original temperature of a black body is 727⁰C. The temperature at which this black
- Two planets at mean distances d₁ and d₂ from the sun have their frequencies n₁ and n₂

Topics: Radioactivity
(83)

Subject: Physics
(2479)

### Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score

18000+ students are using NEETLab to improve their score. What about you?

Solve Previous Year MCQs, Mock Tests, Topicwise Practice Tests, Identify Weak Topics, Formula Flash cards and much more is available in NEETLab Android App to improve your NEET score.

Share this page with your friends

## Leave a Reply