Answer

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**Hint:**Due to free fall, acceleration acting on the body $ a $ is acceleration due to gravity $ g $ . Initial velocity is zero, since the object falls from rest. Half-life is the time required for a quantity to reduce to half of its initial value.

Formula Used: The formulae used in the solution are given here.

$ S = ut + \dfrac{1}{2}a{t^2} $ where $ S $ is the distance, $ u $ is the initial velocity, $ a $ is the acceleration and $ t $ is the time taken.

$ N\left( t \right) = {N_0}{e^{ - \lambda t}} $ where $ {N_0} $ is the initial quantity, $ N\left( t \right) $ is the quantity after time $ t $ , and $ \lambda $ is the decay constant.

**Complete Step by Step Solution**

It has been given that a radioisotope X has a half-life of $ 10s $ . Find the number of nuclei in the sample (if initially there are 1000 isotopes which are following from rest from a height of 3000m) when it is at a height of 1000m from the reference plane.

By Newton’s law of motion, we already know that, $ S = ut + \dfrac{1}{2}a{t^2} $ .

The distance covered is equal to the height. Thus, height $ = S = 2000m. $

Due to free fall, acceleration acting on the body $ a $ is acceleration due to gravity $ g $ .

Thus, $ a = g. $

Initial velocity is zero, since the object falls from rest.

Thus, $ S = \dfrac{1}{2}a{t^2} $ .

Now, $ a = g $ ,

$ 2000 = \dfrac{1}{2} \times 10 \times {t^2} $

$ \Rightarrow t = \sqrt {\dfrac{{2000}}{5}} = 20s $

Thus time taken in falling a height is 20 seconds.

Given that the half-life is 10 seconds. Half-life is the time required for a quantity to reduce to half of its initial value.

$ {t_{1/2}} = 10s $ .

We know that, $ \lambda = \dfrac{{\ln 2}}{{10}} $ .

Again, $ N\left( t \right) = {N_0}{e^{ - \lambda t}} $ where $ {N_0} $ is the initial quantity, $ N\left( t \right) $ is the quantity after time $ t $ , and $ \lambda $ is the decay constant.

At $ t = 20s $ ,

$ N = \dfrac{{{N_0}}}{4} = 250 $ .

Thus, the number of nuclei in the sample is $ 250 $ .

**The correct answer is Option B.**

**Note**

It has been given that, height $ {\text{h = 3000 - 1000 = 2000m}} $

Time taken in falling a height is given as $ t = \sqrt {\dfrac{{2h}}{g}} $ .

Assigning the values, $ g = 10m{s^{ - 2}} $ , $ h = 2000m $ ,

$ t = \sqrt {\dfrac{{2 \times 2000}}{{10}}} = 20s. $

Thus, time taken is 20 seconds.

Number of half-life in this time period is $ n = \dfrac{{20}}{{10}} = 2 $ .

So the number of active nuclei= $ {{initial} \mathord{\left/

{\vphantom {{initial} {{2^n} = {{initial} \mathord{\left/

{\vphantom {{initial} {{2^2}}}} \right.

} {{2^2}}}}}} \right.

} {{2^n} = {{initial} \mathord{\left/

{\vphantom {{initial} {{2^2}}}} \right.

} {{2^2}}}}} $

$ = {{initial} \mathord{\left/

{\vphantom {{initial} {4.}}} \right.

} {4.}} $

Initially there are 1000 isotopes. The number of active nuclei are $ {{1000} \mathord{\left/

{\vphantom {{1000} 4}} \right.

} 4} = 250 $ .

Option B is correct.

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