Question

Initial speed of an alpha particle side a tube a length 4 m is 1 km/s, if it is accelerated in the tube and comes out with a speed of 9 km/s, then the time for which the particle remains inside the tube is
A. $ 8 \times {10^{ - 3}}s $
B. $ 8 \times {10^{ - 4}}s $
C. $ 80 \times {10^{ - 3}}s $
D. $ 800 \times {10^{ - 3}}s $

Answer 1

Hint: In order to find the time taken by the alpha particle we have to find the acceleration of the particle. We know the initial velocity $ u $ ,final velocity $ v $ and the displacement $ s $ .Using the equations of motion we can find the acceleration and thus time needed.
Formula used
 $ v = u + at $ , where $ v $ is the final velocity, $ u $ is the initial velocity, $ t $ is the time taken, $ a $ is the acceleration.
 $ {v^2} = {u^2} + 2as $ , here $ v $ is the final velocity, $ u $ is the initial velocity, $ s $ is the displacement and $ a $ is the acceleration.

Complete step-by-step answer:
It is given that the initial speed of the alpha particle is $ 1km/s $ .The length of the tube is the total displacement of the particle. Therefore $ s = 4m $ .It is also mentioned in the question that the final velocity when it comes out is $ 9km/s $ .Using these values in the equation $ {v^2} = {u^2} + 2as $ .We get
 $ {(9000)^2} = {(1000)^2} + 2 \times a \times 4 $
 $ a = \dfrac{{8.1 \times {{10}^7} - 0.1 \times {{10}^7}}}{8} $
 $ a = 1 \times {10^7}m/{s^2} $
Using this value of acceleration in the equation $ v = u + at $
 $ 9000 = 1000 + (1 \times {10^7}) \times t $
 $ t = \dfrac{{8000}}{{{{10}^7}}}s $
 $ t = 8 \times {10^{ - 4}}s $

The correct option is B

Note: The average of total distance travelled with initial velocity $ u $ and final velocity $ v $ in time $ t $ also gives us the displacement. This can be represented as $ s = \dfrac{{ut + vt}}{2} $ .Here $ {ut} $ shows the displacement with velocity $ u $ in time $ t $ and $ {vt} $ shows the displacement with velocity $ v $ in time $ t $ .Substituting the values given in question, we get
 $ 4 = \dfrac{{1000t + 9000t}}{2} $
 $ t = \dfrac{8}{{10000}}s $
 $ t = 8 \times {10^{ - 4}}s $