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
$4 = \dfrac{{1000t + 9000t}}{2}$
$t = \dfrac{8}{{10000}}s$
$t = 8 \times {10^{ - 4}}s$