2.Motion in Straight Line
medium

An alpha particle enters a hollow tube of $4 \,m$ length with an initial speed of $1 \,km/s$. It is accelerated in the tube and comes out of it with a speed of $9 km/s$. The time for which it remains inside the tube is

A

$8 \times {10^{ - 3}}$s

B

$80 \times {10^{ - 3}}s$

C

$800 \times {10^{ - 3}}s$

D

$8 \times {10^{ - 4}}s$

Solution

(d) ${v^2} = {u^2} + 2as \Rightarrow {(9000)^2} – {(1000)^2}$$ = 2 \times a \times 4$

$ \Rightarrow a = {10^7}m/{s^2}$  Now $t = \frac{{v – u}}{a}$

$ \Rightarrow \;t = \frac{{9000 – 1000}}{{{{10}^7}}} = 8 \times {10^{ – 4}}\;sec$

Standard 11
Physics

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