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Initial speed of an `alpha` particle inside a tube of length 4m is `1 kms^(-1)`, if it is accelerated in the tube and comes out with a speed of `9 kms^(-1)` , then the time for which the particle remains inside the tube isA. `8xx10^(-3)s`B. `8xx10(-4)s`C. `80xx10^(-3)s`D. `800xx10^(-3)s` |
Answer» Correct Answer - B Initial speed, `u=1 kms^(-1)=1000 ms^(-1)` Final speed, `upsilon=9 kms^(-1)=9000 ms^(-1)` By using the relation, `upsilon^(2)=u^(2)+2as` `(9000)^(2)=(1000)^(2)+2xx a xx4 rArr a=10^(7)ms^(-2)` `therefore` The time for which the particle remains in the tube `upsilon=u+at` `rArr t=(upsilon-u)/(a)=(9000-1000)/(10^(7))=8xx10^(-4)s` |
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