1.

The fraction of a radioactive material which reamins active after time t is `9//16`. The fraction which remains active after time `t//2` will be .A. `(4)/(5)`B. `(7)/(8)`C. `(3)/(5)`D. `(3)/(4)`

Answer» Correct Answer - d
`(9)/(16)=((1)/(2))^((t)/(T))`
`(N)/(N_(0))=((1)/(2))^((t)/(2T))`
`((N)/(N_(0)))^(2)=((1)/(2))^(t//T)`
or `(N)/(N_(0)^(2))=(9)/(16)` or `(N)/(N_(0)) =(3)/(4)`
Note the special techaniq used in the problem.


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