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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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