InterviewSolution
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InterviewSolution
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Let I be the purchase value of an equipment and V(t) be the value after ithas been used for t years. The value V(t) depreciates at a rate given bydifferential equation `(d V(t)/(dt)=-k(T-t)`, where `k"">""0`is a constant and T is thetotal life in years of the equipment. Then the scrap value V(T) of theequipment is :(1) `T^2-1/k`(2) `I-(k T^2)/2`(3) `I-(k(T-t)^2)/2`(4) `e^(-k T)`A. `e^(-kT)`B. `T^(2)-I/k`C. `I-(kT^(2))/2`D. `I-(k(T-t)^(2))/(2)` |
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Answer» Correct Answer - C Since total life is T, scrap value is V(T). We have `(dV)(t)=-k(T-t)dt` `rArr int_(I)^(V(t))dV(t) = int_(t=0)^(T)-k(T-t)dt` `rArr V(T)-I = k[((T-t)^(2))/(2)]_(0)^(T)` `rArr V(T)-I = -k[t^(2)/2]` `rArr V(T)=I-(kT^(2))/2` |
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