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If the acceleration due to gravity is `10ms^(-2)` and unit of length and time are changed in kilometer and hour respectively the numerical value of the acceleration is
A. `360000`
B. `72000`
C. `36000`
D. `129600`

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Correct Answer - d
`n_(2) = n_(1)[(L_(1))/(L_(2))]^(1) [(T_(1))/(T_(2))]^(-2) = 10[(meter)/(km)]^(1) [(sec)/(br)]^(-2)`
`n_(2) = 10 [(m)/(10^(3)m)]^(1)[(sec)/(3600 sec)]^(-3) = 129600`.

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