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Under the action of a force, a `2 kg` body moves such that its position x as a function of time is given by `x =(t^(3))/(3)` where x is in metre and t in second. The work done by the force in the first two seconds is .A. 1600 jouleB. 160 jouleC. 16 jouleD. 1.6 joule |
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Answer» Correct Answer - C |
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