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If F=ax+bt^(2)+c where F is force, x is ...

If `F=ax+bt^(2)+c` where F is force, x is distance and t is time. Then what is dimension of `(axc)/(bt^(2))` ?

A

`[ML^(2)T^(-2)]`

B

`[MLT^(-2)]`

C

`[M^(0)L^(0)T^(0)]`

D

`[MLT^(-1)]`

Text Solution

Verified by Experts

The correct Answer is:
B
`[(axc)/(bt^(2))]=(MLT^(-2)xxMLT^(-2))/(MLT^(-2))=MLT^(-2)`
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