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If velocity v acceleration A and force F are chosen as fundamental quantities, then the dimensional formula of angular momentum is terms of v,A and F would beA. `FA^(1)v`B. `Fv^(3)A^(-2)`C. `Fv^(2)A^(-1)`D. `F^(2)v^(2)A^(-1)` |
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Answer» Correct Answer - B `Lpropv^xA^yF^zimpliesLkv^xA^yF^z` putting the dimensions in the above relation `[ML^2T^-1]=k[LT^-1]^x[LT^-2]^y[MLT^-2]^z` `implies[ML^2T^-1]=k[M^zL^(x+y+z)T^(-x-2y-2z)]` Comparing the powers of M,L and T `z=1` ..(i) `x+y+z=2` ..(ii) `-x-2y-2z=-1` ..(iii) On solving (i),(ii)and (iii) `x=3`,`y=-2`,`z=1` So dimension of L in terms of v, A and f `[L]=[Fv^3A^-2]` |
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