Question
If velocity v, acceleration A and force B are chosen as fundamental quantities, then the dimensional formula of angular momentum in terms of v, A and F would be
Answer: Option B
:
B
L ∝vxAyFz⇒ L = kvxAyFz
Putting the dimenstions in the above relation
[ML2T−1]=k[LT−1]x[LT−2]y[MLT−2]z
⇒[ML2T−1]=k[MzLx+y+zT−x−2y−2z]
Comparing the pwers of M,LandT
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,Aandf
[L] = [Fv3A−2]
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:
B
L ∝vxAyFz⇒ L = kvxAyFz
Putting the dimenstions in the above relation
[ML2T−1]=k[LT−1]x[LT−2]y[MLT−2]z
⇒[ML2T−1]=k[MzLx+y+zT−x−2y−2z]
Comparing the pwers of M,LandT
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,Aandf
[L] = [Fv3A−2]
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