U = A×v^3/v+B
[ML^2T^-2]=A×[LT^-1]^3/[LT^-1]+B
According to the principle of homogeneity, similar quantities can be added or subtracted.
Therefore, Dimensions of B are same as Dimensions of velocity
B=[LT^-1]
Also if we shift the given dimensions then
A= U× (B+v)/ v^3
A= [ML^2T^-2]× [LT^-1] / [LT^-1]^3
A= [ML^3T^-2] / [L^3T^-3]
A=[M]
THUS,
A/B = M/ LT^-1
A/B = [ML^-1T]
the potential energy of a particle varies with velocity v as U equals to Av3/v + B where A and B are constant dimensional formula A/B is
(1) ML-1T
(2) M-1LT
(3) L-1T-1
(4) MLT-1
2 answers
U = A×v^3/v+B
[ML^2T^-2]=A×[LT^-1]^3/[LT^-1]+B
According to the principle of homogeneity, similar quantities can be added or subtracted.
Therefore, Dimensions of B are same as Dimensions of velocity
B=[LT^-1]
Also if we shift the given dimensions then
A= U× (B+v)/ v^3
A= [ML^2T^-2]× [LT^-1] / [LT^-1]^3
A= [ML^3T^-2] / [L^3T^-3]
A=[M]
THUS,
A/B = M/ LT^-1
A/B = [ML^-1T]
[ML^2T^-2]=A×[LT^-1]^3/[LT^-1]+B
According to the principle of homogeneity, similar quantities can be added or subtracted.
Therefore, Dimensions of B are same as Dimensions of velocity
B=[LT^-1]
Also if we shift the given dimensions then
A= U× (B+v)/ v^3
A= [ML^2T^-2]× [LT^-1] / [LT^-1]^3
A= [ML^3T^-2] / [L^3T^-3]
A=[M]
THUS,
A/B = M/ LT^-1
A/B = [ML^-1T]
0 answers