if F=ax^1/2+bt^2, where F is force ,x is distance and t is time the dimension of a/b is
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Given,
F=ax^1/2+bt^2
According to law of homogeneity,
Units of F=unit of a×(unit of x)^1/2
=Unit of b× (unit of t)^2
Since,F=m×a=[MLT^(-2)]
MLT^(-2)=[a][L]^1/2
[a]=[ML^(1/2)T^(-2)]-----(1)
[MLT^(-2)]=[b][T^(2)]
[b]=[MLT^(-4)]--------(2)
From (1) and (2)
[a]÷[b]=[M^(0)L^(-1/2)T^(2)]
F=ax^1/2+bt^2
According to law of homogeneity,
Units of F=unit of a×(unit of x)^1/2
=Unit of b× (unit of t)^2
Since,F=m×a=[MLT^(-2)]
MLT^(-2)=[a][L]^1/2
[a]=[ML^(1/2)T^(-2)]-----(1)
[MLT^(-2)]=[b][T^(2)]
[b]=[MLT^(-4)]--------(2)
From (1) and (2)
[a]÷[b]=[M^(0)L^(-1/2)T^(2)]
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