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If velocity, force and time are taken to be fundamental quantities, then the dimensional formula for mass is
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Verified Answer
The correct answer is:
Q = Kv-1 FT
Let the quantity be Q, then,
Q = f (v, F, T)
Assuming that the function is the product of power functions of v, F and T,
Q = Kvx Fy Tz ...(i)
where K is a dimensionless constant of proportionality. The above equation dimensionally becomes
[Q] = [LT-1]x[MLT-2]y[T]z
i.e., [Q] = [My][Lx+yT-x-2y+z] ...(ii)
Now
Q = mass i.e., [Q] = [M]
So Equation (ii) becomes
[M] = [MyLx+yT-x-2y+z]
its dimensional correctness requires
y = 1, x + y = 0 and - x - 2y + z = 0
which on solving yields
x = -1, y = 1 and z = 1
Substituting it in Equation (i), we get
Q = Kv-1 FT
Q = f (v, F, T)
Assuming that the function is the product of power functions of v, F and T,
Q = Kvx Fy Tz ...(i)
where K is a dimensionless constant of proportionality. The above equation dimensionally becomes
[Q] = [LT-1]x[MLT-2]y[T]z
i.e., [Q] = [My][Lx+yT-x-2y+z] ...(ii)
Now
Q = mass i.e., [Q] = [M]
So Equation (ii) becomes
[M] = [MyLx+yT-x-2y+z]
its dimensional correctness requires
y = 1, x + y = 0 and - x - 2y + z = 0
which on solving yields
x = -1, y = 1 and z = 1
Substituting it in Equation (i), we get
Q = Kv-1 FT
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