A particle with mass m and initial speed `V_(0)` is a subject to a velocity-dependent damping force of the form `bV^(n)`.With dimensional analysis determine how the stopping time depends on m, `V_(0)` and b for begin with writing `Deltat=Am^(alpha)b^(beta)V_(0)^(gamma)`, powers `alpha, beta` and `gamma` will be.

Options

A. `alpha=1, beta=-1, gamma=1-n`
B. `alpha=2-n, beta=-1, gamma=2`
C. `alpha=1, beta=1, gamma=1-n`
D. `alpha=-1, beta=-1, gamma=1-n`

Answer

A

Explanation

`F=bv^(n)`
`b=F/v^(n)=[(M^(1)L^(1)T^(-2))/((L^(1)T^(-1)))]^(beta)`
`Deltat=A m^(alpha) b^(beta) v_(0)^(gamma)`
`M^(0) L^(0)T^(1)=A M^(a+b)xxL^(beta-n)T^(-2beta+n)xxL^(gamma)T^(gamma)`
By comparison of power of MLT
`alpha+beta=0`
`beta-n+gamma=0 rArr beta+gamma=n`
`-2beta+n-gamma=1 rArr +2beta+gamma=n-1`
`beta+n=n-1`
`beta=-1`
`alpha=1`
`gamma=1+n`