Question

A force is given by F= at +bt2 where 't is the time. The dimensions of a and b are
a) [MLT-4)] and [MLT1]
b) [MLT-1] and [MLT0]
c) [MLT-3] and [MLT-4]
d) [MLT-3] and [MLT0]​

Answer 1

Answer:

option b is correct by using law of homoginity we can easily solve this problem.

option b is correct by using law of homoginity we can easily solve this problem. Explanation:

option b is correct by using law of homoginity we can easily solve this problem. Explanation:F=at=bt*2

option b is correct by using law of homoginity we can easily solve this problem. Explanation:F=at=bt*2[MLT*-2]=a[T]

option b is correct by using law of homoginity we can easily solve this problem. Explanation:F=at=bt*2[MLT*-2]=a[T]a=[MLT*-2]/[T*1]

option b is correct by using law of homoginity we can easily solve this problem. Explanation:F=at=bt*2[MLT*-2]=a[T]a=[MLT*-2]/[T*1]a=[MLT*-1]

option b is correct by using law of homoginity we can easily solve this problem. Explanation:F=at=bt*2[MLT*-2]=a[T]a=[MLT*-2]/[T*1]a=[MLT*-1]b=[MLT*-2/][T*2]

option b is correct by using law of homoginity we can easily solve this problem. Explanation:F=at=bt*2[MLT*-2]=a[T]a=[MLT*-2]/[T*1]a=[MLT*-1]b=[MLT*-2/][T*2]b=[MLT*0]

option b is correct by using law of homoginity we can easily solve this problem. Explanation:F=at=bt*2[MLT*-2]=a[T]a=[MLT*-2]/[T*1]a=[MLT*-1]b=[MLT*-2/][T*2]b=[MLT*0]hence option

option b is correct by using law of homoginity we can easily solve this problem. Explanation:F=at=bt*2[MLT*-2]=a[T]a=[MLT*-2]/[T*1]a=[MLT*-1]b=[MLT*-2/][T*2]b=[MLT*0]hence optionb) is correct

option b is correct by using law of homoginity we can easily solve this problem. Explanation:F=at=bt*2[MLT*-2]=a[T]a=[MLT*-2]/[T*1]a=[MLT*-1]b=[MLT*-2/][T*2]b=[MLT*0]hence optionb) is correcthope it is correct

Answer 2

By Principal of Homogenety:-

[F] = [at] = [bt²]

[F] = [at]

$[ML{T}^{ - 2} ] = [a ] [T]$

$[a ] = \frac{ [ML{T}^{ - 2} ]}{[T]}$

$\red{[a ] = { [ML{T}^{ - 3} ]}}$

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[F] = [bt²]

$[ML{T}^{ - 2} ] = [b ] [T ^{2} ]$

$[b ] = \frac{ [ML{T}^{ - 2} ]}{[T^{2} ]}$

$\red{[b ] = [ML{T}^{ - 4} ]}$

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c) [MLT-3] and [MLT-4]