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A force is given by \(F = at + b{t^2}\), where \(t\) is time, the dimensions of \(a\) and \(b\) are respectively :( )
A. \(\left[ {ML{T^{ - 4}}} \right]\) and \(\left[ {ML{T^{ - 1}}} \right]\)
B. \(\left[ {ML{T^{ - 1}}} \right]\) and \(\left[ {ML{T^0}} \right]\)
C. \(\left[ {ML{T^{ - 3}}} \right]\) and \(\left[ {ML{T^{ - 4}}} \right]\)
D. \(\left[ {ML{T^{ - 3}}} \right]\) and \(\left[ {ML{T^0}} \right]\)
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A force is given by \(F = at + b{t^2}\), where \(t\) is time, the dimensions of \(a\) and \(b\) are respectively :( )
A. \(\left[ {ML{T^{ - 4}}} \right]\) and \(\left[ {ML{T^{ - 1}}} \right]\)
B. \(\left[ {ML{T^{ - 1}}} \right]\) and \(\left[ {ML{T^0}} \right]\)
C. \(\left[ {ML{T^{ - 3}}} \right]\) and \(\left[ {ML{T^{ - 4}}} \right]\)
D. \(\left[ {ML{T^{ - 3}}} \right]\) and \(\left[ {ML{T^0}} \right]\)

Answer

C
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Solution

Since \(F=at+bt^2---(1)\)
\(\therefore \) Dimension of \( at\) and \(bt^2\) must be equal to force only.
Hence \([F]=[M^{1} L^{1} T^{-2}]---(2)\)
 from \((1)\) and \((2)\)
\(\Rightarrow \ [at]=a[T] =[F]\)
\(\therefore \ a[T] =[M^{1} L^{1} T^{-2}]\)
\(\therefore \ [a]=[M^{1} L^{1} T^{-3}]\)
\(bt^2=[b][T^{2}]=[a][t]=[M^{1} L^{1} T^{-2}]\)
\(\therefore \ \boxed {[b]= [M^{1} L^{1} T^{-4}]}\)
Hence, option \((C)\) is correct.
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