A) \[L{{T}^{-1}},L,T\]
B) \[{{T}^{-1}},L{{T}^{-1}},{{T}^{-1}}\]
C) \[L,L{{T}^{-1}},T\]
D) \[T,T,T\]
Correct Answer: B
Solution :
\[2k=2a\,({{t}_{1}}+{{t}_{2}})\] \[{{t}_{1}}{{t}_{2}}=-1\] \[{{k}^{2}}={{a}^{2}}\left[ \frac{2h}{a}-2 \right]\] gives dimensionless values \[{{y}^{2}}=2a(x-2a)\] \[a=2\] gives dimensionless values \[{{y}^{2}}=4(x-4)\] \[\left| \begin{matrix} 2-1 & 3-4 & 4-5 \\ 1 & 1 & -k \\ k & 2 & 1 \\ \end{matrix} \right|=\,\left| \begin{matrix} 1 & -1 & -1 \\ 1 & 1 & -k \\ k & 2 & 1 \\ \end{matrix} \right|\] powers are dimensionless \[=\,\left| \begin{matrix} 1 & 0 & 0 \\ 1 & 2 & 1-k \\ k & k+2 & 1+k \\ \end{matrix} \right|\]
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