A) The relation is dimensionally correct
B) The relation is dimensionally incorrect
C) The relation may be dimensionally correct
D) None of the above
Correct Answer: A
Solution :
[a] LHS=\[[{{x}_{1}}+{{x}_{2}}]=[{{x}_{1}}]=[{{x}_{2}}]=[{{M}^{0}}L{{T}^{0}}]\] RHS= \[\left[ \frac{{{a}_{1}}{{a}_{2}}{{t}^{2}}}{2({{a}_{1}}+2a)} \right]=\frac{[L{{T}^{-2}}][L{{T}^{-2}}][{{T}^{2}}]}{[L{{T}^{-2}}]}=[{{M}^{0}}L{{T}^{0}}]\] \[\because \] LHS=RHS According to homogeneity principle, equation is dimensionally correct. Hence, option [a] is correct.You need to login to perform this action.
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