Physics

# If $\mu =A+\dfrac { B }{ \lambda } +\dfrac { C }{ { \lambda }^{ 2 } }$ is dimensionally correct then the dimensions of A, B and C respectively are:($\mu$, A, B, C are constants)

No dimensions, $L, { L }^{ 2 }$

##### SOLUTION

$\underset{M^{0}L^{0}T^{0}}{\mu}=A+\dfrac{B}{\lambda }+\dfrac{C}{\lambda^{2}}$

$A = No \ dimension = M^{0}L^{0}T^{0}$

$\mu=\dfrac{B}{\lambda}\Rightarrow B=\mu \lambda =M^{0}L^{1}T^{0}$
Similarly, $C=M^{0}L^{2}T^{0}$

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Single Correct Medium Published on 18th 08, 2020
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