Question

Answer the question in the given attachment.

Answer 1

$[L] \: = {[c]}^{a} \: {[G]}^{b} \: {[ \frac{ {e}^{2} }{4\pi.E _{0} } ]}^{c} \: \: \: \: \: \: \: .......Eq _{1} \\ \\ [L] = {[L {T}^{ - 1}] }^{a} \: {[{M}^{ - 1} {L}^{3} {T}^{ - 2}] }^{b} {[M {L}^{3} {T}^{ - 2} ]}^{c} \\ \\ [L] = {L}^{(a + 3b + 3c)} . {M}^{( - b + c)} . {T}^{ (- a - 2b - 2c)} \\ \\ NOW \\ \\ = > a + 3b + 3c = 1 \\ = > - b + c = 0 \\ = > - a - 2b - 2c = 0 \\ \\ ON \: \: SOLVING \\ \\ a = - 2 \\ b = \frac{1}{2} \\ c = \frac{1}{2} \\ \\ Plug \: The \: Value \: of \: a \: b \: and \: c \: in \: Eq _{1} \\ \\ = > [L]= {[c]}^{ - 2} {[G]}^{ \frac{1}{2} } {[ \frac{ {e}^{2} }{4\pi.E _{0} }] }^{ \frac{1}{2} } \\ \\ = > L \: = \frac{1}{ {c}^{2} } {[G. \frac{ {e}^{2} }{4\pi.E _{0}} ]}^{ \frac{1}{2} } \: \: \: \: \: \: \: ......ANSWER$

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$BE \: \: BRAINLY$