Velocity (v) and acceleration (a) in two syst | vedclass

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(B) Given: $v_{2} = \frac{n}{m^{2}} v_{1}$ and $a_{2} = \frac{a_{1}}{mn}$.Since $v = \frac{L}{T}$ and $a = \frac{L}{T^{2}}$,we have:$\frac{L_{2}}{T_{2

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$L_{1} = \frac{n^{4}}{m^{2}} L_{2}$ and $T_{1} = \frac{n^{2}}{m} T_{2}$

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(B) Given: $v_{2} = \frac{n}{m^{2}} v_{1}$ and $a_{2} = \frac{a_{1}}{mn}$.Since $v = \frac{L}{T}$ and $a = \frac{L}{T^{2}}$,we have:$\frac{L_{2}}{T_{2}} = \frac{n}{m^{2}} \frac{L_{1}}{T_{1}}$  --- $(1)$$\frac{L_{2}}{T_{2}^{2}} = \frac{1}{mn} \frac{L_{1}}{T_{1}^{2}}$ --- $(2)$Dividing $(1)$ by $(2)$:$\frac{T_{1}^{2}}{T_{2}} = \frac{n}{m^{2}} 	imes mn 	imes T_{1} = \frac{n^{2}}{m} T_{1}$$\frac{T_{2}}{T_{1}} = \frac{m}{n^{2}}$ (This implies $T_{1} = \frac{n^{2}}{m} T_{2}$).Substituting $T_{2} = \frac{m}{n^{2}} T_{1}$ into $(1)$:$\frac{L_{2}}{(\frac{m}{n^{2}}) T_{1}} = \frac{n}{m^{2}} \frac{L_{1}}{T_{1}}$$L_{2} = \frac{n}{m^{2}} 	imes \frac{m}{n^{2}} L_{1} = \frac{1}{mn} L_{1}$$L_{1} = mn L_{2}$.Checking the options,the correct relation for time is $T_{1} = \frac{n^{2}}{m} T_{2}$.

List-$I$	List-$II$
$A$. Meter $(L)$	$I$. $\sqrt{\frac{hc}{G}}$
$B$. Second $(S)$	$II$. $\sqrt{\frac{Gh}{c^5}}$
$C$. Kilogram $(M)$	$III$. $\sqrt{\frac{L^2c^3}{Gh}}$
$D$. Kelvin $(K)$	$IV$. $\sqrt{\frac{Gh}{c^3}}$

Velocity $(v)$ and acceleration $(a)$ in two systems of units $1$ and $2$ are related as $v_{2} = \frac{n}{m^{2}} v_{1}$ and $a_{2} = \frac{a_{1}}{mn}$ respectively. Here $m$ and $n$ are constants. The relations for distance $(L)$ and time $(T)$ in the two systems are:

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