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Velocities (V) and accelerations (a) in two systems of units 1 and 2 are related as $V_2=\frac{n}{m^2} V_1$ and $a_2=\frac{a_1}{m n}$ respectively. Here $\mathrm{m}$ and $\mathrm{n}$ are constants. Dimensionally relations between distances $\left(\mathrm{S}_1\right.$ and $\left.\mathrm{S}_2\right)$ and times $\left(\mathrm{t}_1\right.$ and $t_2$ ) in the two systems are respectively.
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The correct answer is:
$S_2=\left(\frac{n}{m}\right)^3 S_1$ and $t_2=\frac{n^2}{m} t_1$
Given, $v_2=\frac{n}{m^2} v_1$ and $a_2=\frac{a_1}{m n}$
$\begin{aligned}
& \frac{v_2}{v_1}=\frac{n}{m^2} \quad \frac{a_2}{a_1}=\frac{1}{m n} \\
& a=\frac{v}{t} \\
& \frac{a_2}{a_1}=\frac{v_2}{t_2} \times \frac{t_1}{v_1} \Rightarrow \frac{1}{m n}=\frac{n}{m^2} \times \frac{t_1}{t_2} \\
& t_2=\frac{n^2}{m} t_1 \\
& \frac{v_2}{v_1}=\frac{S_2}{t_2} \times \frac{t_1}{S_1} \Rightarrow \frac{n}{m^2}=\frac{S_2}{S_1} \times \frac{m}{n^2} \\
& \frac{S_2}{S_1}=\frac{n^3}{m^3} \Rightarrow S_2=\left(\frac{n}{m}\right)^3 S_1
\end{aligned}$
$\begin{aligned}
& \frac{v_2}{v_1}=\frac{n}{m^2} \quad \frac{a_2}{a_1}=\frac{1}{m n} \\
& a=\frac{v}{t} \\
& \frac{a_2}{a_1}=\frac{v_2}{t_2} \times \frac{t_1}{v_1} \Rightarrow \frac{1}{m n}=\frac{n}{m^2} \times \frac{t_1}{t_2} \\
& t_2=\frac{n^2}{m} t_1 \\
& \frac{v_2}{v_1}=\frac{S_2}{t_2} \times \frac{t_1}{S_1} \Rightarrow \frac{n}{m^2}=\frac{S_2}{S_1} \times \frac{m}{n^2} \\
& \frac{S_2}{S_1}=\frac{n^3}{m^3} \Rightarrow S_2=\left(\frac{n}{m}\right)^3 S_1
\end{aligned}$
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