Question

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 and time in two systems respectively are :

• A. \(\frac{n^3}{m_3}L_1=L_2\) and\(\frac{n^2}{m}T_1=T_2\)
• B. \(L_1=\frac{n^4}{m^2}L_2\) and \(T_1=\frac{n^2}{m}T_2\)
• C. \(L_1=\frac{n^2}{m}L_2\) and \(T_1=\frac{n^4}{m^2}T_2\)
• D. \(\frac{n^2}{m}L_1=L_2 \) and \(\frac{n^4}{m^2}T_1=T_2\)

Answer 1

The Correct Option is A

The correct answer is (A) : \(\frac{n^3}{m^3}L_1=L_2\) and \(\frac{n^2}{m}T_1=T_2\)
\([L]=\frac{[v^2]}{[a]}\)
so
\(\frac{[v_2]^2}{[a_2]}=\frac{[\frac{n}{m^2}v_1]^2}{[\frac{a_1}{mn}]}\)
\(\frac{[v_2]^2}{[a_2]}=\frac{n^3}{m^3} \frac{[v_1]^2}{[a_1]}\)
or \([L_2]=\frac{n^3}{m^3}[L_1]\)
Similarly
\([T]=\frac{[v]}{[a]}\)
So,
\([T_2]=\frac{n^2}{m}[T1]\)