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 \(a_2\)= \(\frac {a1}{mn}\) 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 \(\frac{n2}{m}\)\(T_1\)=\(T_2\)
• B. \(L_1\)=n⁴/m²\(L_2\) and \(T_1\)=\(\frac{n^2}{m}\)T₂
• C. \(L_1\)=\(\frac{n^2}{m}\)\(L_2\) and \(T_1\)=\(\frac{n^4}{m_2}\)T₂
• 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

[L]=\(\frac{[v^2]}{[a]}\)

So, [v_²]²[a_²]=\(\frac{[\frac{n}{m^2}v_1]^2}{[\frac{a_1}{mn}]}\)

[v₂]²[a₂]=\(\frac{n^3}{m^3}\)\(\frac{[v_1]^2}{[a_1]}\) or [L₂]=\(\frac{n^3}{m^3}\)[L₁]

Similarly, [T]=\(\frac{[v]}{[a]}\)

So, [T₂]=\(\frac{n^2}{m}\)[T₁]

\(\therefore\) The correct option is (A): \(\frac{n^3}{m^3}\)\(L_1\)=\(L_2\) and \(\frac{n2}{m}\)\(T_1\)=\(T_2\)