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

To solve this problem, we need to understand the relationships between velocity, acceleration, distance, and time in two different systems of units. The given relationships are:

• Velocity: \(v_2 = \frac{n}{m^2}v_1\)
• Acceleration: \(a_2 = \frac{a_1}{mn}\)

Let's use the basic kinematic equations for motion to derive the relationships for distance (\(L\)) and time (\(T\)) in these systems.

Deriving the Relationship for Distance

Consider the equation of motion for constant acceleration:

\(L = vt + \frac{1}{2}at^2\)

For the two systems, we have:

\(L_2 = v_2 \cdot T_2 + \frac{1}{2}a_2 \cdot T_2^2\)

Substitute the given relations \(v_2 = \frac{n}{m^2}v_1\) and \(a_2 = \frac{a_1}{mn}\) into the equation:

\(L_2 = \left(\frac{n}{m^2}v_1\right) \cdot T_2 + \frac{1}{2}\left(\frac{a_1}{mn}\right) \cdot T_2^2\)

To find the relationship between \(L_1\) and \(L_2\), note that the equation maintains homogeneity of dimensions. Thus, equating dimensions:

\(L_2 = \frac{n^3}{m^3}L_1\)

Deriving the Relationship for Time

Now, consider the kinematic equations related to time (\(T\)).

Since \(v = \frac{L}{T}\) and \(a = \frac{v}{T} = \frac{L}{T^2}\), rearrange the relationships for the time:

\(T_2 = \frac{T_1 \cdot mn}{n^2}\cdot\frac{m^2}{n}\)

Simplifying, we find:

\(T_2 = \frac{n^2}{m}T_1\)

Conclusion

The correct relationships for distance and time between the two systems are:

• \(\frac{n^3}{m^3}L_1 = L_2\)
• \(\frac{n^2}{m}T_1 = T_2\)

Thus, the correct answer is:

\(\frac{n^3}{m^3}L_1 = L_2\) and \(\frac{n^2}{m}T_1 = T_2\).