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

To find the relationships for distance \(L\) and time \(T\) in the two systems of units, let's analyze the given velocity and acceleration relations:

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

According to the basic kinematic equation, we have the relation: \(v = \frac{L}{T}\)

From the velocity relation, we can write: \(\frac{L_2}{T_2} = \frac{n}{m^2} \cdot \frac{L_1}{T_1}\)

From this, we can derive: \( L_2 = \frac{n}{m^2} \cdot \frac{L_1 T_2}{T_1} \)

Using the acceleration relation \(a = \frac{L}{T^2}\), substitute for acceleration:

\(\frac{L_2}{T_2^2} = \frac{1}{mn} \cdot \frac{L_1}{T_1^2}\)

This results in: \( L_2 = \frac{1}{mn} \cdot \frac{L_1 T_2^2}{T_1^2} \)

Comparing the two forms for \(L_2\):

\(\frac{n}{m^2} \cdot \frac{L_1 T_2}{T_1} = \frac{1}{mn} \cdot \frac{L_1 T_2^2}{T_1^2}\)

Cancel out the common terms (\(\frac{L_1}{T_1}\)) and rearrange:

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

Simplifying gives: \(T_2 = mn \cdot \frac{T_1}{n}\)

Thus,

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

For distance relationship, equate the final form:

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

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

Option 1: \(\frac{n^3}{m^3} L_1 = L_2\) and \(\frac{n^2}{m} T_1 = T_2\)