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 find the relation for distance \(L\) and time \(T\) between two different systems of units, we will use the given relationships for velocity \(v\) and acceleration \(a\) between the two systems.

The relationship for velocity between the two systems is given by:

\(v_2=\frac{n}{m^2}v_1\) 

The relationship for acceleration between the two systems is given by:

\(a_2=\frac{a_1}{mn}\)

We know that velocity \(v\) is the rate of change of distance \(L\) with respect to time \(T\). Thus, we can write:

\(v = \frac{L}{T}\)

Similarly, acceleration \(a\) is the rate of change of velocity \(v\) with respect to time \(T\). So:

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

For the system of units 1 and 2, the formulas for length \(L\) and time \(T\) can be derived by connecting the given equations for velocity and acceleration.

Equate the expressions of acceleration from both systems using the relation \(a = \frac{L}{T^2}\):

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

Similarly, equate the expressions of velocity:

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

Solving these two equations, we get:

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

Which simplifies to:

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

Thus, the correct relation for distance and time in the two systems is:

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

This matches the given correct option.