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 :
\(\frac{n^3}{m^3}\)\(L_1\)=\(L_2\) and \(\frac{n2}{m}\)\(T_1\)=\(T_2\)
\(L_1\)=n4/m2\(L_2\) and \(T_1\)=\(\frac{n^2}{m}\)T2
\(L_1\)=\(\frac{n^2}{m}\)\(L_2\) and \(T_1\)=\(\frac{n^4}{m_2}\)T2
\(\frac{n^2}{m}\)\(L_1\)=\(L_2\) and \(\frac{n^4}{m^2}\)\(T_1\)=\(T_2\)
The Correct Option is A
Solution and Explanation
[L]=\(\frac{[v^2]}{[a]}\)
So, [v2]2[a2]=\(\frac{[\frac{n}{m^2}v_1]^2}{[\frac{a_1}{mn}]}\)
[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 \(\frac{n2}{m}\)\(T_1\)=\(T_2\)
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Concepts Used:
Motion in a straight line
The motion in a straight line is an object changes its position with respect to its surroundings with time, then it is called in motion. It is a change in the position of an object over time. It is nothing but linear motion.
Types of Linear Motion:
Linear motion is also known as the Rectilinear Motion which are of two types:
- Uniform linear motion with constant velocity or zero acceleration: If a body travels in a straight line by covering an equal amount of distance in an equal interval of time then it is said to have uniform motion.
- Non-Uniform linear motion with variable velocity or non-zero acceleration: Not like the uniform acceleration, the body is said to have a non-uniform motion when the velocity of a body changes by unequal amounts in equal intervals of time. The rate of change of its velocity changes at different points of time during its movement.