Question 118: Units , Dimensions and Error Analysis

Question

Velocity (v) and acceleration (a) in two systems of units 1 and 2 are related as $${v_2} = {n \over {{m^2}}}{v_1}$$ and $${a_2} = {{{a_1}} \over {mn}}$$ respectively. Here m and n are constants. The relations for distance and time in two systems respectively are :

Accepted Answer

$$[L] = {{[{v^2}]} \over {[a]}}$$
so $${{{{[{v_2}]}^2}} \over {[{a_2}]}} = {{{{\left[ {{n \over {{m^2}}}{v_1}} \right]}^2}} \over {\left[ {{{{a_1}} \over {mn}}} \right]}}$$
$${{{{[{v_2}]}^2}} \over {[{a_2}]}} = {{{n^3}} \over {{m^3}}}{{{{[{v_1}]}^2}} \over {[{a_1}]}}$$
or $$[{L_2}] = {{{n^3}} \over {{m^3}}}[{L_1}]$$
Similarly,
$$[T] = {{[v]} \over {[a]}}$$
So, $$[{T_2}] = {{{n^2}} \over m}[{T_1}]$$

Answer

$${{{n^3}} \over {{m^3}}}{L_1} = {L_2}$$ and $${{{n^2}} \over m}{T_1} = {T_2}$$

Answer

$${L_1} = {{{n^4}} \over {{m^2}}}{L_2}$$ and $${T_1} = {{{n^2}} \over m}{T_2}$$

Answer

$${L_1} = {{{n^2}} \over m}{L_2}$$ and $${T_1} = {{{n^4}} \over {{m^2}}}{T_2}$$

Answer

$${{{n^2}} \over m}{L_1} = {L_2}$$ and $${{{n^4}} \over {{m^2}}}{T_1} = {T_2}$$

Units , Dimensions and Error Analysis

Velocity (v) and acceleration (a) in two systems of units 1 and 2 are related as ${v_2} = {n \over {{m^2}}}{v_1} and {a_2} = {{{a_1}} \over {mn}}$ respectively. Here m and n are constants. The relations for distance and time in two systems respectively are :

Choose the correct answer: