Verified Solution • 2025–26 Session moderate

QJEEM22Ph7fc3: JEE Main 2022 Physics - Class 11 JEEM22Ph7fc3 Solution: Velocity (v) and acceleration (a) in two systems of units 1 and …

Velocity (v) and acceleration (a) in two systems of units 1 and 2 are related as v2=nm2v1{v_2} = {n \over {{m^2}}}{v_1} and a2=a1mn{a_2} = {{{a_1}} \over {mn}} respectively. Here m and n are constants. The relations for distance and time in two systems respectively are :

Multiple Choice Options

  • A n3m3L1=L2{{{n^3}} \over {{m^3}}}{L_1} = {L_2} and n2mT1=T2{{{n^2}} \over m}{T_1} = {T_2}
  • B L1=n4m2L2{L_1} = {{{n^4}} \over {{m^2}}}{L_2} and T1=n2mT2{T_1} = {{{n^2}} \over m}{T_2}
  • C L1=n2mL2{L_1} = {{{n^2}} \over m}{L_2} and T1=n4m2T2{T_1} = {{{n^4}} \over {{m^2}}}{T_2}
  • D n2mL1=L2{{{n^2}} \over m}{L_1} = {L_2} and n4m2T1=T2{{{n^4}} \over {{m^2}}}{T_1} = {T_2}

Step-by-Step Verified Solution

[L]=[v2][a][L] = {{[{v^2}]} \over {[a]}}

so [v2]2[a2]=[nm2v1]2[a1mn]{{{{[{v_2}]}^2}} \over {[{a_2}]}} = {{{{\left[ {{n \over {{m^2}}}{v_1}} \right]}^2}} \over {\left[ {{{{a_1}} \over {mn}}} \right]}}

[v2]2[a2]=n3m3[v1]2[a1]{{{{[{v_2}]}^2}} \over {[{a_2}]}} = {{{n^3}} \over {{m^3}}}{{{{[{v_1}]}^2}} \over {[{a_1}]}}

or [L2]=n3m3[L1][{L_2}] = {{{n^3}} \over {{m^3}}}[{L_1}]

Similarly,

[T]=[v][a][T] = {{[v]} \over {[a]}}

So, [T2]=n2m[T1][{T_2}] = {{{n^2}} \over m}[{T_1}]

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