Question

I the following dimensionally consistent equation

\({\rm{F}} = \frac{{\rm{X}}}{{{\rm{Linear\;density}}}} + {\rm{Y}}\)

where F is the force, the dimensional formulae for X and Y are given as –

1. \(\left[ {{M^2}{L^0}{T^{ - 2}}} \right],\left[ {ML{T^{ - 2}}} \right]\)
2. \(\left[ {{M^2}{L^{ - 2}}{T^{ - 2}}} \right],\left[ {ML{T^{ - 2}}} \right]\)
3. \(\left[ {ML{T^{ - 2}}} \right],\left[ {M{L^2}{T^{ - 2}}} \right]\)
4. \(\left[ {{M^0}{L^0}{T^0}} \right],\left[ {M{L^0}{T^0}} \right]\)

Answer 1

Correct Answer - Option 1 : \(\left[ {{M^2}{L^0}{T^{ - 2}}} \right],\left[ {ML{T^{ - 2}}} \right]\)

The dimension for linear mass density is \(M{L^{ - 1}}\), while that for force is \(ML{T^{ - 2}}\). The dimension for Y will be \(ML{T^{ - 2}}\), while that for X will be  \(\left[ {ML{T^{ - 2}}} \right]\left[ {M{L^{ - 1}}} \right] = {M^2}{L^0}{T^{ - 2}}\;\).