Question

A quantity x is given by \[\left( {\dfrac{{IF{v^2}}}{{W{L^4}}}} \right)\]in terms of the moment of inertia I, force F, velocity v, work W, and Length L. Then find the dimensional formula for x which is the same as.
A. Coefficient of viscosity
B. Energy density
C. Force constant
D. Planck’s constant

Answer 1

Hint: Before we start addressing the problem, we need to know about the energy density and dimensional formula. Energy density is the amount of energy stored in a given system of space per unit volume. The dimensional formula is one in which it denotes the units in their base or fundamental form.

Complete step by step solution:
Consider a quantity x which is given by \[\left( {\dfrac{{IF{v^2}}}{{W{L^4}}}} \right)\]in terms of the moment of inertia I, force F, velocity v, work W, and Length L. Then we need to find the dimensional formula for x. That is,
\[x = \left( {\dfrac{{IF{v^2}}}{{W{L^4}}}} \right)\]…………… (1)

Let’s find the dimensional formula for the moment of inertia I, force F, velocity v, work W, and Length L.
The dimensional formula for the moment of inertia, force, velocity, work, and Length are as follows,
The S.I. unit of moment of inertia is \[kg{m^2}\]and the dimensional formula is, \[\left[ I \right] = \left[ {M{L^2}} \right]\]
The S.I. unit of force is N that is, \[kgm{s^{ - 2}}\]and the dimensional formula is, \[\left[ F \right] = \left[ {ML{T^{ - 2}}} \right]\]
The S.I. unit of velocity is \[m{s^{ - 1}}\]and the dimensional formula is, \[\left[ v \right] = \left[ {L{T^{ - 1}}} \right]\]
The S.I. unit of work is \[kg{m^2}{s^{ - 2}}\]and the dimensional formula is, \[\left[ W \right] = \left[ {M{L^2}{T^{ - 2}}} \right]\]
The S.I. unit of length is m and the dimensional formula is, \[\left[ L \right] = \left[ L \right]\]

Now, substitute this dimensional formula in equation (1) we get,
\[x = \left[ {\dfrac{{\left[ {M{L^2}} \right]\left[ {ML{T^{ - 2}}} \right]{{\left[ {L{T^{ - 1}}} \right]}^2}}}{{\left[ {M{L^2}{T^{ - 2}}} \right]{{\left[ L \right]}^4}}}} \right]\]
\[x = \dfrac{{\left[ {{M^2}{L^5}{T^{ - 4}}} \right]}}{{\left[ {M{L^6}{T^{ - 2}}} \right]}}\]
\[\therefore x = \left[ {M{L^{ - 1}}{T^{ - 2}}} \right]\]
Therefore, the dimensional formula for x which is the same as energy density.

Hence, Option B is the correct answer

Note:Remember the S.I. units for all the given terms, then it will be easy to find the dimensional formula and you will get the correct answer.