If `I` is moment of inertia, `F` is force, `v` is velocity, `E` is energy and `L` is length then, dimension of `(IFv^2/(EL^4))` will be:

Options

A. Energy density
B. viscosity
C. young modulus
D. Torque

Answer

To find the dimensions of the expression \(\frac{IFv^2}{EL^4}\), we will analyze the dimensions of each variable involved:

1. **Moment of Inertia (I)**:
- The moment of inertia \(I\) is given by the formula \(I = mR^2\), where \(m\) is mass and \(R\) is the radius.
- Therefore, the dimension of \(I\) is:
\[ [I] = [m][R^2] = M L^2 T^0 \]

2. **Force (F)**:
- The force \(F\) is defined as \(F = ma\), where \(a\) is acceleration.
- The dimension of force is:
\[ [F] = [m][a] = M L T^{-2} \]

3. **Velocity (v)**:
- The dimension of velocity \(v\) is:
\[ [v] = L T^{-1} \]
- Therefore, \(v^2\) will have the dimension:
\[ [v^2] = (L T^{-1})^2 = L^2 T^{-2} \]

4. **Energy (E)**:
- Energy is defined as work done, which is force times displacement. Thus:
\[ [E] = [F][d] = [F][L] = (M L T^{-2})(L) = M L^2 T^{-2} \]

5. **Length (L)**:
- The dimension of length \(L\) is simply:
\[ [L] = L \]

Now, substituting these dimensions into the expression \(\frac{IFv^2}{EL^4}\):

\[ \frac{IFv^2}{EL^4} = \frac{(M L^2 T^0)(M L T^{-2})(L^2 T^{-2})}{(M L^2 T^{-2})(L^4)} \]

Calculating the numerator:

\[ IFv^2 = (M L^2 T^0)(M L T^{-2})(L^2 T^{-2}) = M^2 L^5 T^{-4} \]

Calculating the denominator:

\[ EL^4 = (M L^2 T^{-2})(L^4) = M L^6 T^{-2} \]

Now, substituting back into the expression:

\[ \frac{IFv^2}{EL^4} = \frac{M^2 L^5 T^{-4}}{M L^6 T^{-2}} = \frac{M^{2-1} L^{5-6} T^{-4 - (-2)}}{1} = M^1 L^{-1} T^{-2} \]

Thus, the final dimension of the expression \(\frac{IFv^2}{EL^4}\) is:

\[ [M^1 L^{-1} T^{-2}] \]

### Final Answer:
The dimensions of \(\frac{IFv^2}{EL^4}\) are \([M^1 L^{-1} T^{-2}]\).