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 dimension of the expression \((IFv^2/(EL^4))\), we need to determine the dimensions of each variable involved: moment of inertia \(I\), force \(F\), velocity \(v\), energy \(E\), and length \(L\).

### Step-by-Step Solution:

1. **Identify the dimensions of each variable:**
- Moment of Inertia \(I\):
\[ [I] = ML^2 \]
- Force \(F\):
\[ [F] = MLT^{-2} \]
- Velocity \(v\):
\[ [v] = LT^{-1} \]
- Energy \(E\):
\[ [E] = ML^2T^{-2} \]
- Length \(L\):
\[ [L] = L \]

2. **Substitute the dimensions into the expression:**
The expression we need to evaluate is:
\[ \frac{IFv^2}{EL^4} \]
Substituting the dimensions we found:
\[ [IFv^2] = [I][F][v^2] = (ML^2)(MLT^{-2})((LT^{-1})^2) \]
First, calculate \(v^2\):
\[ [v^2] = (LT^{-1})^2 = L^2T^{-2} \]
Now, substituting back:
\[ [IFv^2] = (ML^2)(MLT^{-2})(L^2T^{-2}) = M^2L^6T^{-4} \]

3. **Calculate the dimensions of the denominator \(EL^4\):**
\[ [EL^4] = [E][L^4] = (ML^2T^{-2})(L^4) = ML^6T^{-2} \]

4. **Combine the dimensions:**
Now substitute the dimensions of the numerator and denominator into the expression:
\[ \frac{IFv^2}{EL^4} = \frac{M^2L^6T^{-4}}{ML^6T^{-2}} \]
Simplifying this:
\[ = \frac{M^2}{M} \cdot \frac{L^6}{L^6} \cdot \frac{T^{-4}}{T^{-2}} = M^{2-1}L^{6-6}T^{-4+2} = ML^0T^{-2} = MT^{-2} \]

5. **Final Result:**
The final dimension of the expression \(\frac{IFv^2}{EL^4}\) is:
\[ [IFv^2/(EL^4)] = MT^{-2} \]