Question

A quantity $x$ is given by $(IFv^2 / WL^{4})$ in terms of moment of inertia $I, $ force $ F$, velocity $v ,$ work $W$ and Length $L$. The dimensional formula for $x$ is same as that of :

• A. Planck's constant
• B. Force constant
• C. Energy density
• D. Coefficient of viscosity

Answer 1

The Correct Option is C

To find the dimensional formula of quantity x given by \left( \frac{IFv^2}{WL^{4}} \right), where I is the moment of inertia, F is the force, v is the velocity, W is the work, and L is the length, we need to find the dimensional formulas for each physical quantity.

1. Moment of Inertia (I): The dimensional formula of the moment of inertia is the same as mass times length squared, i.e., [M^1 L^2 T^0].
2. Force (F): The dimensional formula of force is derived from Newton's second law, F = ma, giving [M^1 L^1 T^{-2}].
3. Velocity (v): The dimensional formula of velocity is length per time, i.e., [M^0 L^1 T^{-1}].
4. Work (W): Work is the product of force and displacement, hence [M^1 L^2 T^{-2}].
5. Length (L): The dimensional formula of length is [L^1].

Calculate the dimensional formula of x:

The expression for x is \left( \frac{IFv^2}{WL^{4}} \right). Finding the dimensions of each component:

• I = [M^1 L^2 T^0]
• F = [M^1 L^1 T^{-2}]
• v^2 = [M^0 L^2 T^{-2}]
• W = [M^1 L^2 T^{-2}]
• L^4 = [L^4]

Substituting these into the expression:

x = \frac{[M^1 L^2 T^0] \cdot [M^1 L^1 T^{-2}] \cdot [L^2 T^{-2}]}{[M^1 L^2 T^{-2}] \cdot [L^4]}

Breaking it down:

x = \frac{[M^1 L^2 T^0] \cdot [M^1 L^3 T^{-4}]}{[M^1 L^6 T^{-2}]}

Simplifying:

x = [M^{1+1-1} L^{2+3-6} T^{0-4+2}] = [M^1 L^{-1} T^{-2}]

The final dimensional formula of x is [M^1 L^{-1} T^{-2}].

The quantity which has the same dimensions as [M^1 L^{-1} T^{-2}] is 'Energy density'.

Conclusion: Therefore, the correct answer is Energy density.