An expression of energy density is given by $$u = \alpha \beta \sin\left(\frac{\alpha x}{kt}\right)$$, where $$\alpha, \beta$$ are constants, $$x$$ is displacement, $$k$$ is Boltzmann constant and $$t$$ is the temperature. The dimensions of $$\beta$$ will be

We need to find the dimensional formula of the constant $$\beta$$ from the given expression for energy density.

The given equation is:

$$u = \alpha \beta \sin\left(\frac{\alpha x}{kt}\right)$$

1. Analyze the Argument of the Sine Function

Trigonometric functions like sine take dimensionless arguments. Therefore, the term inside the parenthesis must be dimensionless ($$[M^0 L^0 T^0]$$):

$$\left[\frac{\alpha x}{kt}\right] = [1]$$

Let's find the dimensions of the variables involved:

• Displacement ($$x$$) = $$[L]$$
• The product of the Boltzmann constant and temperature ($$kt$$) represents thermal energy, which has the dimensions of energy/work:

  $$[kt] = [M L^2 T^{-2}]$$

Equating the dimensions to find the dimensional formula for $$\alpha$$:

$$[\alpha] \cdot \frac{[L]}{[M L^2 T^{-2}]} = [1]$$

$$[\alpha] = \frac{[M L^2 T^{-2}]}{[L]} = [M L T^{-2}]$$

2. Analyze the Dimensions of Energy Density ($u$)

Energy density is defined as energy per unit volume:

$$[u] = \frac{[\text{Energy}]}{[\text{Volume}]} = \frac{[M L^2 T^{-2}]}{[L^3]} = [M L^{-1} T^{-2}]$$

3. Determine the Dimensional Formula of $$\beta$$

Since the sine function itself is completely dimensionless, the dimensions of the left side of the equation must equal the dimensions of the coefficients on the right side:

$$[u] = [\alpha] \cdot [\beta]$$

Substitute the dimensions of $$u$$ and $$\alpha$$ to isolate $$[\beta]$$:

$$[M L^{-1} T^{-2}] = [M L T^{-2}] \cdot [\beta]$$

$$[\beta] = \frac{[M L^{-1} T^{-2}]}{[M L T^{-2}]}$$

Simplifying the terms by canceling mass ($$M$$) and time ($$T$$):

$$[\beta] = [L^{-1 - 1}] = [L^{-2}]$$

In standard base form, this is written as:

$$[\beta] = [M^0 L^{-2} T^0]$$

Correction Note

(Option D) contains a sign typo regarding the exponent of length ($$L$$). Following proper dimensional analysis rules, the true mathematical output is $$[M^0 L^{-2} T^0]$$.