Step 1: Identify the dimensions of each variable. • Velocity ($v$): The dimension of velocity is length per unit time. $$[v] = [L][T]^{-1}$$ • Tension ($T$): Tension is a force. Force is mass times acceleration. Acceleration is length per unit time squared. $$[T] = [M][L][T]^{-2}$$ • Length ($L$): The dimension of length is length. $$[L] = [L]$$ • Mass ($M$): The dimension of mass is mass. $$[M] = [M]$$ Step 2: Substitute the dimensions into the right-hand side (RHS) of the given formula $v = \sqrt{\frac{TL}{M}}$. $$ \text{RHS} = \sqrt{\frac{[T][L]}{[M]}} $$ $$ \text{RHS} = \sqrt{\frac{([M][L][T]^{-2})([L])}{[M]}} $$ Step 3: Simplify the dimensions on the RHS. $$ \text{RHS} = \sqrt{\frac{[M][L]^2[T]^{-2}}{[M]}} $$ Cancel out $[M]$ from the numerator and denominator: $$ \text{RHS} = \sqrt{[L]^2[T]^{-2}} $$ Take the square root: $$ \text{RHS} = ([L]^2[T]^{-2})^{\frac{1}{2}} $$ $$ \text{RHS} = [L]^{2 \times \frac{1}{2}}[T]^{-2 \times \frac{1}{2}} $$ $$ \text{RHS} = [L][T]^{-1} $$ Step 4: Compare the dimensions of the RHS with the dimensions of the LHS ($v$). The dimension of the LHS is $[v] = [L][T]^{-1}$. The dimension of the RHS is $[L][T]^{-1}$. Since the dimensions on both sides of the equation are the same, the formula is dimensionally valid. The formula is dimensionally valid because: $$ \text{LHS} = [L][T]^{-1} $$ $$ \text{RHS} = [L][T]^{-1} $$ $$ \boxed{\text{LHS} = \text{RHS}} $$