If E and E_0 denote energies at time t and t_ | vedclass

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(D) Dimensional analysis requires that both sides of an equation have the same dimensions and that the argument of any exponential function must be di

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$C, D$ only

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(D) Dimensional analysis requires that both sides of an equation have the same dimensions and that the argument of any exponential function must be dimensionless.$(A) E = 2 E_0 \frac{L}{L_0}$$LHS$: $[M L^2 T^{-2}]$$RHS$: $[M L^2 T^{-2}] 	imes \frac{[L]}{[L]} = [M L^2 T^{-2}]$Status: Dimensionally correct.$(B) E = E_0 e^{-\frac{2 L}{L_0}}$Exponent: $\frac{[L]}{[L]} = [1]$ (Dimensionless). Correct.$LHS$: $[M L^2 T^{-2}]$$RHS$: $[M L^2 T^{-2}] 	imes [1] = [M L^2 T^{-2}]$Status: Dimensionally correct.$(C) E = 2 L e^{-\frac{L}{E_0}}$Exponent: $\frac{[L]}{[M L^2 T^{-2}]} = [M^{-1} L^{-1} T^2]$. Since the exponent is not dimensionless, the expression is invalid.$LHS$: $[M L^2 T^{-2}]$Pre-factor: $[L]$Status: Dimensionally incorrect.$(D) E = 2 \left( \frac{E_0}{L_0} ight) e^{-\frac{L}{L_0}}$Exponent: $\frac{[L]}{[L]} = [1]$. Correct.$LHS$: $[M L^2 T^{-2}]$$RHS$: $\frac{[M L^2 T^{-2}]}{[L]} = [M L T^{-2}]$Status: Dimensionally incorrect.Therefore, equations $(C)$ and $(D)$ are incorrect.

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