TS EAMCET 18-July-2022 Shift 1 Solved Paper

We know, for a formula to be correct only when dimensions of LHS and dimensions of RHS must be equal.$E=\frac{2 E_{0} L}{L_{0}}$$\Rightarrow \frac{E}{E_{0}} = \frac{2 L}{L_{0}}$.$\Rightarrow$ Both LHS and RHS of Eq. (i), are dimensionless. Hence, the formula is dimensionally correct.- In $E=E_{0} e^{-\frac{2 L}{L_{0}}}$, clearly the exponent is dimensionless because, $L$ and $L_{0}$ will have same dimension and dimensions of $E$ and $E_{0}$ will be equal, since both $E$ and $E_{0}$ will have dimensions of energy. Thus, this formula is dimensionally correct.- In $E=L e^{-\frac{L}{E_{0}}}$, clearly the power of exponential $\frac{-L}{E_{0}}$ will have dimensions and hence the power of. exponent is not dimensionless, hence this equation is dimensionally incorrect.- In $E=2 \left(\frac{E_{0}}{L_{0}}\right) \times e^{-\frac{L}{L_{0}}}$, here $\frac{E_{0}}{L_{0}}$ will have different dimensions from $E$, hence this formula is also dimensionally incorrect.