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=‌

2E0L

L0

⇒‌‌‌

E

E0

=‌

2L

L0

.
⇒ Both LHS and RHS of Eq. (i), are dimensionless. Hence, the formula is dimensionally correct.
- In E=E0e−‌

2L

L0

, clearly the exponent is dimensionless because, L and L0 will have same dimension and dimensions of E and E0 will be equal, since both E and E0 will have dimensions of energy. Thus, this formula is dimensionally correct.
- In E=Le−L∕E0, clearly the power of exponential ‌

−L

E0

will have dimensions and hence the power of. exponent is not dimensionless, hence this equation is dimensionally incorrect.
- In E=2(‌

E0

L0

)×e−L∕L0, here ‌

E0

L0

will have different dimensions from E, hence this formula is also dimensionally incorrect.