NEET Practice Questions (MCQs) with Answers & Solutions

Using the extended formula for parallel combination (Eq. 3.59): $\epsilon_{eq}/r_{eq} = \epsilon_1/r_1 + ... + \epsilon_n/r_n$. If all cells are identical ($\epsilon_1 = ... = \epsilon = \epsilon$) and ($r_1 = ... = r_n = r$), then $1/r_{eq} = n/r$, so $r_{eq} = r/n$. Substituting back: $\epsilon_{eq}/(r/n) = n(\epsilon/r) \Rightarrow \epsilon_{eq} = (r/n) \times n(\epsilon/r) = \epsilon$. Thus, for identical cells in parallel, the equivalent emf is equal to the emf of a single cell.

The NCERT text states: 'V = $\epsilon - Ir$' (Eq. 3.38). This equation represents the terminal voltage of a cell when current I is flowing. The term 'Ir' is the potential drop across the internal resistance (r) of the cell, and it subtracts from the electromotive force ($\epsilon$) to give the actual potential difference available at the terminals.

The NCERT text states: 'Thus, emf $\epsilon$ is the potential difference between the positive and negative electrodes in an open circuit, i.e., when no current is flowing through the cell.' (after Eq. 3.37). In an open circuit, the external resistance R is infinite, so no current flows.

The NCERT text clarifies: 'If in the combination, the current leaves any cell from the negative electrode, the emf of the cell enters the expression for e_eq with a negative sign, as in Eq. (3.47).' This signifies that the cell is connected in opposition, effectively reducing the overall emf.

From Eq. 3.54 in the NCERT text: '$e_{eq} = (\epsilon_1 r_2 + \epsilon_2 r_1) / (r_1 + r_2)$'. This formula is derived by comparing the terminal voltage expression for the combination with that of a single equivalent cell.

The NCERT text states: 'In practical calculations, internal resistances of cells in the circuit may be neglected when the current I is such that $\epsilon >> Ir$.' This means if the voltage drop across the internal resistance is very small compared to the emf, the internal resistance has a minimal effect on the terminal voltage.

The NCERT text states: 'If the negative terminal of the second is connected to positive terminal of the first, Eqs. (3.56) and (3.57) would still be valid with $\epsilon_2 \rightarrow -\epsilon_2$.' This indicates that one cell is acting in opposition to the other, which is accounted for by changing the sign of its emf in the parallel combination formulas.

The context states: 'The German chemist, Johann Dobereiner in early 1800’s was the first to consider the idea of trends among properties of elements.'

According to the context: 'In each case, he noticed that the middle element of each of the triads had an atomic weight about halfway between the atomic weights of the other two' (Table 3.1 also exemplifies this).

The text mentions: 'Newlands’s Law of Octaves seemed to be true only for elements up to calcium.'