Given below are two statements :
Statement-I : The equivalent emf of two nonideal batteries connected in parallel is smaller than either of the two emfs.
Statement-II : The equivalent internal resistance of two nonideal batteries connected in parallel is smaller than the internal resistance of either of the two batteries. In the light of the above statements, choose the correct answer from the options given below.

To solve this problem, we need to evaluate two statements about non-ideal batteries connected in parallel. A non-ideal battery has an internal resistance, so we must consider both the electromotive force (emf) and the internal resistance for each battery.

Consider two batteries:

• Battery 1: emf = $$E_1$$, internal resistance = $$r_1$$
• Battery 2: emf = $$E_2$$, internal resistance = $$r_2$$

When connected in parallel, the equivalent emf $$E_{eq}$$ and equivalent internal resistance $$r_{eq}$$ are given by the formulas:

$$E_{eq} = \frac{E_1 r_2 + E_2 r_1}{r_1 + r_2}$$

$$r_{eq} = \frac{r_1 r_2}{r_1 + r_2}$$

We will analyze each statement separately.

Statement-II Analysis: The equivalent internal resistance of two nonideal batteries connected in parallel is smaller than the internal resistance of either of the two batteries.

Using the formula for equivalent internal resistance:

$$r_{eq} = \frac{r_1 r_2}{r_1 + r_2}$$

Since $$r_1 > 0$$ and $$r_2 > 0$$, we can compare $$r_{eq}$$ with $$r_1$$:

$$\frac{r_{eq}}{r_1} = \frac{\frac{r_1 r_2}{r_1 + r_2}}{r_1} = \frac{r_2}{r_1 + r_2} < 1$$

because $$r_2 < r_1 + r_2$$. Therefore, $$r_{eq} < r_1$$.

Similarly, comparing $$r_{eq}$$ with $$r_2$$:

$$\frac{r_{eq}}{r_2} = \frac{\frac{r_1 r_2}{r_1 + r_2}}{r_2} = \frac{r_1}{r_1 + r_2} < 1$$

so $$r_{eq} < r_2$$.

Thus, $$r_{eq}$$ is always less than both $$r_1$$ and $$r_2$$. Therefore, Statement-II is true.

Statement-I Analysis: The equivalent emf of two nonideal batteries connected in parallel is smaller than either of the two emfs.

Using the equivalent emf formula:

$$E_{eq} = \frac{E_1 r_2 + E_2 r_1}{r_1 + r_2}$$

This expression is a weighted average of $$E_1$$ and $$E_2$$:

$$E_{eq} = E_1 \cdot \frac{r_2}{r_1 + r_2} + E_2 \cdot \frac{r_1}{r_1 + r_2}$$

Since $$\frac{r_2}{r_1 + r_2} + \frac{r_1}{r_1 + r_2} = 1$$ and both fractions are positive, $$E_{eq}$$ is a convex combination of $$E_1$$ and $$E_2$$. This means $$E_{eq}$$ lies between the minimum and maximum of $$E_1$$ and $$E_2$$.

Without loss of generality, assume $$E_1 \leq E_2$$. Then:

$$E_1 \leq E_{eq} \leq E_2$$

Therefore, $$E_{eq}$$ is not smaller than both emfs; it is at least as large as the smaller emf and at most as large as the larger emf.

For example, let $$E_1 = 10 \text{ V}$$, $$r_1 = 1 \Omega$$, $$E_2 = 5 \text{ V}$$, $$r_2 = 1 \Omega$$. Then:

$$E_{eq} = \frac{10 \times 1 + 5 \times 1}{1 + 1} = \frac{15}{2} = 7.5 \text{ V}$$

Here, $$7.5 \text{ V}$$ is greater than $$5 \text{ V}$$ ($$E_2$$) and less than $$10 \text{ V}$$ ($$E_1$$). So it is not smaller than both emfs.

If the emfs are equal, say $$E_1 = E_2 = E$$, then:

$$E_{eq} = \frac{E r_2 + E r_1}{r_1 + r_2} = E \frac{r_1 + r_2}{r_1 + r_2} = E$$

which is equal to both emfs, not smaller.

Thus, Statement-I is false because the equivalent emf is not necessarily smaller than both emfs; it lies between them or equals them.

Conclusion:

• Statement-I is false.
• Statement-II is true.

Therefore, the correct option is B: Statement-I is false but Statement-II is true.