To understand why the correct choice is "Statement-I is false but Statement-II is true", we need to examine each statement individually.
Statement-I: "The equivalent emf of two nonideal batteries connected in parallel is smaller than either of the two emfs."
When two nonideal batteries are connected in parallel, the equivalent emf (Eeq) is not simply smaller than either emf but rather an average based on their internal resistances. The formula for the equivalent emf is:
Eeq= (E1R2 + E2R1)/(R1+R2)
Here, E1 and E2 are the emfs of the two batteries, while R1 and R2 are their internal resistances. The Eeq is essentially a weighted average and can be greater than or equal to the smaller emf and smaller than or equal to the larger emf. Therefore, Statement-I is false.
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."
When internal resistances are combined in parallel, the equivalent resistance (Req) is smaller than either of the individual resistances. The formula for the equivalent internal resistance is:
1/Req = 1/R1 + 1/R2
As a result, Req is indeed smaller than either R1 or R2. Therefore, Statement-II is true.
Thus, the correct answer is: Statement-I is false but Statement-II is true.
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.
Choose the correct answer from the options given below.
Step 1 — Concept of parallel connection of nonideal batteries:
Let two batteries with emfs $E_1$, $E_2$ and internal resistances $r_1$, $r_2$ be connected in parallel.
The equivalent emf $E_{eq}$ is given by:
$$ E_{eq} = \frac{E_1/r_1 + E_2/r_2}{1/r_1 + 1/r_2} $$
and the equivalent internal resistance $r_{eq}$ is given by:
$$ r_{eq} = \frac{r_1 r_2}{r_1 + r_2} $$
Step 2 — Compare internal resistances:
Since $r_{eq} = \dfrac{r_1 r_2}{r_1 + r_2}$, it is always less than either $r_1$ or $r_2$.
Therefore, Statement-II is true.
Step 3 — Compare emfs:
- If $E_1 = E_2$, then $E_{eq} = E_1 = E_2$.
- If $E_1 \neq E_2$, then $E_{eq}$ is a weighted average of $E_1$ and $E_2$, and hence lies between $E_1$ and $E_2$.
Thus, $E_{eq}$ is not smaller than both; it lies between them.
Hence, Statement-I is false.
Step 4 — Conclusion:
Statement-I is false but Statement-II is true.
Correct Option: 4
The graph shows the variation of current with voltage for a p-n junction diode. Estimate the dynamic resistance of the diode at \( V = -0.6 \) V.

Sliding contact of a potentiometer is in the middle of the potentiometer wire having resistance \( R_p = 1 \, \Omega \) as shown in the figure. An external resistance of \( R_e = 2 \, \Omega \) is connected via the sliding contact.
The current \( i \) is : 
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is:
Two blocks of masses \( m \) and \( M \), \( (M > m) \), are placed on a frictionless table as shown in figure. A massless spring with spring constant \( k \) is attached with the lower block. If the system is slightly displaced and released then \( \mu = \) coefficient of friction between the two blocks.
(A) The time period of small oscillation of the two blocks is \( T = 2\pi \sqrt{\dfrac{(m + M)}{k}} \)
(B) The acceleration of the blocks is \( a = \dfrac{kx}{M + m} \)
(\( x = \) displacement of the blocks from the mean position)
(C) The magnitude of the frictional force on the upper block is \( \dfrac{m\mu |x|}{M + m} \)
(D) The maximum amplitude of the upper block, if it does not slip, is \( \dfrac{\mu (M + m) g}{k} \)
(E) Maximum frictional force can be \( \mu (M + m) g \)
Choose the correct answer from the options given below:
Let \( f : \mathbb{R} \to \mathbb{R} \) be a twice differentiable function such that \[ (\sin x \cos y)(f(2x + 2y) - f(2x - 2y)) = (\cos x \sin y)(f(2x + 2y) + f(2x - 2y)), \] for all \( x, y \in \mathbb{R}. \)
If \( f'(0) = \frac{1}{2} \), then the value of \( 24f''\left( \frac{5\pi}{3} \right) \) is: