The determination of why the correct choice is "Statement-I is false but Statement-II is true" necessitates an individual examination of each statement.
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, their equivalent emf (Eeq) is determined by a weighted average based on their internal resistances, not simply a value smaller than either individual emf. The formula for the equivalent emf is:
Eeq= (E1R2 + E2R1)/(R1+R2)
In this equation, E1 and E2 represent the emfs of the batteries, and R1 and R2 are their internal resistances. The resulting Eeq is a weighted average that can be greater than or equal to the smaller emf and less than or equal to the larger emf. Consequently, Statement-I is classified as 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 consistently smaller than either of the individual resistances. The governing formula for the equivalent internal resistance is:
1/Req = 1/R1 + 1/R2
This formula demonstrates that Req will always be smaller than both R1 and R2. Therefore, Statement-II is classified as true.
Accordingly, the correct conclusion is: Statement-I is false but Statement-II is true.
Assuming in forward bias condition there is a voltage drop of \(0.7\) V across a silicon diode, the current through diode \(D_1\) in the circuit shown is ________ mA. (Assume all diodes in the given circuit are identical) 

