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**Hint:**When bulbs are connected in such a manner that one terminal of one bulb is kept open and another terminal is connected to one terminal of another bulb then they are considered to be connected as in series combination. Formula used to solve the problem: \[P = \dfrac{{{V^2}}}{R}\] and \[P = {I^2}R\] where P is the power, R is resistance and V is the Voltage or potential applied across the bulb.

**Complete step by step solution:**

Given

Calculate the resistance of each bulb

Since, Power \[P = \dfrac{{{V^2}}}{R}\]

\[ \Rightarrow R = \dfrac{{{V^2}}}{P}\]

So, \[{R_1} = \dfrac{{{V^2}}}{{{P_1}}} = \dfrac{{{V^2}}}{{25}}\]

and \[{R_2} = \dfrac{{{V^2}}}{{{P_2}}} = \dfrac{{{V^2}}}{{100}}\]

Calculate the equivalent resistance

\[R{}_{eq} = {R_1} + {R_2}\]

\[ \Rightarrow R{}_{eq} = {V^2}\left( {\dfrac{1}{2} + \dfrac{1}{{100}}} \right)\]

\[\therefore R{}_{eq} = \dfrac{{{V^2}}}{{20}}\]

Since, In series current is constant

So, when connected to supply of voltage 440 v the current in the circuit would be

\[\Rightarrow I = \dfrac{{V'}}{{R{}_{eq}}}\]

Since, V’ = 2V

\[ \Rightarrow I = \dfrac{2V}{\dfrac{V^2}{20}}\]

\[\therefore I = \dfrac{{40}}{{{V_0}}}\]

Now

Power generated in the bulb 1 will be

\[\Rightarrow P{}_1 = {I^2}{R_1} = {\left( {\dfrac{{40}}{V}} \right)^2} \times \left( {\dfrac{{{V^2}}}{{25}}} \right)\]

\[\therefore P{}_1 = 64W\]

Power generated in the bulb 2 will be

\[\Rightarrow P{}_2 = {I^2}{R_2} = {\left( {\dfrac{{40}}{V}} \right)^2} \times \left( {\dfrac{{{V^2}}}{{100}}} \right)\]

\[\therefore P{}_2 = 16W\]

Here it is clearly seen that

$64W>25W$

Therefore, Only bulb 1 or of 25 watt will get fused.

**Hence option (B) is the correct answer.**

**Note:**Any resistive electronic device consumes electrical power and it does not depend upon the direction of current.

In both the above cases, power is only consumed and this power consumed is given by

\[P = \dfrac{{{V^2}}}{R} = {I^2}R = VI\]

In any electrical circuit, law of conservation of energy is followed. i.e.

Net power supplied by all the batteries of the circuit =net power consumed by all the resistors in the circuit.

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