Solution:

Bulbs in Series:
When bulbs are connected in series, they share the same current but have different voltages across them. To find out which bulb will fuse, we need to calculate the current flowing through the circuit and then use Ohm's Law to find the voltage across each bulb.

Step 1: Calculate the total resistance of the circuit
Since the bulbs are connected in series, their resistances add up.
Rtotal = R1 + R2 = (V/I1) + (V/I2) = V(1/I1 + 1/I2)

For bulb 1: R1 = (110/25) = 4.4 ohms
For bulb 2: R2 = (110/100) = 1.1 ohms
For the circuit: Rtotal = V(1/4.4 + 1/1.1) = 68.75 ohms

Step 2: Calculate the current flowing through the circuit
I = V/Rtotal = 220/68.75 = 3.2 A

Step 3: Calculate the voltage across each bulb
V1 = IR1 = 3.2 x 4.4 = 14.08 V
V2 = IR2 = 3.2 x 1.1 = 3.52 V

Since the voltage across bulb 1 is greater than its rated voltage of 110 V, it will fuse. Bulb 2 will remain lit but at a lower brightness because of the reduced voltage across it.

Bulbs in Parallel:
When bulbs are connected in parallel, they have the same voltage across them but share the current. To find out what happens when the bulbs are connected in parallel, we need to calculate the total resistance of the circuit and then use Ohm's Law to find the current flowing through each bulb.

Step 1: Calculate the total resistance of the circuit
Since the bulbs are connected in parallel, the reciprocal of their resistances adds up.
1/Rtotal = 1/R1 + 1/R2 = (1/V)(1/P1 + 1/P2)

For bulb 1: R1 = (110/25) = 4.4 ohms
For bulb 2: R2 = (110/100) = 1.1 ohms
For the circuit: 1/Rtotal = (1/110)(1/625 + 1/12100) = 0.000169
Rtotal = 5915.7 ohms

Step 2: Calculate the current flowing through the circuit
I = V/Rtotal = 220/5915.7 = 0.037 A

Step 3: Calculate the current flowing through each bulb
I1 = V/R1 = 220/4.4 = 50 A
I2 = V/R2 = 220/1.1 = 200 A

Since the current flowing through bulb 1 is much greater than its rated current of 25 W, it will fuse. Bulb 2 will remain lit but at a higher brightness because of the increased current flowing through it.

Answer : option (b) 2.5 W bulb will fuse.

explanation : first of all, find resistance of each bulb, R = V²/P

for first bulb ( P = 2.5 watt, V = 110 volts)

so, R1 = (110)²/(2.5) = 12100/2.5

= 4840 ohm.

for second bulb (100 watt, 110 volts)

so, R2 = (110)²/(100)

= 121 ohm

both bulbs are joined in series combination. so, Req = R1 + R2

= 4840 + 121

= 4961 ohm

so, current passing through circuit , I = (220 volts)/(4861 ohm)

= 220/4861 ≈ 0.045 A

now current passing through first bulb, I1 = P1/V1 = (2.5 watt)/110 volts

≈ 0.02272 A

and current through second bulb, I2 = P2/V2 = (100 watt)/110 volts

≈ 0.909A

here I > I1

The total current is bigger than the allowed value on the first bulb, so the 2.5watt bulb will fuse.

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