**Analysis:**

When two lamps are connected in series, they share the same current. The potential difference across each lamp depends on their respective resistances. The brightness of a lamp is directly proportional to the power it consumes. Therefore, the lamp with higher power consumption will glow more brightly.

**Calculations:**

To determine the brightness of each lamp, we need to calculate the current passing through them and the potential difference across each lamp.

Given:
- Lamp 1: Power (P1) = 60 W, Voltage (V1) = 220 V
- Lamp 2: Power (P2) = 40 W, Voltage (V2) = 220 V

**Step 1: Calculate the resistance of each lamp:**

The resistance of a lamp can be calculated using Ohm's law: R = V^2 / P, where R is resistance, V is voltage, and P is power.

- Resistance of Lamp 1 (R1) = (220^2) / 60 = 806.67 Ω
- Resistance of Lamp 2 (R2) = (220^2) / 40 = 1210 Ω

**Step 2: Calculate the total resistance of the circuit:**

In a series circuit, the total resistance is the sum of individual resistances.

- Total Resistance (RTotal) = R1 + R2 = 2016.67 Ω

**Step 3: Calculate the current passing through the circuit:**

The current passing through the circuit can be calculated using Ohm's law: I = V / R, where I is current, V is voltage, and R is resistance.

- Total Current (ITotal) = 220 / 2016.67 = 0.109 A

**Step 4: Calculate the potential difference across each lamp:**

The potential difference across each lamp can be calculated using Ohm's law: V = IR, where V is voltage, I is current, and R is resistance.

- Voltage across Lamp 1 (V1) = 0.109 * 806.67 = 88.00 V
- Voltage across Lamp 2 (V2) = 0.109 * 1210 = 132.19 V

**Conclusion:**

Based on the calculations, the lamp with a power rating of 40 W will glow more brightly. This is because it has a higher voltage across it compared to the lamp with a power rating of 60 W. The voltage across a lamp determines its brightness, and in this case, Lamp 2 has a higher voltage, resulting in a brighter glow.

We have given two lamps in such a way :
Power of 1st lamp, P₁ = 40W , voltage of 1st lamp, V₁ = 220V
power of 2nd lamp , P₂ = 60W , voltage of 2nd lamp ,V₂ = 220V

we know, one things ,
Power = V²/R [ when potential difference is same then consider P = V²/R ]
R = V²/P
so, resistance of 1st lamp , R₁ = V₁²/P₁ = (220)²/60 = 4840/6 = 2420/3Ω
resistance of 2nd lamp , R₂ = V₂²/P₂ = (220)²/40 = 48400/40 = 1210Ω
Now, question said ,
(a) Both the lamps are in parallel ,
So, 1/Req = 1/R₁ + 1/R₂
1/Req = 3/2420 + 1/1210 = 5/2420 = 1/484
Req = 484Ω

Now, Current drawn from electrical supply ,i = potential difference/Req
= 220V/484 = 20/44 = 5/11 A

(b) energy consumed by lamps in one hour = Energy consumed by 1st lamps in one hour + energy consumed by 2nd lamps in one hour
= P₁ × 1hour + P₂ × 1 hour
=(P₁ + P₂) × 1 hour
= (60W + 40W) × 1 hour
= 100 Wh
= 100/1000 KWh [ ∵ 1 KWh = 10²Wh ]
= 0.1 KWh