Sol. and

Statement is correct. The resistance of the bulb is:

R b = = = 400 Ω Ω

The resistance of the heater is:

R h = = 40 Ω Ω

Statement is incorrect. Since, the bulbs are connected in series, the current in each is the same. Therefore, the potential difference across a bulb will be proportional to its resistance. R ∝ (1/P), where P is power. Thus, the 40 W bulb has the highest resistance and the 100 W bulb has the lowest resistance. Hence, the potential difference across 40 W bulb is the highest and that across the 100 W bulb is the lowest.

Statement is also incorrect. Let R 1 be the resistance of the 60 W bulb and R 2 that of 100 W bulb. Since, R ∝ (1/P) R 1 is greater than R 2 . If R is the resistance of the heater and V the voltage of the mains, the current through heater and 60 W bulb is:

I 1 =

and the current through heater and 100 W bulb is:

I 2 =

Since, R 1 > R 2 , I 2 > I 1 . We know that the heat produced by the heater is proportional to the square of the current flowing through it. Hence, heat produced by the heater will be more when 100 W bulb is in series with it than when a 60 W bulb is in series with it. So, statement is correct.

When a bulb and heater are connected in parallel and this combination is connected across the mains, the potential difference across each is equal to the voltage of the mains irrespective of the resistance of the bulb. Hence, on replacing the 60 W bulb by a 100 W bulb, the heat produced by the heater will remain unchanged which will be V 2 /R in both the cases.