For a bulb of rating 4V, 6W, we can calculate the resistance using the formula: \[ P = \frac{V^2}{R} \] Thus, \( R = \frac{V^2}{P} = \frac{4^2}{6} = \frac{16}{6} \approx 2.67 \, \Omega \). Now, to operate safely with a 240V mains supply, the total resistance in the series connection must be: \[ R_{{total}} = \frac{V_{{total}}^2}{P_{{total}}} \] For total power 6W per bulb, and for the total power at 240V, the number of bulbs required can be calculated to be 40.
Thus, the correct answer is option (b).

In the circuit shown, the galvanometer (G) has an internal resistance of $100 \Omega$. The galvanometer current $I_G$ is ________ $\mu A$ (rounded off to the nearest integer).

With the help of the given circuit, find out the total resistance of the circuit and the current flowing through the cell.
Two batteries of emf's \(3V \& 6V\) and internal resistances 0.2 Ω \(\&\) 0.4 Ω are connected in parallel. This combination is connected to a 4 Ω resistor. Find:
(i) the equivalent emf of the combination
(ii) the equivalent internal resistance of the combination
(iii) the current drawn from the combination
On the day of her examination, Riya sharpened her pencil from both ends as shown below. 
The diameter of the cylindrical and conical part of the pencil is 4.2 mm. If the height of each conical part is 2.8 mm and the length of the entire pencil is 105.6 mm, find the total surface area of the pencil.
