The question asks for the amount of work done by an electric bulb given its power rating and the time duration.
Power ($P$) is defined as the rate at which work ($W$) is done, or energy is transferred, over time ($t$). The formula is:
$ P = \frac{W}{t} $
To find the work done ($W$), we can rearrange the formula:
$ W = P \times t $
We are given:
Substitute these values into the rearranged formula:
$ W = 90 \text{ W} \times 10 \text{ s} $
$ W = 900 \text{ J} $
The unit of work is Joules (J).
Therefore, the electric bulb does 900 Joules of work in 10 seconds. This matches Option A.
Consider two cells of emf ε1 and ε2 with internal resistances r1 and r2, respectively. The two cells are connected in parallel by connecting their positive terminals together and connecting their negative terminals together. The combination is equivalent to a single cell with emf given by:
The cost of energy to operate an industrial refrigerator that consumes 5 kW power working 10 hours per day for 30 days will be (Given that the charge k.W.h of energy = Rs. 4)
A battery of EMF 6.0 V and internal resistance 1.0 Ω is connected to a resistor of 11 Ω. The terminal potential difference for the battery is: