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Question

A 90 W electric bulb does how much work in 10 seconds?

The correct answer is
900 J

Work Calculation for Electric Bulb

The question asks for the amount of work done by an electric bulb given its power rating and the time duration.

Physics Formula

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 $

Applying the Values

We are given:

  • Power ($P$) = 90 W
  • Time ($t$) = 10 s

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).

Conclusion

Therefore, the electric bulb does 900 Joules of work in 10 seconds. This matches Option A.

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Important Questions from Work, Energy, and EMF

  1. 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:

  2. What is the work required to increase the speed of a 0.4 kg ball from 1 m/s to 3 m/s?
  3. Two batteries E 1 (emf: 6 V, internal resistance: 0.5 Ω ) and E 2  (emf: 12 V, internal resistance: 1.0  Ω ) are  connected in parallel by connecting their positive terminals to point A and negative terminals to point B. A  third battery E 3  [emf: 6 V, internal resistance: ( \(\frac{2}{3}\) )  Ω ] is connected in series with this combination by  connecting its positive terminal to B. The equivalent emf of this combination is
  4. 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)

  5. 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: 

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