Answer:
[tex]15\%[/tex].
Explanation:
The efficiency of a machine is the percentage of energy input that was turned into useful energy.
The power rating of this lamp is [tex]40\; \rm W[/tex] (same as [tex]40\; \rm J \cdot s^{-1}[/tex],) meaning that [tex]40\; \rm J[/tex] of energy is supplied to this lamp every second.
The question states that [tex]34\; \rm J[/tex] out of that [tex]40\; \rm J[/tex] of energy input would be turned into heat, which is not useful energy output in this scenario. Assuming that all other forms of energy loss is negligible. The rest of the [tex](40\; \rm J - 34\; \rm J) = 6\; \rm J[/tex] of energy supplied to this lamp would be turned into useful energy output.
Thus, every second, this lamp would receive [tex]40\; \rm J[/tex] of energy input and would outputs [tex]6\; \rm J[/tex] of useful work. The efficiency of this lamp would be:
[tex]\begin{aligned}& \text{Efficiency} \\ =\; & \frac{\text{Useful energy out}}{\text{Total energy in}} \times 100\% \\ =\; & \frac{6\; \rm J}{40\; \rm J} \times 100\%\\ =\; &15\% \end{aligned}[/tex].
