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this statement is correct, but you don't consider only this factor. The fact is when there is higher resistance, the overall current flow would become less. This would lead to less voltage, and therefore less energy would be dissipated per charge while going through the bulb. As a result, overall less heat would be produced. (just read some of it from google to be sure )
You see, voltage remains the same since the source(battery, mains etc) provides a definite amount that remains constant. So, if larger resistance equals more brightness, N03/Q25/P1 should be D.