The bulb of a motorcycle head-lamp is marked 40W, 6V. The resistance of the filament when it is switched on is
A
\(\frac{6^2}{40^2}\)ohms
B
\(\frac{40}{6^2}\)ohms
C
\(\frac{40}{6}\)ohms
D
6 x 40 ohms
E
\(\frac{6^2}{40}\)ohms
Verified Correct Answer: E.
Physical Principles & Numerical Solution
Power \(\frac{V^2}{R}\)
40 = \(\frac{6 \times 6}{r}\)
R = \(\frac{36}{40}\)\(\Omega\)
Source: Myschool Past Questions Archive (Record #29205)