Q. If the length of filament of a heater is reduced by $10\%$, the power of the heater will
A) Increase by about $9\%$
B) Increase by about $11\%$
C) Increase by about $19\%$
D) Decrease by about $10\%$
Explanation
(b) $P \propto \frac{1}{R}\ and\ R \propto l \Rightarrow P \propto \frac{1}{l}$
$\Rightarrow \frac{P_{1}}{P_{2}} = \frac{l_{2}}{l_{1}} \Rightarrow \frac{P_{1}}{P_{2}} = \frac{(100 - 10)}{100} = \frac{90}{100} \Rightarrow P_{2} = 1.11P_{1}$
% change in power $= \ \frac{P_{2} - P_{1}}{P_{1}} \times 100 = 11\%$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.