If the length of the filament of a heater is | vedclass

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Question

$P \propto \frac{1}{R}$  and  $ R \propto l \Rightarrow P \propto \frac{1}{l}$

$ \Rightarrow \frac{{{P_1}}}{{{P_2}}} = \frac{{{l_2}}}{{{l

Vedclass · Accepted Answer

Increase by about $11\%$

Vedclass · Answer

$P \propto \frac{1}{R}$  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} = 1.11\,{P_1}$

$\%$ change in power = $\frac{{{P_2} - {P_1}}}{{{P_1}}} 	imes 100 = 11\% $

If the length of the filament of a heater is reduced by $10\%$, the power of the heater will

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