Question

if the length of the filament of a heater is reduced by 10 the power of the heater will

Answer 1

we know that P=V^2/R

=V^2/ρl/A

P'=1/(l-10/100l)

P'/P=l/(l-l/10)=10/9

 

P'-P/P*100=10/9*100

=11%

SO it increases by 11%

Answer 2

Answer:

the power of the heater will increase by 11.11 %

Explanation:

if the length of the filament of a heater is reduced by 10% the power of the heater will

Let say  R is the resitance of Filament & Length of filament = L

R = ρ L/A

L is decreased by 10 %

so New L = L - (10/100)L = 0.9L

New R = ρ 0.9 L/A = 0.9R

Earlier Power of Heater = P  = V²/R

New Power of Heater = V²/0.9R  = P/0.9  = 10P/9

Power increased by 10P/9 - P  = P/9

% increase in Power =  ((P/9)/P) * 100  = 100/9 %  = 11.11 %

the power of the heater will increase by 11.11 %