A resistor of 20 CM length and resistance 5 cm is stretched to a uniform wire of 40 cm length .The resistance now isplz solve the steps.too​

Answers (1)

R=\rho \frac{l}{A}

Here Here \ R_1=5 \Omega , l_1=20 \ cm , l_2= 40 \cm

Since the volume of wire will be constat so v=lA=constant

So 

R=\rho \frac{l}{A}=\frac{\rho l^2}{V}

So 

\frac{R_1}{R_2}=\frac{l^2_1}{l^2_2}=(\frac{20}{40})^2=\frac{1}{4}\\\Rightarrow R_2=4R_1=20 \ \Omega

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