Question

If a wire of resistance R is stretched to double of its length keeping the diameter same, then new resistance will be:

• A. $R/2$
• B. $2R$
• C. $4R$
• D. $16R$

Answer 1

The correct option is C $4R$

let the initial Resistance be $R_i=\rho \frac{L}{A}$

Now the final length is doubled $L_1=2L$, so $LA=L_1{A}_1$

So $A_1=\frac{A}{2}$

Final resistance $R_f=\rho \frac{L_1}{A_1}=\rho \frac{4L}{A}=4R$