Question

A resistance of 15 ohm is drawn so that its length becomes double

Answer 1

Explanation:

Your question is incomplete but I think it may be asking the final resistance.

So here is the answer,

AS PER QN THE WIRE DRAWN TO MAKE THE LENGTH DOUBLE.

L2=2L1

BY STRETCHING THE CROSS SECTION AREA REDUED BY 2 SO  

A2= A1 / 2

RESISTIVITY REMAIN CONSTANT FOR MATERIAL

NOW

$R_2=\rho\frac{L_2}{A_2} \\R_2= \rho \frac{2L_1} {A_1/2}\\R_2 = 4\rho \frac{L_1} {A_1}\\R_2 = 4 R_1\\R_2 = 4 * 15 = 60$ Ω

HOPE I HELPED YOU

COMMENT ME FOR FURTHER QUERY.

PLZ LIKE AND MARK MY ANSWER AS BRAINLIEST.