Question

Let a wire of a certain length be stretched such that its diameter is reduced to two third of its original value . Then the resistance of the new wire will become approximately

• A. five times
• B. two times
• C. three times
• D. nine times

Answer 1

The correct option is A five times

Given, $l=$ length

$D=$ diameter

$A=$ area

Resistance of wire, $R=\rho \frac{l}{A}$

$D^{\prime }=\frac{2}{3}D$

$A^{\prime }=\frac{4}{9}A$

As mass and density remains constant after stretching, volume is also remains constant

$lA=l^{\prime }A$

$Al=l^{\prime }\frac{4}{9}A$

$l^{\prime }=\frac{9l}{4}$

New resistance after stretching,

$R^{\prime }=\rho \frac{l^{\prime }}{A^{\prime }}$

$R^{\prime }=\frac{81}{16}R$

$R^{\prime }=5R$

The correct option is A.