Question

A wire of uniform cross-section has a resistance of $R$ . If it is drawn to three times the length, but the volume remains constant, what will be its resistance?

• A. $3R\Omega$
• B. $5R\Omega$
• C. $7R\Omega$
• D. $9R\Omega$

Answer 1

Answer: (D)

$9R\Omega$

Wire of uniform cross-section has a resistance of $R$.
Let the length of the wire be $l$.
Cross sectional Area be $A$:
And volume, $V=Al$
Now the resistance of a wire can be expressed as: $R=\rho \frac{l}{A}$
Where resistivity is $\rho$.

Now the wire is stretched such that final length ($l_1$) becomes three times the initial length (l), but the volume (V) remains constant.
$Al=3A_{1}l$
$A_1=\frac{A}{3}$
The new resistance will be:
$R_1=\rho \frac{l_1}{A_1}$
Putting the value of $l_1=3l$ and $A_1$=$\frac{A}{3}$

We have, $R_1=\rho \frac{3l}{\frac{A}{3}}=9\rho \frac{l}{A}$
$R_1=9R$ ohm