Question

A wire of radius $r$ has resistance $R$. If it is stretched to a radius $r/2$, its resistance becomes:

• A. $0.5R$
• B. $2R$
• C. $4R$
• D. $16R$

Answer 1

The correct option is D $16R$

As we know,

$R=\frac{\rho L}{A}..........\left(i\right)$

If you stretch the wire till it's radius is halved the previous value, then it's length will increase 4 times to keep the volume constant.

So, new values are,

$L_2=4L$

$A_2=\frac{A}{4}$

where, L and A are the values of original length and area

Resistivity remains the same as it depends upon the material of the wire.

So, the new resistance will be,

$R_1=\frac{\rho L}{A}$

$R_1=\frac{\rho 4L}{A/2}$

$R_1=\frac{16\rho L}{A}$

$R_1=16R$