When a wire is stretched and its radius becomes `r//2` then its resistance will be

Options

A. 16R
B. 2R
C. 4R
D. 0

Answer

A

Explanation

Initial radius of wire , ` r_(1)= and ` final radius of wire ` ,r_2=r//2 =0.5 r `
Since volume of the wire after stretch remains constant , therefore `l_(1)A_(1)=l_(2)A_(2)`
`or (l_(1))/(l_(2))=(A_(2))/(A_(1))=(r_(2)^(2))/(r_(1)^(2))=((0.5r)/(r))^(2)=(1)/(4)` .
The resistance (R) `=rho""(l)/(A) prop (l)/(A)` .
Therefore ` (R_(1))/(R_(2))=(l_(1))/(l_(2)) xx (A_(2))/(A_(1))=(1)/(4) xx (1)/(4) =(1)/(16)` or ` R_(2)=16R_(1)=16R `