Question

A wire with $15\Omega$. resistance is stretched by one tenth of its original length and volume of wire is kept constant. Then its resistance will be

• A. $15.18\Omega$
• B. $81.15\Omega$
• C. $51.18\Omega$
• D. $18.15\Omega$

Answer 1

Answer: (D)

If the wire is stretched by $(1/10{)}^{th}$ of its original length then the new length of wire become
$l_2=l+\frac{l}{10}=\frac{11l}{10}\dots \left(i\right)$
As the volume of wire remains constant then
$\pi r_1^{2}l=\pi r_2^2{l}_2=\pi r_2^2\left(\frac{11l}{10}\right)$ (using $(i)$)
$\Rightarrow r_2^2=\frac{10}{11}{r}^2\dots \left(ii\right)$
Now the resistance of stretched wire.
$R_2=\frac{\rho \left(\frac{11}{10}l\right)}{\pi r_2^2}=\frac{\left(\frac{11}{10}\right)\rho l}{\pi \times \frac{10}{11}{r}_1^2}$
$={\left(\frac{11}{10}\right)}^2\times \frac{\rho l}{\pi r_1^2}$
$\left(\because R_1=\frac{\rho l}{\pi r_1^2}=15\Omega \right)$
$\therefore R_2={\left(\frac{11}{10}\right)}^2\times 15=18.15\Omega$