Question

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

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

Answer 1

Answer: (D)

Answer 2

: If the wire is stretched by $( 1 / 10 ) ^ { \mathrm { th } }$ of its original length then the new length of wire become.

As the volume of wire remains constant then $\pi r _ { 1 } ^ { 2 } l = \pi r _ { 2 } ^ { 2 } l _ { 2 } = \pi r _ { 2 } ^ { 2 } \left( \frac { 11 l } { 10 } \right)$ (using (i))

$\Rightarrow \mathrm { r } _ { 2 } ^ { 2 } = \frac { 10 } { 11 } \mathrm { r } _ { 1 } ^ { 2 }$

Now the resistance of stretched wire.,

$\mathrm { R } _ { 2 } = \frac { \rho \left( \frac { 11 } { 10 } \mathrm { l } \right) } { \pi \mathrm { r } _ { 2 } ^ { 2 } } = \frac { \left( \frac { 11 } { 10 } \right) \rho \mathrm { l } } { \pi \times \frac { 10 } { 11 } \mathrm { r } _ { 1 } ^ { 2 } } = \left( \frac { 11 } { 10 } \right) ^ { 2 } \times \frac { \rho \mathrm { l } } { \pi \mathrm { r } _ { 1 } ^ { 2 } }$

$\left( \because \mathrm { R } _ { 1 } = \frac { \rho \mathrm { l } } { \pi \mathrm { r } _ { 1 } ^ { 2 } } = 15 \Omega \right)$

$\therefore \mathrm { R } _ { 2 } = \left( \frac { 11 } { 10 } \right) ^ { 2 } \times 15 = 18.15 \Omega$