Question

A cylindrical rod is reformed to half of its original length keeping volume constant. If its resistance before this change were R, then the resistance after reformation of rod will be

• A. R
• B. R/4
• C. 3R/4
• D. R/2

Answer 1

Answer: (B)

R/4

Answer 2

: The resistance of rod before reformation

$\mathrm { R } _ { 1 } = \mathrm { R } = \frac { \rho \mathrm { l } _ { 1 } } { \pi \mathrm { r } _ { 1 } ^ { 2 } }$

Now the rod is reformed such that,

$\therefore \pi \mathrm { r } _ { 1 } ^ { 2 } \mathrm { l } _ { 1 } = \pi \mathrm { r } _ { 2 } ^ { 2 } \mathrm { l } _ { 2 }$ (Volume remains constant,

or …(i)

Now the resistance of the rod after reformation

$\therefore \frac { \mathrm { R } _ { 1 } } { \mathrm { R } _ { 2 } } = \frac { \rho \mathrm { l } _ { 1 } } { \pi \mathrm { r } _ { 1 } ^ { 2 } } / \frac { \rho \mathrm { l } _ { 1 } } { \pi \mathrm { r } _ { 2 } ^ { 2 } } = \frac { \mathrm { l } _ { 1 } } { 1 _ { 2 } } \times \frac { \mathrm { r } _ { 2 } ^ { 2 } } { \mathrm { r } _ { 1 } ^ { 2 } }$

or, (using (i))

$\frac { \mathrm { R } _ { 1 } } { \mathrm { R } _ { 2 } } = 4$

$\therefore \mathrm { R } _ { 2 } = \frac { \mathrm { R } } { 4 }$