Question

A wire extends by 1 mm when a force is applied. Double the force is applied to another wire of same material and length but half the radius of cross-section. The elongation of the wire in mm will be

• A. 8
• B. 4
• C. 2
• D. 1

Answer 1

Answer: (A)

8

Answer 2

$l = \frac { F L } { \pi r ^ { 2 } r }$⇒ $l \propto \frac { F } { r ^ { 2 } }$ (Y and L are constant)

$\frac { l _ { 2 } } { l _ { 1 } } = \frac { F _ { 2 } } { F _ { 1 } } \times \left( \frac { r _ { 1 } } { r _ { 2 } } \right) ^ { 2 } = 2 \times ( 2 ) ^ { 2 } = 8$ ∴