MCQ
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 ........
- ✓$8$
- B$4$
- C$2$
- D$1$
✓
Answer
Correct option: A.
$8$
a
(a)$l = \frac{{FL}}{{\pi {r^2}r}}$==> $l \propto \frac{F}{{{r^2}}}$ (Y and L are constant)
$\frac{{{l_2}}}{{{l_1}}} = \frac{{{F_2}}}{{{F_1}}} \times {\left( {\frac{{{r_1}}}{{{r_2}}}} \right)^2} = 2 \times {(2)^2} = 8$
${l_2} = 8{l_1} = 8 \times 1 = 8mm$
(a)$l = \frac{{FL}}{{\pi {r^2}r}}$==> $l \propto \frac{F}{{{r^2}}}$ (Y and L are constant)
$\frac{{{l_2}}}{{{l_1}}} = \frac{{{F_2}}}{{{F_1}}} \times {\left( {\frac{{{r_1}}}{{{r_2}}}} \right)^2} = 2 \times {(2)^2} = 8$
${l_2} = 8{l_1} = 8 \times 1 = 8mm$
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