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Question: A copper wire of length 4.0m and area of cross-section \(1.2 \mathrm {~cm} ^ { 2 }\) is stretched wi...

A copper wire of length 4.0m and area of cross-section 1.2 cm21.2 \mathrm {~cm} ^ { 2 } is stretched with a force of 4.8×1034.8 \times 10 ^ { 3 } N. If Young’s modulus for copper is 1.2×1011 N/m21.2 \times 10 ^ { 11 } \mathrm {~N} / \mathrm { m } ^ { 2 } the increase in the length of the wire will be

A

1.33 mm

B

1.33 cm

C

2.66 mm

D

2.66 cm

Answer

1.33 mm

Explanation

Solution

l=FLAY=4.8×103×41.2×10−4×1.2×1011=1.33 mml = \frac { F L } { A Y } = \frac { 4.8 \times 10 ^ { 3 } \times 4 } { 1.2 \times 10 ^ { - 4 } \times 1.2 \times 10 ^ { 11 } } = 1.33 \mathrm {~mm}