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Question: A wire of length 50 cm and cross sectional area of 1 sq. mm is extended by 1 mm. The required work w...

A wire of length 50 cm and cross sectional area of 1 sq. mm is extended by 1 mm. The required work will be

(Y=2×1010Nm−2)\left( Y = 2 \times 10 ^ { 10 } \mathrm { Nm } ^ { - 2 } \right)

A

6×10−2J6 \times 10 ^ { - 2 } J

B

4×10−2J4 \times 10 ^ { - 2 } J

C

2×10−2J2 \times 10 ^ { - 2 } J

D
Answer

2×10−2J2 \times 10 ^ { - 2 } J

Explanation

Solution

W=YAl22L=2×1010×10−6×(10−3)22×50×10−2W = \frac { Y A l ^ { 2 } } { 2 L } = \frac { 2 \times 10 ^ { 10 } \times 10 ^ { - 6 } \times \left( 10 ^ { - 3 } \right) ^ { 2 } } { 2 \times 50 \times 10 ^ { - 2 } }