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Q.

On increasing the length by 0.5 mm in a steel wire of length 2 m and area of cross-section  2 mm2, the force required is [Y for steel =2.2×1011 N/m2]]

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a

1.1×105 N

b

1.1×104 N

c

1.1×103 N

d

1.1×102 N

answer is D.

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Detailed Solution

F=YAlL=2.2×1011×2×10−6​×5​ ×​10−42=1.1×102N
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