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]]
1.1×105 N
1.1×104 N
1.1×103 N
1.1×102 N
F=YAlL=2.2×1011×2×10−6×5 ×10−42=1.1×102N