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1. A weight of 1.0 kg is suspended from the lower endof a wire of cross-section 10 sq. mm. Find the stressproduced in it. |
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Answer» Given - m = 1kg A = 10 mm2 = 10^-5 m2 g = 9.8 m/s2 F = ? T = ? Solution - For suspended weight, F = mg F = 1×9.8 F = 9.8 N Stress produced is given by, T = F/A T = 9.8/(10⁻⁵ T = 9.8 × 10⁵ Pa T = 980 kPa Stress produced is 980 kPa in vertically upward direction. |
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