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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.

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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