Mumbai university > FE > SEM 1 > Applied Physics 1

Marks: 5M

Year: Dec 2013

Given:-

Thickness, b = 50 cm = 0.5 m

Pulse arrival time for a defect, $t_d = 40 μs = 40 × 10^{-6}s$

Pulse arrival time for the other end of the bar, $t_b = 90 μs = 90 × 10^{-6}s$

To Find: - The location of fault (x)

Solution:-

The thickness of the Steel bar can be expressed as,

$ b = \frac{vt_b}{2}$

$ ∴ v = \frac{2b}{t_b}$

The fault position can be determined by,

$ x = \frac{vt_d}{2}$

$ ∴ x = \frac{t_d}{2} × \frac{2b}{t_b}$

$ ∴ x = \frac{40 × 10^{-6}}{2} × \frac{2 × 0.5}{90 × 10^{-6}}$

$ ∴ x = 0.2222 m$

$ ∴ x = 22.22 cm$