Given –

$d_2=50cm$

$\Delta t_1=40 \mu sec, \Delta t_2=90 \mu sec$

Solution –

Let $d_1$ be the location of defect & $d_2$ be the thickness of steel bar.

Let $\Delta t_1$ be echo time from defect & $\Delta t_2$ be the echo time for rod.

Now, we know that,

$d=\dfrac {V\cdot \Delta t}{2}$

$\therefore V=\dfrac {2d}{\Delta t}$

∵ Velocity of ultrasonic waves is constant.

$\therefore V_{1} =V_{2} $

$\dfrac {2d_{1}}{\Delta t_{1}} = \dfrac {2d_{2}}{\Delta t_{2}}$

$\dfrac {d_{1}}{40}=\dfrac {d_{2}}{90}$

$d_{1}=\dfrac {4}{9}\times d_{2}\ = \ \dfrac{4}{9}\times 50$

$d_{1}=22.22cm$

∴ The position of defect is at 22.22 cm.