Question

A 18.5 cm long probe used in a diagnostic process consists of a thin 16.0 cm long plastic rod with a cross-sectional area of and a 2.50 cm long nylon tip. Both the rod and tip have a cross-sectional area of 0.955×10−6 m2.

The (compressive) Young's modulus of plastic is 3.51 GPa, and the compressive Young's modulus of the nylon used is 2.02 GPa.

The instrument is pressed against a bone with a compressive force of 105 N. What is the change in length of the instrument? (to 2 s.f and in mm)

Answer 1

As per the question,

Length of the probe $(l)=18.5cm$

Length of the plastic rod $(l_1)=16cm$

$\Delta l_1$ is the change in the length of the plastic rod

Length of the nylon tip $(l_2)=2.5cm$

$\Delta l_2$ is the change in the length of the nylon tip

cross sectional area of the plastic rod $=0.955\times 10^{-6}m^2$

Young's modulus of plastic $Y=3.51 \times 10^9 Pa$

Young's modulus of nylon $Y=2.02 \times 10^9Pa$

Compressive force =105N

$Y=\dfrac{Fl_1}{A\Delta l_1}$

$\Delta l_1=\dfrac{Fl_1}{YA}=\dfrac{16\times 105}{0.955\times 10^{-6}\times 3.51\times10^9}=0.501cm$

Similarly,

$\Delta l_2=\dfrac{Fl_2}{AY}=\dfrac{(2.5)\times 105}{0.955\times 10^{-6}\times 2.02\times 10^9}=0.136cm$

Hence, net increase in length = $\Delta l_1 +\Delta l_2=0.501+0.136=0.637cm$

or $=6.4mm$