A 3.2-N force is applied to the plunger of a hypodermic needle. If the diameter of the plunger is 1.3 cm and that of the needle is 0.20 mm, (a) with what force does the fluid leave the needle? (b) What force on the plunger would be needed to push fluid into a vein where the gauge pressure is 75 mm-Hg? Answer for the instant just before the fluid starts to move.

The force applied to the plunger is\({F_{plunger}} = 3.2\;{\rm{N}}\).

The diameter of the plunger is\(D = 1.3\;{\rm{cm}}\).

The radius of the plunger will be\(R = 6.5\;{\rm{cm}} = 6.5 \times {10^{ - 2}}\;{\rm{m}}\).

The diameter of the needle is\(d = 0.20\;{\rm{mm}}\).

The radius of the needle will be \(r = 0.10\;{\rm{mm}} = 0.1 \times {10^{ - 3}}\;{\rm{m}}\).

The pressure of the fluid is given as,

\(P = \frac{F}{A}\)

The Pascal’s principle states that if an external pressure is applied to a confined fluid, the pressure at every point within the fluid increases by that amount.

Therefore, we know

\(\begin{array}{l}{P_{plunger}} = {P_{needle}}\\\frac{{{F_{plunger}}}}{{{A_{plunger}}}} = \frac{{{F_{needle}}}}{{{A_{needle}}}}\end{array}\)

As, Area is given by,

\(A = \pi {r^2}\)

We have,

\(\begin{array}{l}\frac{{{F_{plunger}}}}{{\pi {R^2}}} = \frac{{{F_{needle}}}}{{\pi {r^2}}}\\\frac{{{F_{plunger}}}}{{{R^2}}} = \frac{{{F_{needle}}}}{{{r^2}}}\\{F_{needle}} = \frac{{{F_{plunger}} \times {r^2}}}{{{R^2}}}\end{array}\)

Substituting the given values,

\(\begin{array}{l}{F_{needle}} = \frac{{3.2 \times {{(0.1 \times {{10}^{ - 3}})}^2}}}{{{{(6.5 \times {{10}^{ - 2}})}^2}}}\\{F_{needle}} = \frac{{3.2 \times 1 \times {{10}^{ - 8}}}}{{42.25 \times {{10}^{ - 4}}}}\;{\rm{N}}\\{F_{needle}} = 0.0757 \times {10^{ - 8 + 4}}\;{\rm{N}}\\{F_{needle}} = 7.57 \times {10^{ - 6}}\;{\rm{N}}\end{array}\)

The magnitude of force the fluid leaves the needle is \(7.57 \times {10^{ - 6}}\;{\rm{N}}{\rm{.}}\)

The the gauge pressure in the vein is\({P_G} = 75\;{\rm{mm - Hg}} = 75 \times 0.133\;{\rm{kPa}} = 9.975\;{\rm{kPa}}\).

Also, the atmospheric pressure is\({P_ \circ } = 101.3\;{\rm{kPa}}\)

Therefore, total pressure will be,

\(\begin{array}{l}P = {P_G} + {P_ \circ }\\P = 9.975 + 101.3\;{\rm{kPa}}\\P = 111.275\;{\rm{kPa}}\end{array}\).

Thus, we have \({P_{needle}} = P = 111.275\;{\rm{kPa}}\).

Substituting the values in following expression,

\(\begin{array}{l}\frac{{{F_{plunger}}}}{{{R^2}}} = \frac{{{F_{needle}}}}{{{r^2}}}\\{F_{plunger}} = \frac{{{F_{needle}} \times {R^2}}}{{{r^2}}}\end{array}\)

We have,

\(\begin{array}{l}{F_{plunger}} = \frac{{111.275 \times {{(6.5 \times {{10}^{ - 2}})}^2}}}{{{{(0.1 \times {{10}^{ - 3}})}^2}}}\\{F_{plunger}} = \frac{{111.275 \times 42.25 \times {{10}^{ - 4}}}}{{1 \times {{10}^{ - 8}}}}\;{\rm{N}}\\{F_{plunger}} = 47.01 \times {10^{ - 2 + 8}}\;{\rm{N}}\\{F_{plunger}} = 47.01 \times {10^6}\;{\rm{N}}\end{array}\)

Hence, force on the plunger required to push fluid into a vein where the gauge pressure is 75 mm-Hg is \(47.01 \times {10^6}\;{\rm{N}}{\rm{.}}\)