91Ó°ÊÓ

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 \(47.01 \times {10^6}\;{\rm{N}}{\rm{.}}\)