Chapter 10, Problem 64GP

(a)

To determine

To Calculate: The force.

Answer to Problem 64GP

  $0.0057\text{N}$.

Explanation of Solution

Given info:

Force, $\mathrm{F}=2.4\text{N}$

Plunger’s diameter, d= ${\mathrm{d}}_1=1.3\text{cm}=0.013\text{m}$

Needle’s diameter, ${\mathrm{d}}_2=0.20\times {10}^{-3}\text{m}$

Pressure in second case, $\mathrm{P}=18\text{mm-Hg}=2400\text{Pa}$

Formula used:

• According to Pascal’s law

  ${\mathrm{P}}_1={\mathrm{P}}_2$

Here, ${\mathrm{P}}_1$ is the pressure at Plunger,

  ${\mathrm{P}}_2$ is the pressure at needle,

• $\frac{{\mathrm{F}}_1}{{\mathrm{A}}_1}=\frac{{\mathrm{F}}_2}{{\mathrm{A}}_2}$

Here, ${\mathrm{F}}_1$ is the force at plunger,

  ${\mathrm{F}}_2$ is the force at needle,

  ${\mathrm{A}}_1$ is the area of cross-section of plunger,

  ${\mathrm{A}}_2$ is the area of cross-section of needle.

Calculation:

Write the expression for the area at plunger and needle.

  $\begin{array}{l}{\mathrm{A}}_1=\mathrm{\pi }{\mathrm{r}}^2=\mathrm{\pi }\frac{{\mathrm{d}}_1{}^2}{4}\\ =\frac{\mathrm{\pi }\times {\left(1.3\times {10}^{-2}\right)}^2}{4}\end{array}$

And,

  $\begin{array}{c}{\mathrm{A}}_2=\mathrm{\pi }\frac{{\mathrm{d}}_2{}^2}{4}\\ =\frac{\mathrm{\pi }{\left(0.2\times {10}^{-3}\right)}^2}{4}\end{array}$

Now using the pascal relation,

  ${\mathrm{P}}_1={\mathrm{P}}_2$

  $\frac{{\mathrm{F}}_1}{{\mathrm{A}}_1}=\frac{{\mathrm{F}}_2}{{\mathrm{A}}_2}$

  $\begin{array}{l}{\mathrm{F}}_2=\frac{{\mathrm{A}}_1}{{\mathrm{A}}_2}{\mathrm{F}}_1\\ =\frac{\frac{\mathrm{\pi }\times {\left(0.013\right)}^2}{4}\times 2.4\text{N}}{\frac{\mathrm{\pi }{\left(0.0002\right)}^2}{4}}=\frac{{\left(0.013\right)}^2\times 2.4\text{N}}{{\left(0.0002\right)}^2}=0.0057\text{N}\end{array}$

Conclusion:

The force is $0.0057\text{N}$ .

(b)

To determine

To calculate: The force.

Answer to Problem 64GP

The force is $7.54\times {10}^{-5}\text{N}$.

Explanation of Solution

Given info:

Force, $\mathrm{F}=2.4\text{N}$

Plunger’s diameter, d= ${\mathrm{d}}_1=1.3\text{cm}=0.013\text{m}$

Needle’s diameter, ${\mathrm{d}}_2=0.20\times {10}^{-3}\text{m}$

Pressure in second case, $\mathrm{P}=18\text{mm-Hg}=2400\text{Pa}$

Formula used:

• According to Pascal’s law

  ${\mathrm{P}}_1={\mathrm{P}}_2$

Here, ${\mathrm{P}}_1$ is the pressure at Plunger,

  ${\mathrm{P}}_2$ is the pressure at needle,

• $\frac{{\mathrm{F}}_1}{{\mathrm{A}}_1}=\frac{{\mathrm{F}}_2}{{\mathrm{A}}_2}$

Here, ${\mathrm{F}}_1$ is the force at plunger,

  ${\mathrm{F}}_2$ is the force at needle,

  ${\mathrm{A}}_1$ is the area of cross-section of plunger,

  ${\mathrm{A}}_2$ is the area of cross-section of needle.

Calculation:

Pressure is given as,

  $\mathrm{P}=\frac{\mathrm{F}}{\mathrm{A}}$

  $\begin{array}{l}\mathrm{F}=\mathrm{P}\mathrm{A}=2400\text{Pa}\times \left(\mathrm{\pi }{\mathrm{r}}^2\right)\\ =2400\text{Pa}\times \left(3.14\times {\left(\frac{0.20\times {10}^{-3}\text{m}}{2}\right)}^2\right)\\ =7.54\times {10}^{-5}\text{N}\end{array}$

Conclusion:

The force is $7.54\times {10}^{-5}\text{N}$ .