Respuesta :
ANSWER:
(a) 6.3 cm
(b) 0.42 cm
STEP-BY-STEP EXPLANATION:
Given:
Density of oil = 0.9 g/cm^3
Density of mercury = 13.6 g/cm^3
Rise in the height of the mercury = 8.33 mm of Hg = 0.833 cm of Hg
(a)
The change in pressure is:
[tex]\Delta P=\rho\cdot g\cdot\Delta h[/tex]The change in pressure when oil is used is:
[tex]\Delta P=\rho_{\text{oil}}\cdot g\cdot d[/tex]Equating the above two equation
[tex]\begin{gathered} \rho\cdot g\cdot\Delta h=\rho_{\text{oil}}\cdot g\cdot d \\ \text{ solving for d:} \\ d=\frac{\rho\cdot\Delta h}{\rho_{\text{oil}}} \\ d=\frac{13.6\cdot0.833}{0.9} \\ d=12.59\text{ cm} \end{gathered}[/tex]The amount of fluid rise is half the difference of level, so the height of rise in the oil
[tex]\begin{gathered} \Delta h_{oil}=\frac{d}{2} \\ \text{ replacing} \\ \Delta h_{oil}=\frac{12.59}{2}=6.295\cong6.3 \\ \Delta h_{oil}=6.3\text{ cm} \end{gathered}[/tex](b) If the manometer uses mercury then
[tex]\begin{gathered} \Delta h^{\prime}=\Delta h_{oil}\cdot\mleft(\frac{\rho_{oil}}{\rho}\mright) \\ \text{ replacing:} \\ \Delta h^{\prime}=6.3\cdot\mleft(\frac{0.9}{13.6}\mright) \\ \Delta h^{\prime}=0.42\text{ cm} \end{gathered}[/tex]