Respuesta :
Answer:
A long U-tube contains mercury (density = 14 × 103 kg/m^3). When 10 cm of water (density = 1.0 × 103 kg/m^3) is poured into the left arm, the mercury in the right arm rises above its original level by:
A. 0.36 cm B. 0.72 cm C. 14 cm D. 35 cm
Option B is the right choice as the rise in the mercury level is of 0.72 cm.
Explanation:
Given:
Density of the mercury, [tex]\rho_m[/tex] = [tex]14\times 10^3\ kg.m^-^3[/tex]
Density of the water, [tex]\rho_w[/tex] = [tex]1\times 10^3\ kg.m^-^3[/tex]
Water poured on the left arm of the U-tube, [tex]h[/tex] = [tex]10\ cm[/tex]
According to the question:
The mercury in the right arm rises above its original.
We have to find this rise, [tex]h_2[/tex] .
Lets take a horizontal reference line on the U-tube.
Note:
In a static fluid there is no horizontal variation of pressure.
So,
Considering [tex]P_1[/tex] = [tex]P_2[/tex] on left and right of the manometer.
Hydro-static pressure = [tex]\rho gh[/tex]
⇒ [tex]P_1=\rho_w\times g\times h[/tex]
⇒ [tex]P_2=\rho_m\times g\times h_2[/tex]
⇒ Equating both.
⇒ [tex]\rho_w\times g\times h =\rho_m\times g\times h_2[/tex]
⇒ [tex]h_2=\frac{\rho_w\times g\times h}{\rho_m\times g}[/tex]
⇒ [tex]h_2=\frac{\rho_w\times h}{\rho_m}[/tex]
⇒ Plugging the values.
⇒ [tex]h_2=\frac{1\times 10^3\times 10}{14\times 10^3}[/tex]
⇒ [tex]h_2=\frac{10}{14}[/tex]
⇒ [tex]h_2=0.714[/tex] cm ≈ [tex]0.72[/tex] cm
So the rise of the mercury in the right arm is of 0.72 cm.
The distance of the mercury will be "0.36 cm".
Given:
Pressure due to depth of water,
- [tex]d_w = 10 \ cm[/tex]
Now,
→ [tex]P_1 = P_{atm}+ \rho_w gd_w[/tex]
When the mercury is incompressible, the pressure due to depth's mercury will be:
- [tex]d_{Hg} = 2h[/tex]
then,
→ [tex]P_2 = P_{atm}+ \rho_{Hg} gd_{Hg}[/tex]
[tex]= P_{atm}+2 \rho_{Hg} gh[/tex]
By equating first as well as the second pressure we get
→ [tex]P_{atm}+ \rho_w gd_w = P_{atm}+ 2 \rho_{Hg} gh[/tex]
[tex]\rho_w d_w = 2 \rho_Hg h[/tex]
[tex]h = \frac{1}{2} \frac{\rho_w d_w}{\rho_{Hg}}[/tex]
By substituting the values, we get
[tex]= 0.5\times \frac{1000\times 0.10}{14000}[/tex]
[tex]= 0.36 \ cm[/tex]
Thus the above answer is correct.
Learn more about mercury here:
https://brainly.com/question/16267969
