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A long U-tube contains mercury (density = 14×103 kg/m3). When 10cm of water (density =1 .0 ×103 kg/m3) is poured into the left arm, the mercury in the right arm rises above its original level by

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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.

Cricetus

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

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