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 merc

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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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Xavia 5 years 2021-08-10T13:30:38+00:00 1 Answers 17 views 0

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  1. thugiang
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    2021-08-10T13:31:53+00:00
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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, \rho_m = 14\times 10^3\ kg.m^-^3  

    Density of the water, \rho_w = 1\times 10^3\ kg.m^-^3

    Water poured on the left arm of the U-tube, h = 10\ cm

    According to the question:

    The mercury in the right arm rises above its original.

    We have to find this rise, h_2 .

    Lets  take a horizontal reference line on the U-tube.

    Note:

    In a static fluid there is no horizontal variation of pressure.

    So,

    Considering P_1 = P_2 on left and right of the manometer.

    Hydro-static pressure = \rho gh

    ⇒ P_1=\rho_w\times g\times h

    ⇒ P_2=\rho_m\times g\times h_2

    ⇒ Equating both.

    ⇒ \rho_w\times g\times h =\rho_m\times g\times h_2

    ⇒ h_2=\frac{\rho_w\times g\times h}{\rho_m\times g}

    ⇒ h_2=\frac{\rho_w\times h}{\rho_m}

    ⇒ Plugging the values.

    ⇒ h_2=\frac{1\times 10^3\times 10}{14\times 10^3}

    ⇒ h_2=\frac{10}{14}

    ⇒ h_2=0.714 cm ≈ 0.72 cm

    So the rise of the mercury in the right arm is of 0.72 cm.

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