A U-tube whose limbs are vertical and of equal length contains mercury of relative density 13.6. The surface of the mercury is 15 cm from the top of each limb. The cross-sectional area of the U-tube is 10 cm². 100 cm³ of oil, of relative density 0.68, is poured into one limb. The surface of the oil is x cm from the top of the limb. Find the value of x. A uniform rod, of length ℓ and weight Wₗ, is freely hinged at the point P. The rod is free to move about a horizontal axis through P. The other end of the rod is immersed in water. The relative density of the rod is 0.64. The rod is in equilibrium and is inclined as shown in the diagram. Find i) the length of the immersed part of the rod in terms of ℓ ii) the reaction at the hinge in terms of W. [Density of water = 1000 kg m⁻³]

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A U-tube whose limbs are vertical and of equal length contains mercury of relative density 13.6 - Leaving Cert Applied Maths - Question 9 - 2020

Question 9

A U-tube whose limbs are vertical and of equal length contains mercury of relative density 13.6. The surface of the mercury is 15 cm from the top of each limb. The c... show full transcript

Worked Solution & Example Answer:A U-tube whose limbs are vertical and of equal length contains mercury of relative density 13.6 - Leaving Cert Applied Maths - Question 9 - 2020

Step 1

Find the value of x

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Answer

To find the value of x in the U-tube, we need to set up the balance of pressures caused by the mercury and oil in the tube.

Calculate the height of the oil column. The mercury level difference in the two limbs can be expressed in terms of the heights. Use the relation for pressures at equal levels: $P_A = P_B$ where: $P_A = ext{Density}_{ ext{mercury}} imes g imes h_{A}$ and $P_B = ext{Density}_{ ext{oil}} imes g imes h_{B}$ Given:
- Density of mercury = $13.6 imes 1000 ext{ kg/m}^3$
- Density of oil = $0.68 imes 1000 ext{ kg/m}^3$
- The height difference of mercury is 15 cm (0.15 m), so $h_A = 10 ext{ cm} (0.1 m)$ and $h_B = x ext{ cm}$ This leads us to: $680 imes 10^{-2} = 13600 imes 2 imes 10^{-2}$
Solving for x From the balance of heights, we have: $x + 10 - 15 = y$ Solving yields: $x = 5.25 ext{ cm}$

Step 2

the length of the immersed part of the rod in terms of ℓ

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Answer

Let the length of the immersed part of the rod be x. Using the relative densities, we know:

For the rod: $B = rac{W_L}{0.64}$ Using equations of balance of forces:

Using the sine relation: $B \cdot \left( ℓ - \frac{x}{2} \right) \sin(\alpha) = W \cdot \frac{ℓ}{2} \sin(\alpha)$ $B \cdot (ℓ - \frac{x}{2}) sin (α) = W \cdot \frac{ℓ}{2} sin (α)$
- This simplifies to: $x = 0.4ℓ$ Thus, the length of the immersed part can be simplified to $0.4ℓ$ .

Step 3

the reaction at the hinge in terms of W

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Answer

To find the reaction force at the hinge, we can use the equilibrium equations. Using the equation:

The total vertical reactions: $R + B + \frac{x \cdot W}{0.64} = W$ From the previously calculated B, we can substitute: $R = \frac{W}{3}$ Thus, the reaction at the hinge in terms of W is: $R = \frac{W}{3}$

A U-tube whose limbs are vertical and of equal length contains mercury of relative density 13.6 - Leaving Cert Applied Maths - Question 9 - 2020

Worked Solution & Example Answer:A U-tube whose limbs are vertical and of equal length contains mercury of relative density 13.6 - Leaving Cert Applied Maths - Question 9 - 2020

Find the value of x

the length of the immersed part of the rod in terms of ℓ

the reaction at the hinge in terms of W

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