Question

In a u-tube as shown in figure, water and oil are in the left side and right side of the tube respectively. The heights from the bottom for water and oil columns are 15 cm and 20 cm respectively. The density of the oil is :- [take \(\rho_{water}\) = 1000 kg/m\(^3\)]

• A. 1200 kg/m$^3$
• B. 750 kg/m$^3$
• C. 1000 kg/m$^3$
• D. 1333 kg/m$^3$

Answer 1

The Correct Option is B

To find the density of the oil in the U-tube setup, we need to consider the principle of hydrostatic equilibrium, which states that the pressure exerted by a fluid column at a specific height is equal across the U-tube.

Given:

• Height of water column, h_{\text{water}} = 15 \text{ cm} = 0.15 \text{ m}
• Density of water, \rho_{\text{water}} = 1000 \text{ kg/m}^{3}
• Height of oil column, h_{\text{oil}} = 20 \text{ cm} = 0.20 \text{ m}
• Density of oil, \rho_{\text{oil}} (to be determined)

In equilibrium, the pressure at the bottom of both sides must be the same:

P_{\text{water}} = P_{\text{oil}}

The pressure due to a fluid column is given by P = \rho g h.

So for the water column:

P_{\text{water}} = \rho_{\text{water}} \cdot g \cdot h_{\text{water}}

For the oil column:

P_{\text{oil}} = \rho_{\text{oil}} \cdot g \cdot h_{\text{oil}}

Setting these pressures equal gives:

\rho_{\text{water}} \cdot g \cdot h_{\text{water}} = \rho_{\text{oil}} \cdot g \cdot h_{\text{oil}}

Cancelling out g from both sides, we get:

\rho_{\text{water}} \cdot h_{\text{water}} = \rho_{\text{oil}} \cdot h_{\text{oil}}

Substituting the given values:

1000 \cdot 0.15 = \rho_{\text{oil}} \cdot 0.20

Simplifying, we find:

\rho_{\text{oil}} = \frac{1000 \cdot 0.15}{0.20}

\rho_{\text{oil}} = \frac{150}{0.20}

\rho_{\text{oil}} = 750 \text{ kg/m}^{3}

Therefore, the density of the oil is 750 kg/m³.

Answer: 750 kg/m³