A U-tube with both ends open to the atmosphere is partially filled with water. Oil, which is immiscible with w

Question

A U-tube with both ends open to the atmosphere is partially filled with water. Oil, which is immiscible with water, is poured into one side until it stands at a distance of $10 	ext{ mm}$ above the water level on the other side. Meanwhile, the water rises by $65 	ext{ mm}$ from its original level (see diagram). The density of the oil is:

Accepted Answer

1. **Identify the Principle:** In a U-tube containing static fluids, the pressure at the same horizontal level in a continuous fluid must be equal . Let's choose the interface between the oil and water in the left arm as our reference level.
2. **Determine Heights:**
   - The water rises by $65 	ext{ mm}$ from its original level in the right arm. Since the liquid is incompressible, the water level must have gone down by $65 	ext{ mm}$ in the left arm. Therefore, the total height of the water column above the reference level (interface) in the right arm is $h_{water} = 65 	ext{ mm} + 65 	ext{ mm} = 130 	ext{ mm}$.
   - The oil column in the left arm stands $10 	ext{ mm}$ above the final water level in the right arm. So, the total height of the oil column is $h_{oil} = h_{water} + 10 	ext{ mm} = 130 + 10 = 140 	ext{ mm}$.
3. **Set up the Pressure Equation:** At the reference level (interface), the pressure due to the oil column equals the pressure due to the water column.
   $$ P_{oil} = P_{water} $$
   $$ ho_{oil} g h_{oil} = ho_{water} g h_{water} $$
   $$ ho_{oil} h_{oil} = ho_{water} h_{water} $$
4. **Calculate Density:**
   - $ho_{water} = 1000 	ext{ kg m}^{-3}$ 
   - $h_{water} = 130 	ext{ mm}$
   - $h_{oil} = 140 	ext{ mm}$
   $$ ho_{oil} = ho_{water} 	imes \frac{h_{water}}{h_{oil}} $$
   $$ ho_{oil} = 1000 	imes \frac{130}{140} = 1000 	imes \frac{13}{14} \approx 928.57 	ext{ kg m}^{-3} $$
   Rounding to the nearest integer, we get $928 	ext{ kg m}^{-3}$.

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