Question

A U-tube is partially filled with water. Oil which does not mix with water is next poured into one side of the U-tube until entire water rises by 25 cm on the other side. If the density of oil is 0.8 g cm\(^{-3}\), the oil level will stand higher than the water level by:

• A. 6.25 cm
• B. 12.50 cm
• C. 31.75 cm
• D. 63.50 cm

Hint:

In fluid statics, the balance of pressures in connected columns of different fluids can be used to determine the heights of the fluid columns.

Answer 1

The Correct Option is B

Given:
Density of oil, \( \rho_{{oil}} = 0.8 \, {g cm}^{-3} \)
Density of water, \( \rho_{{water}} = 1 \, {g cm}^{-3} \)
Rise in water level on one side, \( h_{{water}} = 25 \, {cm} \)
Step 1: Determine the height of the oil column The pressure at the bottom of the U-tube must be the same on both sides. 
Therefore, the pressure due to the oil column must balance the pressure due to the water column. \[ \rho_{{oil}} \times g \times h_{{oil}} = \rho_{{water}} \times g \times h_{{water}} \] \[ 0.8 \times h_{{oil}} = 1 \times 25 \] \[ h_{{oil}} = \frac{25}{0.8} = 31.25 \, {cm} \] Step 2: Calculate the difference in height The oil level will stand higher than the water level by: \[ \Delta h = h_{{oil}} - h_{{water}} = 31.25 - 25 = 6.25 \, {cm} \] However, considering the rise in water level on the other side, the total difference in height is: \[ \Delta h_{{total}} = 2 \times 6.25 = 12.50 \, {cm} \] Final Answer:  12.50 cm