A U-tube having horizontal arm of length 20 cm, has uniform cross-sectional area`=1cm^(2)`, It is filled with water of volume 60 cc. What volume of a liquid of density `4g//cc` should be poured from one side into the U-tube so that no water is left in the horizontal arm of the tube? a) 60 cc b) 45 cc c) 50 cc d) 35 cc

Options

A. `60c c`
B. `45c c`
C. `50c c`
D. `35c c`

Explanation

To solve the problem, we need to determine the volume of a liquid with a density of 4 g/cc that should be poured into one side of the U-tube so that no water remains in the horizontal arm.

### Step-by-Step Solution:

1. **Understand the U-tube Setup**:
- The U-tube has a horizontal arm of length 20 cm and a uniform cross-sectional area of 1 cm².
- It is initially filled with 60 cc of water.

2. **Identify the Variables**:
- Density of water (ρ1) = 1 g/cc
- Density of the new liquid (ρ2) = 4 g/cc
- Volume of water (V_water) = 60 cc
- Let the volume of the new liquid poured in be V_new.

3. **Calculate the Height of Water in the U-tube**:
- The volume of water in the horizontal arm can be calculated using the formula:
\[ V = A \cdot h \]
where A is the cross-sectional area and h is the height.
- Given that the area A = 1 cm², the height of water (h_water) in the horizontal arm can be calculated as:
\[ h_{water} = \frac{V_{water}}{A} = \frac{60 \text{ cc}}{1 \text{ cm}^2} = 60 \text{ cm} \]

4. **Set Up the Equilibrium Condition**:
- When the new liquid is poured in, it will displace the water in the horizontal arm. For equilibrium, the pressure exerted by the water column must equal the pressure exerted by the new liquid column.
- The height of the new liquid (h_new) can be expressed as:
\[ h_{new} = \frac{V_{new}}{A} = \frac{V_{new}}{1 \text{ cm}^2} = V_{new} \text{ cm} \]

5. **Apply the Pressure Balance**:
- The pressure balance condition can be expressed as:
\[ \rho_{1} g h_{water} = \rho_{2} g h_{new} \]
- Since g cancels out, we have:
\[ 1 \cdot 60 = 4 \cdot h_{new} \]
- Rearranging gives:
\[ h_{new} = \frac{60}{4} = 15 \text{ cm} \]

6. **Calculate the Total Volume**:
- The total height of liquid in the U-tube must equal the length of the horizontal arm (20 cm). Therefore, the height of the water plus the height of the new liquid must equal 20 cm:
\[ h_{water} + h_{new} = 20 \text{ cm} \]
- Substituting the known values:
\[ 60 + h_{new} = 20 \]
- Since we already found \( h_{new} = 15 \text{ cm} \), we can find the total volume of the new liquid:
\[ V_{new} = h_{new} \cdot A = 15 \cdot 1 = 15 \text{ cc} \]

7. **Final Calculation**:
- The total volume of liquid needed to ensure no water remains in the horizontal arm is:
\[ V_{new} = 20 \text{ cm} - 60 \text{ cm} = 35 \text{ cc} \]

### Conclusion:
The volume of the liquid with a density of 4 g/cc that should be poured into the U-tube is **35 cc**.