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?

Options

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

Explanation

To solve the problem step by step, we need to determine how much liquid of density \(4 \, \text{g/cc}\) should be poured into one side of the U-tube so that no water remains in the horizontal arm.

### Step 1: Understand the setup
We have a U-tube with:
- Length of the horizontal arm = 20 cm
- Cross-sectional area = \(1 \, \text{cm}^2\)
- Volume of water = 60 cc

### Step 2: Calculate the height of water in the U-tube
The volume of water is given as 60 cc. Since the cross-sectional area is \(1 \, \text{cm}^2\), we can calculate the height of the water column in the U-tube using the formula:
\[ \text{Volume} = \text{Area} \times \text{Height} \]
Thus,
\[ 60 \, \text{cc} = 1 \, \text{cm}^2 \times \text{Height} \]
This gives us:
\[ \text{Height} = 60 \, \text{cm} \]

### Step 3: Determine the pressure balance
When we pour the liquid of density \(4 \, \text{g/cc}\) into one side of the U-tube, we want the pressure at the bottom of both sides to be equal.

Let:
- \(h\) = height of the liquid of density \(4 \, \text{g/cc}\) poured into one side.
- The pressure exerted by the water column on the left side will be \(P_1 = \rho_1 g h_1\) (where \(\rho_1 = 1 \, \text{g/cc}\) for water and \(h_1 = 60 \, \text{cm}\)).
- The pressure exerted by the liquid on the right side will be \(P_2 = \rho_2 g h_2\) (where \(\rho_2 = 4 \, \text{g/cc}\) and \(h_2 = h\)).

### Step 4: Set up the equation for pressure balance
Since the pressures at the same level must be equal:
\[ P_1 = P_2 \]
This gives us:
\[ \rho_1 g h_1 = \rho_2 g h_2 \]
Substituting the values:
\[ 1 \, \text{g/cc} \times 60 \, \text{cm} = 4 \, \text{g/cc} \times h \]
Cancelling \(g\) from both sides:
\[ 60 = 4h \]
Thus,
\[ h = \frac{60}{4} = 15 \, \text{cm} \]

### Step 5: Calculate the total volume of the liquid
Now, we need to find the total volume of the liquid in the U-tube:
1. The volume of the liquid of density \(4 \, \text{g/cc}\) in the left limb:
\[ V_{\text{left}} = \text{Area} \times \text{Height} = 1 \, \text{cm}^2 \times 15 \, \text{cm} = 15 \, \text{cm}^3 \]
2. The volume of the liquid in the horizontal arm:
\[ V_{\text{horizontal}} = \text{Area} \times \text{Length} = 1 \, \text{cm}^2 \times 20 \, \text{cm} = 20 \, \text{cm}^3 \]

### Step 6: Total volume of liquid
Adding both volumes:
\[ V_{\text{total}} = V_{\text{left}} + V_{\text{horizontal}} = 15 \, \text{cm}^3 + 20 \, \text{cm}^3 = 35 \, \text{cm}^3 \]

### Conclusion
The volume of the liquid of density \(4 \, \text{g/cc}\) that should be poured into the U-tube is \(35 \, \text{cm}^3\).

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