A bucket water filled upto a height = 15 cm. The bucket is tied to a rope which is passed over a frictionless light pulley and the other end of the rope is tied to a weight of mass which is half of that of the (bucket + water). The water pressure above atmospheric pressure at the bottom is

Options

A. 0.5 kPa
B. 1 kPa
C. 5kPa
D. None of these

Answer

To solve the problem, we need to find the water pressure above atmospheric pressure at the bottom of a bucket filled with water. The bucket is tied to a rope that goes over a frictionless pulley, with a weight on the other end that is half the mass of the bucket plus water.

### Step-by-Step Solution:

1. **Identify the Variables:**
- Let \( h = 15 \, \text{cm} = 0.15 \, \text{m} \) (height of water in the bucket).
- Let \( m \) be the mass of the bucket filled with water.
- The mass of the weight tied to the rope is \( \frac{m}{2} \).

2. **Determine the Forces Acting on the Bucket:**
- The forces acting on the bucket are:
- The gravitational force acting downwards: \( mg \)
- The tension in the rope acting upwards: \( T \)
- According to Newton's second law, for the bucket:
\[ mg - T = ma \]
- Here, \( a \) is the acceleration of the bucket.

3. **Determine the Forces Acting on the Weight:**
- The forces acting on the weight are:
- The gravitational force acting downwards: \( \frac{m}{2}g \)
- The tension in the rope acting upwards: \( T \)
- For the weight, we can write:
\[ T - \frac{m}{2}g = -\frac{m}{2}a \]

4. **Solve the Equations:**
- From the first equation:
\[ T = mg - ma \]
- From the second equation:
\[ T = \frac{m}{2}g - \frac{m}{2}a \]
- Setting the two expressions for \( T \) equal:
\[ mg - ma = \frac{m}{2}g - \frac{m}{2}a \]
- Simplifying this gives:
\[ mg - \frac{m}{2}g = ma - \frac{m}{2}a \]
- Factoring out \( m \):
\[ \frac{1}{2}mg = \frac{1}{2}ma \]
- Thus, we find:
\[ g = a \]

5. **Find the Effective Gravitational Force:**
- The effective gravitational force acting on the water in the bucket is:
\[ g_{\text{eff}} = g - \frac{g}{3} = \frac{2g}{3} \]

6. **Calculate the Pressure at the Bottom of the Bucket:**
- The pressure at the bottom of the bucket is given by:
\[ P = h \cdot \rho \cdot g_{\text{eff}} = h \cdot \rho \cdot \frac{2g}{3} \]
- Substituting \( h = 0.15 \, \text{m} \):
\[ P = 0.15 \cdot \rho \cdot \frac{2g}{3} \]
- Assuming \( \rho = 1000 \, \text{kg/m}^3 \) (density of water) and \( g \approx 9.81 \, \text{m/s}^2 \):
\[ P = 0.15 \cdot 1000 \cdot \frac{2 \cdot 9.81}{3} = 0.15 \cdot 1000 \cdot 6.54 \approx 981 \, \text{Pa} = 0.981 \, \text{kPa} \]

7. **Final Result:**
- The pressure above atmospheric pressure at the bottom of the bucket is approximately \( 1 \, \text{kPa} \).