To analyze the three statements provided in the question, we will evaluate each statement step by step.
### Statement 1:
**Statement:** When an orifice is made in the middle of the wall of a vessel, the range of the liquid coming out of the orifice is equal to the height of the liquid.
**Solution:**
1. **Identify the height of the liquid (h):** Let the height of the liquid in the vessel be \( h \).
2. **Determine the velocity of the liquid exiting the orifice:** The velocity \( V \) of the liquid coming out of the orifice can be calculated using Torricelli’s theorem, which states:
\[
V = \sqrt{2gh}
\]
where \( g \) is the acceleration due to gravity.
3. **Calculate the time taken to fall a distance \( h/2 \):** The time \( t \) taken for the liquid to fall from the height \( h/2 \) to the ground can be found using the equation of motion:
\[
s = ut + \frac{1}{2}gt^2
\]
Here, \( s = \frac{h}{2} \), \( u = 0 \) (initial velocity), and rearranging gives:
\[
\frac{h}{2} = \frac{1}{2}gt^2 \implies t^2 = \frac{h}{g} \implies t = \sqrt{\frac{h}{g}}
\]
4. **Calculate the range (R):** The horizontal range \( R \) can be calculated as:
\[
R = V \cdot t = \sqrt{2gh} \cdot \sqrt{\frac{h}{g}} = \sqrt{2h^2} = h
\]
5. **Conclusion for Statement 1:** The range of the liquid coming out of the orifice is indeed equal to the height of the liquid, thus **Statement 1 is true**.
### Statement 2:
**Statement:** Liquid is flowing through two identical pipes A and B. Volume of liquid flowing per second through A and B are \( v_0 \) and \( 2v_0 \) respectively. Flow in A is turbulent and steady in B.
**Solution:**
1. **Understand the flow characteristics:** The flow in pipe A is turbulent, which typically occurs when the Reynolds number \( Re > 4000 \). For pipe B, the flow is stated to be steady.
2. **Relate flow rate to Reynolds number:** The Reynolds number is given by:
\[
Re = \frac{\rho V D}{\eta}
\]
where \( V \) is the velocity, \( D \) is the diameter, \( \rho \) is the density, and \( \eta \) is the viscosity.
3. **Calculate velocities for both pipes:** The flow rate \( Q \) is related to the velocity \( V \) and cross-sectional area \( A \) of the pipe:
\[
Q = A \cdot V
\]
For pipe A, let \( A \) be the area, then:
\[
v_0 = A \cdot V_A \quad \text{and for pipe B,} \quad 2v_0 = A \cdot V_B
\]
This implies \( V_B = 2V_A \).
4. **Determine the Reynolds numbers:** Since the flow rate in pipe B is greater than in A, the Reynolds number for B will also be greater than that of A. Therefore, if A is turbulent, B must also be turbulent.
5. **Conclusion for Statement 2:** Since it is stated that flow in A is turbulent and in B is steady, this is contradictory. Thus, **Statement 2 is false**.
### Statement 3:
**Statement:** Rate of flow of a viscous liquid through a pipe is directly proportional to the fourth power of the radius of the pipe.
**Solution:**
1. **Use Poiseuille’s Law:** The rate of flow \( Q \) for a viscous fluid through a pipe is given by:
\[
Q = \frac{\pi r^4 (P_1 - P_2)}{8 \eta L}
\]
where \( P_1 - P_2 \) is the pressure difference, \( \eta \) is the viscosity, and \( L \) is the length of the pipe.
2. **Identify the relationship:** From the equation, we can see that the flow rate \( Q \) is proportional to \( r^4 \):
\[
Q \propto r^4
\]
3. **Conclusion for Statement 3:** Therefore, the statement that the rate of flow of a viscous liquid through a pipe is directly proportional to the fourth power of the radius of the pipe is **true**.
### Final Conclusion:
- **Statement 1:** True
- **Statement 2:** False
- **Statement 3:** True
Thus, the final answer is **TFT** (True, False, True).
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