To determine which of the statements provided in the question are correct, we will analyze each statement one by one.
### Step 1: Analyze Statement A
**Statement A:** A uniform capillary is held vertically with its lower end dipped in water. The water rises to 10 cm. If the capillary is now broken at a level of 6 cm above the water level in the vessel, a fountain will be observed.
**Analysis:**
- When the capillary is broken at 6 cm, the water column above this break will no longer be supported by atmospheric pressure, and thus, it will not rise to form a fountain.
- This scenario would imply a perpetual motion machine, which is not possible.
**Conclusion:** Statement A is incorrect.
### Step 2: Analyze Statement B
**Statement B:** Water is flowing through a capillary in streamline fashion at rate Q. For a capillary of 2-fold radius, other factors remaining unchanged, the flow rate will be 2q.
**Analysis:**
- The flow rate \( I \) is given by \( I = A \cdot v \), where \( A \) is the cross-sectional area and \( v \) is the velocity.
- The area \( A \) is proportional to the square of the radius \( r \). If the radius is doubled (2-fold), then the new area \( A' = \pi (2r)^2 = 4\pi r^2 = 4A \).
- Since the area increases by a factor of 4, the flow rate will also increase by a factor of 4, not 2.
**Conclusion:** Statement B is incorrect.
### Step 3: Analyze Statement C
**Statement C:** For a soap bubble of outer surface area S and tension T, the surface energy will be 2Ts.
**Analysis:**
- The surface energy \( SE \) of a soap bubble is given by \( SE = 4T \times S \) because a soap bubble has two surfaces (inner and outer).
- Therefore, the surface energy cannot be simply \( 2Ts \).
**Conclusion:** Statement C is incorrect.
### Step 4: Analyze Statement D
**Statement D:** When a sprayer changes one drop of liquid into \( 10^6 \) droplets, the surface energy is increased by a factor of \( 10^4 \).
**Analysis:**
- Let the volume of one drop be \( V_1 = \frac{4}{3} \pi R^3 \) and the volume of one droplet be \( V_2 = \frac{4}{3} \pi r^3 \).
- If one drop is converted into \( 10^6 \) droplets, then \( V_1 = 10^6 V_2 \).
- This implies \( R^3 = 10^6 r^3 \) or \( R = 10^{-2} r \).
- The surface area \( A \) is proportional to the square of the radius, so \( A_1/A_2 = R^2/r^2 = (10^{-2})^2 = 10^{-4} \).
- The initial surface energy \( SE_1 = 2T \cdot A_1 \) and the final surface energy for all droplets combined \( SE_2 = 10^6 \cdot 2T \cdot A_2 \).
- Thus, \( SE_2 = 10^6 \cdot 2T \cdot (A_1/10^4) = 10^2 \cdot SE_1 \).
**Conclusion:** Statement D is incorrect.
### Final Conclusion
None of the statements A, B, C, or D are correct.
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