A tube of length 1m is filled completely with an ideal liquid of mass 2M, and closed at both ends. The tube is rotated uniformly in horizontal plane about one of its ends. If the force exerted by the liquid at the other end is \( F \) and the angular velocity of the tube is \( \omega \), then the value of \( \alpha \) is ______ in SI units.
A tube of length 1m is filled completely with an ideal liquid of mass 2M, and closed at both ends. The tube is rotated uniformly in horizontal plane about one of its ends. If the force exerted by the liquid at the other end is \( F \) and the angular velocity of the tube is \( \omega \), then the value of \( \alpha \) is ______ in SI units.
Show Hint
Correct Answer: 1
Approach Solution - 1
- Tube length \( L = 1 \) m
- Mass of liquid \( m = 2M \)
- Angular velocity \( \omega \)
For incremental mass \( dm = \rho \cdot dx = \frac{2M}{1} \cdot dx = 2M \cdot dx \).
Now, substituting in the expression for \( dF \):
\( dF = \omega^2 \cdot x \cdot (2M \cdot dx) = 2M \omega^2 x \, dx \).
The total force \( F \) exerted at the other end of the tube is obtained by integrating \( dF \) from 0 to 1 (length of the tube):
\( F = \int_0^1 2M \omega^2 x \, dx = 2M \omega^2 \left[\frac{x^2}{2}\right]_0^1 \)
\( F = 2M \omega^2 \cdot \frac{1^2}{2} = M \omega^2 \).
Thus, the value of \( \alpha \) is the coefficient of \( M \omega^2 \) in the expression for \( F \), which is 1. Therefore, the value of \( \alpha \) is 1. This solution fits the expected range of 1 to 1, confirming the accuracy of our calculation.
Approach Solution -2
Step 1: Concept.
When a tube filled with liquid rotates about one end in a horizontal plane,
each element of the liquid experiences a centrifugal force directed outward.
This produces a pressure variation along the tube.
Step 2: Consider a small element of liquid.
Let the tube have:
\[
\text{Length} = L = 1\,\text{m}, \quad \text{Total mass of liquid} = 2M.
\]
Thus, the linear mass density is:
\[
\lambda = \frac{2M}{L} = 2M.
\]
Consider a small element of length \( dx \) at a distance \( x \) from the axis of rotation. The centrifugal force on this element is: \[ dF = \lambda \omega^2 x \, dx. \]
Step 3: Pressure variation.
The differential pressure on this element is related by:
\[
\frac{dp}{dx} = \lambda \omega^2 x.
\]
Integrating from \( x = 0 \) (axis of rotation) to \( x = L \) (free end):
\[
p = \int_0^L \lambda \omega^2 x \, dx = \frac{1}{2}\lambda \omega^2 L^2.
\]
Step 4: Force on the closed end.
Since pressure \( p \) acts uniformly over the cross-sectional area \( A \) of the tube:
\[
F = pA = \frac{1}{2}\lambda \omega^2 L^2 A.
\]
But total mass of the liquid \( m = \lambda L = 2M \Rightarrow \lambda = \frac{2M}{L}.
\]
Substitute:
\[
F = \frac{1}{2} \left(\frac{2M}{L}\right) \omega^2 L^2 A = M \omega^2 L A.
\]
For \( L = 1\,\text{m} \):
\[
F = M \omega^2 A.
\]
Hence, the constant \( \alpha \) in \( F = \alpha \omega^2 \) is:
\[
\boxed{\alpha = M}.
\]
Final Answer:
\[ \boxed{\alpha = M} \]
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