Question

A tube of length 50 cm is filled completely with an incompressible liquid of mass 250 g and closed at both ends. The tube is then rotated in horizontal plane about one of its ends with a uniform angular velocity x√F rad s⁻¹. If F be the force exerted by the liquid at the other end then the value of x will be______.

Answer 1

Correct Answer: 4

To solve this problem, we need to compute the angular velocity \(x\) such that the rotational force exerted by the liquid at the other end of the tube matches \(F\). The tube length \(L\) is 50 cm (0.5 m), and the mass \(m\) of the liquid is 250 g (0.25 kg). Using the concept of rotating systems and centrifugal force, we establish the following:

The centrifugal force exerted by a rotating liquid column of small element \(dm\) at a distance \(r\) from the axis of rotation is given by \(dF = r \cdot d(mr) \cdot \omega^2\), where \(\omega\) is the angular velocity. By integration over the length of the tube, the total force \(F\) at the other end when the system rotates is:

\(F = \int_0^L r \cdot \omega^2 \cdot \frac{dm}{dr} \cdot dr\)

Given that \(dm = \frac{m}{L} \cdot dr\) (uniform mass distribution), substitute into the equation:

\(F = \int_0^L r \cdot \omega^2 \cdot \frac{m}{L} \cdot dr = \frac{m \cdot \omega^2}{L} \cdot \int_0^L r \cdot dr\)

The integration simplifies as:

\(F = \frac{m \cdot \omega^2}{L} \cdot \left[\frac{r^2}{2}\right]_0^L = \frac{m \cdot \omega^2}{L} \cdot \frac{L^2}{2} = \frac{m \cdot \omega^2 \cdot L}{2}\)

Substitute the known values to solve for \(\omega\):

\(F = \frac{0.25 \cdot \omega^2 \cdot 0.5}{2} \Rightarrow F = 0.0625 \cdot \omega^2\)

Since \( \omega = x\sqrt{F} \), substitute and solve for \(x\):

\(F = 0.0625 \cdot (x\sqrt{F})^2 \Rightarrow F = 0.0625 \cdot x^2 \cdot F\)

Simplifying gives \(1 = 0.0625 \cdot x^2\) which leads to:

\(x^2 = \frac{1}{0.0625} = 16 \Rightarrow x = 4\)

Thus, the value of \(x\) is 4, confirming it falls within the expected range of 4 to 4.