Question

If the diameter of fat globule in a cream separator is reduced to half and the rotational speed of the centrifuge increased to three times, the terminal settling velocity of fat globule is:

• A. decreased to 0.44 times
• B. increased to 0.44 times
• C. decreased to 2.25 times
• D. increased to 2.25 times

Hint:

Settling velocity in centrifuges depends strongly on both particle diameter (\(d^2\)) and rotational speed (\(\omega^2\)). Always square both effects before combining.

Answer 1

The Correct Option is D

Step 1: Recall terminal velocity relation. For a centrifuge, the terminal settling velocity of a particle (fat globule) is given by Stokes' law (modified for centrifugal acceleration): \[ V \propto d^2 \cdot \omega^2 \] where \(d =\) diameter of globule, and \(\omega =\) angular velocity of centrifuge.

Step 2: Apply given changes. - Diameter is reduced to half: \[ d_{\text{new}} = \frac{d}{2} \Rightarrow (d_{\text{new}})^2 = \frac{d^2}{4} \] - Rotational speed is increased 3 times: \[ \omega_{\text{new}} = 3\omega \Rightarrow (\omega_{\text{new}})^2 = 9\omega^2 \]

Step 3: Net effect on velocity. \[ V_{\text{new}} \propto \frac{d^2}{4} \times 9\omega^2 = \frac{9}{4}(d^2 \omega^2) \] \[ V_{\text{new}} = 2.25 \, V_{\text{old}} \]

Step 4: Final result. Hence, the terminal settling velocity increases 2.25 times. \[ \boxed{\text{Option (D)}} \]