MHT CET · Physics · Mechanical Properties of Fluids
A can filled with water is revolved in a vertical circle of radius \(r\) with constant speed and water just does not fall down. The time period of revolution is \((g=\) acceleration due to gravity)
- A \(2 \pi \sqrt{r g}\)
- B \(2 \pi \sqrt{5 r g}\)
- C \(2 \pi \sqrt{\frac{r}{g}}\)
- D \(2 \pi \sqrt{\frac{g}{r}}\)
Answer & Solution
Correct Answer
(C) \(2 \pi \sqrt{\frac{r}{g}}\)
Step-by-step Solution
Detailed explanation
For water not to fall off the can, then centrifugal force must balance the weight of water,
\(\therefore \frac{m v^2}{r} \geq m g\)
Minimum speed of the can so that water does not fall down is given by,
\(v=\sqrt{r g}\)
Time period of revolution,
\(T=\frac{2 \pi r}{v}=2 \pi \sqrt{\frac{r}{g}}\)
\(\therefore \frac{m v^2}{r} \geq m g\)
Minimum speed of the can so that water does not fall down is given by,
\(v=\sqrt{r g}\)
Time period of revolution,
\(T=\frac{2 \pi r}{v}=2 \pi \sqrt{\frac{r}{g}}\)
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