Question

A bucket of water is being revolved in vertical circle of radius $1\, m$. Minimum frequency required to prevent the water from getting down the path is $ (g=10\,m/s^{2}) $

• A. $ \frac{2\pi }{\sqrt{10}} $
• B. $ \frac{2\pi }{\sqrt{5}} $
• C. $ \frac{\sqrt{10}}{2\pi } $
• D. $ \frac{\sqrt{5}}{2\pi } $

Answer 1

Answer: (C)

As radius is doubled hence, force will be halved.
$v_{\min }=r \omega_{\min }$
$v_{\min }=r \times 2 \pi n_{\min }$
$n_{\min }=\frac{v_{\min }}{2 \pi r}=\frac{\sqrt{r g}}{2 \pi r}$
$\therefore n_{\min }=\frac{\sqrt{1 \times 10}}{2 \pi \times 1}=\frac{\sqrt{10}}{2 \pi}$