A bucket full of water is revolved in vertical circle of radius 2,m . What should be maximum time-pe

English

Gujarati

Hindi

5.Work, Energy, Power and Collision

medium

A bucket full of water is revolved in vertical circle of radius $2\,m$. What should be the maximum time-period of revolution so that the water doesn't fall off the bucket ......... $\sec$.

A

$1 $

B

$2 $

C

$3$

D

$4 $

Solution

(c) Minimum angular velocity ${\omega _{\min }} = \sqrt {g/R} $

$\therefore \,\,{T_{\max }} = \frac{{2\pi }}{{{\omega _{\min }}}} = 2\pi \sqrt {\frac{R}{g}} $

$ = 2\pi \sqrt {\frac{2}{{10}}} = 2\sqrt 2 \cong 3\,s$

Standard 11

Physics

Share

Similar Questions

The maximum velocity at the lowest point, so that the string just slack at the highest point in a vertical circle of radius $l$

medium

The maximum and minimum tension in the string whirling in a circle of radius $2.5 \,m$ with constant velocity are in the ratio $5 : 3$ then its velocity is

hard

A small bob tied at one end of a thin string of length $1\, m$ is describing a vertical circle so that the maximum and minimum tension in the string are in the ratio $5: 1 .$ The velocity of the bob at the height position is …………. $m / s$ (Take $\left.g=10 \,m / s ^{2}\right)$

hard

(JEE MAIN-2021)

A stone of mass $m$, tied to a string is being whirled in a vertical circle with a uniform speed. The tension in the string is

medium

(JEE MAIN-2022)

A stone of mass $900 \mathrm{~g}$ is tied to a string and moved in a vertical circle of radius $1 \mathrm{~m}$ making $10\ \mathrm{rpm}$. The tension in the string, when the stone is at the lowest point is (if $\pi^2=9.8$ and $g=9.8 \mathrm{~m} / \mathrm{s}^2$ )

hard

(JEE MAIN-2024)