- Home
- Standard 11
- Physics
3-2.Motion in Plane
easy
A stone of mass $1\,kg$ is tied to end of a massless string of length $1\,m$. If the breaking tension of the string is $400\,N$, then maximum linear velocity, the stone can have without breaking the string, while rotating in horizontal plane, is $.......\,ms^{-1}$
A$20$
B$40$
C$400$
D$10$
(JEE MAIN-2023)
Solution
$T \cos \theta=m g$$T \sin \theta=\frac{m v^2}{I^2 \sin \theta}$
$\cos \theta=\frac{m g}{T}$
$\sin 2 \theta=\frac{m v^2}{T I^2}$
From (1) and (2)
$I=\left(\frac{m g}{T}\right)^2+\frac{m v^2}{T l^2}$
$I=\left(\frac{10}{400}\right)^2+\frac{v^2}{400}$
$v^2=399.78$
$v=20 m / s$
Standard 11
Physics
