Start Free Mock LoginAlready registered? Login here

Physics - W.E.P. Question with Solution | TestHub

PhysicsW.E.P.Vertical circular motionEasy2 minQB

PhysicsEasysingle choice

A stone hanging from a massless string of length $15 m$ is projected horizontally with speed $12 m s^{- 1}$ . The speed of the particle, at the point where the tension in the string is equal to the weight of the particle, is close to

Options:

Answer:

Solution:

$u = 12 m s^{- 1}$

$T - m g \cos θ = \frac{m v^{2}}{l}$

As, $T = m g$ we get

$m g (1 - \cos θ) = \frac{m v^{2}}{l}$

Also we know that

$\frac{1}{2} m v^{2} - \frac{1}{2} m u^{2} = - m g l (l - \cos θ)$

$\frac{1}{2} m v^{2} - \frac{1}{2} m u^{2} = - m v^{2}$

$\Rightarrow v = \frac{u}{\sqrt{3}} = \frac{12}{\sqrt{3}} = 6.93 m s^{- 1} \approx 7 m s^{- 1}$

Stream:JEESubject:PhysicsTopic:W.E.P.Subtopic:Vertical circular motion

⏱ 2mℹ️ Source: QB

Doubts & Discussion

Loading discussions...