A stone hanging from a massless string of length 15 m is projected horizontally with speed 147 m s - 1 . Then the speed of the particle, at the point where tension in string equals the weight of particle, is

T - m g cos ⁡ θ = m v 2 l and v 2 - u 2 = - 2 g l 1 - cos ⁡ θ also T = m g

Start Free Mock LoginAlready registered? Login here

Physics - Circular Motion Question with Solution | TestHub

PhysicsCircular MotionMiscellaneousEasy2 minPYQ_2020

PhysicsEasysingle choice

A stone hanging from a massless string of length $15 m$ is projected horizontally with speed $\sqrt{147} m s^{- 1}$ . Then the speed of the particle, at the point where tension in string equals the weight of particle, is

Options:

Answer:

Solution:

$T - m g \cos θ = \frac{m v^{2}}{l}$

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

also $T = m g$

Stream:NTA_ABHYAS_NEETSubject:PhysicsTopic:Circular MotionSubtopic:Miscellaneous

⏱ 2mℹ️ Source: PYQ_2020

Doubts & Discussion

Loading discussions...