A positively charged particle of charge q and mass m is suspended from a point by a string of length $l$ in a vertical plane. In the entire space a uniform horizontal electric field E exists. The particle is drawn aside so that string becomes vertical and then particle is projected horizontally with velocity v such that particle starts to move along a circle with some constant speed v. Then choose the correct statements.

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

Tension in the string is $m g + \frac{q^{2} E^{2}}{m g}$

Time period of revolution is independent of mass of particle.

Time period of revolution is $2 π \sqrt{\frac{l g}{g^{2} + {(\frac{q E}{m})}^{2}}}$

Speed of the particle is $v = \frac{q E}{m} \sqrt{\frac{l}{g}}$

answer is A, B, C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$g e f f = \sqrt{g^{2} + {(\frac{q E}{m})}^{2}} T \cos θ = m g_{e f f} T \cos θ = m R ω^{2} \frac{R ω^{2}}{g_{e f f}} = \tan θ ω^{2} = g_{e f f} \frac{\tan θ}{l \sin θ} \Rightarrow \frac{g_{e f f}}{l \cos θ} ω = \sqrt{\frac{g_{e f f}}{l \cos θ}} \cos θ = \frac{I}{g_{e f f}} \sin θ = \frac{q E}{m g_{e f f}}$

Time period $- \frac{2 π}{ω} - \frac{2 π}{g_{e f f}} \sqrt{l g} - 2 π \sqrt{\frac{l g}{g^{2} + {(\frac{q E}{m})}^{2}}}$

$v - R w - l \sin θ \sqrt{\frac{g_{e f f}}{l \cos θ}} v = l \frac{q E}{m g_{e f f}} . \sqrt{\frac{{g^{2}}_{e f f}}{l g}} v = \frac{q E}{m} . \sqrt{\frac{l}{g}} T = \frac{m {g^{2}}_{e f f}}{g} \Rightarrow \frac{m}{g} [g^{2} + {(\frac{q E}{m})}^{2}] T = [m g + \frac{q^{2} E^{2}}{m g}]$